STATISTICS AND PROBABILITY-veritas

Circular arrangement

Fix one spot and arrange the rest 5 ( 6 available spots). =5!

PERMUTATIONS: Permutations with Repeating Elements -YOU MUST KNOW THE FORMULA WHEN ELEMENTS REPEAT This problem is another superb example of abstraction. Simply memorizing a formula and applying it are never enough for the GMAT. You must truly understand how the formula works so that you can properly apply it to problems like this. Understand that the difficulty here comes not from any calculations, but rather from applying the formula to a confusing situation in which it is not crystal clear what the values are for the variables in the formula (N, A, B, etc.).

How many ways can you arrange the letters ABC in a row? It is simply 3!: 3 x 2 x 1. The answer is 6. Here they are: ABC, ACB, BAC, BCA, CAB, CBA. But what if the question asks: How many ways can you arrange the letters ABB in a row? As you learned from this drill earlier, the answer is no longer the 3! If you take the list from above and change every C to a B, you can see that they will no longer all be unique from each other: ABB, ABB, BAB, BBA, BAB, BBA Now there are only three unique arrangements—ABB, BAB, and BBA—so the answer is 3. For this example, you could simply write out the possibilities, but how would you deal with the following question: In how many unique ways can the letters in MISSISSIPPI be arranged in a row? It would take the better part of a lifetime to try to write them all out, so there must be a better way. On any permutation problem in which certain elements repeat (or you can think of it as any permutation problem in which certain elements are identical), you should apply the following formula: N! = # of ways to arrange N elements in which a certain number of those A!B!... elements repeat a certain number of times Each repeating element gets its own factorial in the denominator, and the variable (A, B, etc.) represents the number of times that element repeats. Let's apply it to Mississippi: N = 11 and there are three separate elements that repeat, so there will be three separate factorials in the denominator. The "i" repeats four times, the "s" repeats four times, and the"p"repeatstwice.Thusthecalculationwouldbe: 11! /4!2! = (cancel the 4! from the top and bottom and cancel 8 and 6 from the numerator with 4 x 2 and 3 x 2 in denominator) = 11 x 10 x 9 x 7 x 5 = 990 x 35 = 35,000 - 350 = 34,650. Applying the Formula for Permutations with Identical Elements N! Use the formula A!B!... first with the drill problem below and then on the more conceptually difficult GMAT problem that follows: In how many ways can the letters AAABBC be uniquely arranged in a row? 13. There are four identical copies each of five different magazines. In how many ways can the magazines be arranged in a row on a shelf, if nothing else is arranged with them? (A) 5!/ 4! (B) 5! /(4!)5 (C) 20!/ (5!)4 (D) 20! /(4!)^5 (E) 20! /5(4!) The drill problem is typically not problematic if you have learned the equation. There are six total letters, so N = 6. The A repeats three times and B repeats twice, so there will be two factorials in the denominator: 3! and 2!. Inserting that information, the formula looks like this: 6! /3!2!=60 There are 60 possible arrangements for AAABBC. Question #13 is more conceptual and tests whether you really understand how to apply the formula. To start, figure out the value of N. If there are four identical copies of five different magazines, then there are 20 total magazines. If there are five different magazines each repeating four times, then there must be five separate factorials in the denominator. Since they each repeat four times there must be 4! multiplied 5 times in the denominator, as shown below: 20! /4!4!4!4!4!The correct answer choice is D.

Probability of One Event or Another

In the previous examples, you were dealing with the probability of multiple events—of one event and another. What happens, however, when you are asked for one event or another? Let's return to the lottery and lightning example and rephrase the question: If on a given day the probability of winning the lottery is one in a million, and the probability of getting struck by lightning is one in a million (assume they are independent events), what is the probability that Bill wins the lottery or gets struck by lightning on that day? The original example was asking for the probability of one event and another, so the resulting probability was astronomically unlikely ( 1 /(10^6)x 1/(10^6) = 1/10^12) . However, when the question is asking for the probability of one event or another, that probability must be greater than the probability of either event alone (unless the individual probabilities are each 0 or 1, in which case the resulting"or"probability would be the same). To calculate this probability, you must use the general case formula for probability: General Case Formula for Probability: P(A or B) = P(A) + P(B) - P(A and B) So, probability of lottery or lightning = 1 (10^6)+1(10^6)-1/(10^12) The common mistake that almost everyone will make on this problem is to simply add the probabilities but not subtract the overlapping probability (the probability that they will both occur). With mutually exclusive events, that overlapping probability is zero, so you do not need to subtract anything. In this example, the two events are not mutually exclusive, so you must subtract the infinitesimal probability of both occurring. Why? The answer is not intuitive, but over the next few pages you will come to understand it conceptually. NOTE: Understanding deeply the general case formula for probability is not essential to success on the GMAT. However, for questions above the 80th percentile, it could matter. Simply memorizing the formula and remembering to subtract any overlapping probability will suffice for most problems. Understanding the General Case Formula To better understand the general case formula, let's first apply it to mutually exclusive events. If you are asking for the probability of one event or another, and those events are mutually exclusive, then you simply add the two probabilities, because the value of P(A and B) on the right side of the equation will always be 0. Imagine that two people have bought one ticket each in a lottery with only one prize that does not allow ticket numbers to be repeated. If there are a million tickets in the lottery, each person has a one in a million chance of winning. If one person wins, the other person cannot win. What is the probability that one or the other of them wins? This is an example of when you should add probabilities using the general case formula. The probability that one of the two people wins is thus two in a million: Inthiscase: P(AorB) =1/(10^6)+1/(10^6)-0=2(10^6) Consider several other question stems of "or" probability problems with mutually exclusive events: 1. What is the probability of getting a head or a tail on the flip of a coin? 2. What is the probability of getting a 2, 3, or 4 on one roll of a fair six-sided die? 3. What is the probability of the Yankees or the Red Sox winning the World Series? Visually, mutually exclusive events can be represented in the following manner when there is no overlap between the events: Event A Event B P(A) P(B) General Case Formula with Events that Are Not Mutually Exclusive Some problems that you face on the GMAT will involve events that are not mutually exclusive. Consider the following problem: What is the probability that Amanda or Bill is born on a Monday? Many students are tempted to answer 2/7 , but that would only be true if the events were mutually exclusive. It is possible for both people to be born on a Monday, so P(A and B) is not equal to 0. To make the point, imagine if the question asked about seven people rather than two people. If you simply added the probabilities, it would lead to an illogical conclusion: There would be a 100% probability that out of a group of seven people someone was born on a Monday. The problem with adding 1/7 and 1/7 is that you are double-counting the probability that both people were born on the same weekday. The correction we have to make is subtracting the probability that they are both born on the same day. The probability that they were both born on a Monday is simply P(A and B) = 1 /7x 1/7 = 1/49 . Plugging that into the general case formula, P(A or B) = P(A) + P(B) - P(A and B), you get the probability of one or the other being born on aMondayas 1/7 + 1/7 - 1/49 =13/49. Visually, events that are not mutually exclusive can be represented with a Venn Diagram: Event A Event B P(A) P(A and B) P(B) Understanding Overlapping Probability with "Or" Questions To understand the difference between "or" probability with events that are mutually exclusive and with events that are not mutually exclusive, consider the two different diagrams on the past several pages. Imagine you are dropping a ball into a box with each of the previous two diagrams in the bottom of the box. On the first example (the two separate circles representing mutually exclusive events), it is clear that the ball cannot land in both circles simultaneously. Since you want the probability that the ball lands in either circle, you can simply add the value of the two separate circles. In the case with events that are not mutually exclusive, you can see that the ball can land in both simultaneously. Since you are looking for the probability that the ball lands somewhere within the two circles, you cannot simply add the value of the two circles, as you are double-counting the probability that it lands in the middle region. As you will learn in the Word Problems lesson, the general case formula is the same as the formula for general two-set Venn Diagram problems. Regardless of how well you understand this concept it is important to remember this: You must subtract the probability that both events can occur when you are calculating the probability of one event or the other event occurring. Before you wrap up probability with a set of drills, consider a couple example question stems of "or" probability questions when you have to remember to subtract the overlapping probability: What is the probability of getting an A in Calculus or in English? What is the probability of winning the Mega Millions or the Powerball? What is the probability of being accepted to Harvard Business School or Wharton School of Business?

takeaway in all of this: Do not let mean questions waste your time with ugly calculations.

The standard definition of the mean (sum of the terms divided by the number of terms) is labor-intensive, but the application of it can be made much more efficient by thinking critically and by practicing with a variety of question setups. Be efficient, and bring your mean-time-per-question down significantly!

Factorial Value

0! 1 1! 1 2! 2 3! 6 4! 24 5! 120 6! 720

2. John has a 1/10 chance of being accepted at Stanford Graduate School of 1/10 Management and a chance of being accepted at Harvard Business School. What is the probability that he is accepted at either Stanford or Harvard?

. In this example you must recognize that John could be accepted at both, and when you add the two probabilities you are double-counting that possibility. Apply the general case formula and remember to subtract the probability that he is accepted at both (which is 1 /100): P(AorB)=P(A)+P(B)-P(Aand B) = 1/10 + 1/10 − (1/10*1/10)= 19 or19%.

2. Dan has applied to three business schools. Which of the following choices represent possible probabilities for his acceptance to at least one? (A) 88% b 6/ 5 (C) 1.5 (D) .15 (E) .0001

2. Since all probability values must lie between 0 and 1, the only possible probability values are answer choices A, D, and E. It is impossible to have probability values of 6/5 or 1.5, as they are both greater than 1.

3. There are 10 black balls and 10 red balls in a bowl. If you reach in and pick two balls simultaneously, what is the probability that one will be black and the other red?

3. Remember that with so-called simultaneous events you'll still calculate them using sequences. Here there are two sequences that give you the desired outcome: Red, then black; and Black, then red. The probability of each sequence will be the same, so you only have to calculate one, then just double it: Red, then black (10/20)(10/19)=( 1 /2)(10/19)=5/19 And since the probability the other way (black, then red) is the same, double 5 /19 to get the correct answer: 5/19 +5/19 = 10/19

7. John flips a coin 50 times in a row and gets heads on the first 49 attempts. The probability that he gets heads on the 50th attempt will be: (A) < 50% (B) >50% (C) 50%

7. Since coin flips are independent events, the answer is 50%. What has happened previously, regardless of how unusual it might seem, has no impact on the probability of the next event. Many people think that because it has been heads for so long that the probability would be higher than 50%. This error, called the gambler's fallacy, is quite common in probability.

7. There are three red balls and two black balls in a jar. What is the probability that John gets at least one red ball when he picks two balls at random out of the jar at the same time?

7. This problem is bit trickier and more like what you will see in the lesson portion of this book. When you see the words "at least one red ball," that should be a trigger to consider the complementary event of getting no red balls. It is much easier to calculate the probability of finding no red balls: On the first pick there are two balls (the black ones) that are not red ( 2/5 ), and on the second pick there is one ball that is not red ( 1/4 ). Multiply those probabilities together and you have the probability of getting no red balls: 2/5 x 1 /4= 1/10 . Since getting at least one red is complementary to that probability, the answer is 1 - 1/10 = 9/10 . The answer is 90%.

COMBINATORICS there are two broad categories of Combinatorics questions: permutations and combinations. Permutation questions are ones in which a finite number of things are arranged in order; combination questions are ones in which a finite number of things are arranged in groups (and thus order does not matter). The only permutation problems in which you need to use a formula are those that involve identical or repeating elements. For all other permutations, you should simply write out the number of spots and the choices for each spot, and then multiply those values for the total number of arrangements. The formula for permutations (which many people have learned) is more time-consuming and error- prone than just writing out and multiplying.

9. Bill is renting DVDs from the video store. He must choose three DVDs from a list of eight new releases and then decide in what order to watch them. How many different schedules of DVDs can he create, if he cannot watch the same movie more than once? (A) 24 (B) 56 (C) 72 (D) 336 (E) 512 10. There is a set of 10 letters: A, B, C, D, E, F, G, H, I, J. There are five-character and four-character codes to be made out of these letters. Each of the codes can use any of the 10 letters without repeating them in any one code. What is the ratio of the total number of possible five-character codes to the number of possible four-character codes? (A) 1:6 (B) 1:5 (C) 5:1 (D) 6:1 (E) 10:1 For any permutation question in which there are no repeating or identical elements, you can use logic and basic counting principles to solve. In the first question (#9), you should note that the order matters (because it is asking for a schedule of videos) and that you cannot repeat any video. For any basic permutation question, you must first determine the number of unique spots in which you will place items (in this example, there are three spots) and then determine the choices for each spot (in this example, you have eight choices for the first, seven for the second, and six for the third). You then multiply those choices together to get the total number of possibilities. When determining the number of arrangements in a counting problem, you always multiply. As you will see shortly, you must sometimes sum together the results of several individual permutation problems within a bigger problem, but you never add to determine the actual number of arrangements. For problem #9, consider the calculation below: 8 possibilities x 7 possibilities x 6 possibilities = 336 (# choices for 1st video # choices for 2nd video # choices for 3rd video) The correct answer choice is D. For the second problem (#10), you must map out the possibilities for both four- and five-character codes, and then insert them into a ratio. 5-character codes = 10 x 9 x 8 x 7 x 6 (Don't multiply out as parts will cancel in the ratio.) 4-character codes = 10 x 9 x 8 x 7 (Don't multiply out as parts will cancel in the ratio.) Ratioof 5-charactercodes = 10x9x8x7x6 = 6 AnswerchoiceDiscorrect.

permutations

As you learned in the introduction to this section, permutations are arrangements in which order matters. In the strict mathematical definition, they are also counting problems in which items cannot be reused. Before we look at the formula for permutations, let's look at a simple example of a classic permutation: Permutations in Which N and K Are the Same In any case when you are determining a permutation in which the number of available items (defined as N) equals the number of spots in which to arrange those items (defined as K), the answer is simply N!. NOTE: This formula only works in true permutations where items cannot be reused. Most of the permutations that you will be dealing with on the GMAT involve picking from a larger pool of items (called N) to arrange in a smaller number of spots (called K). In an example such as the previous one, N and K are the same, so the answer is simply N!. F N O T E : There is a formula that many people are taught for solving permutation problems. In problems where the order of elements is relevant, the elements can't be reused, and N items are to be arranged in K spots, the number of different permutations can be calculated by a formula: Total number of permutations = N ! /(N − K)! NOTE: As you can see, using the formula is more time-consuming than using basic counting methods. Permutations with repeating elements:- What if things are identical. They are true permutations because not reusing items but they are identical. If N items are distinct- permutation for N BUT here some are identical =reduced by no of permutations that could be created from the identical items. Formula(N=k) N!/a!b! PERMUTATIONS WHERE 'k' IS CHANGING:- Calculate values for all possibilities and then multiply or add When you see trigger words: "at least " " at most " " or ". In any permutation when k=n and k=n-1, no of arrangements will always be the same. PERMUTATIONS WITH RESTRICTIONS: in approaching these there are two ways: 1. Add up all the ways that are allowed by the restrictions 2. Get the total without any restrictions and then subtract the no of ways that break the restrictions COMMON RESTRICTION PROBLEMS: -circular arrangements -couples or groups that must sit together - alternating seats (boy, girl, etc) -limited seating ( John must sit between 2 friends)

Independent Events

Events A and B are independent if the occurrence of one event does not affect the probability of the occurrence of another. In the case of sequential events, the initial probability is restored before each subsequent event. Examples of independent events: • Flips of a coin • Rolls of a die • Winning the lottery and getting struck by lightning • Picking balls out of a jar with replacement With probability involving independent events, the individual probabilities are not affected by what has occurred previously.

Complementary Events

Events are called complementary if one or the other must occur. Since only one of the complementary events can occur, all complementary events are mutually exclusive. In other words, events complement one another if one or the other must occur, but they can never occur together. As a result, the sum of the probabilities of complementary events always equals 1. Examples of complementary events: • Getting a heads or tails on a flip of a coin • Getting a speeding ticket or not getting a speeding ticket on the way home from class • Getting into Harvard Business School or not getting into Harvard Business School • Getting at least one tail on three flips of a coin or getting no tails Events A and B are complementary if one and only one of them must occur: P(A or B) = 1.

2. Over a four-game stretch, Dennis's bowling score averaged 240. By what percent would his score have had to have been higher in order for him to average a perfect game (300)?

In this question, calculating either sum is unnecessary, as the percent change in his average would be the same as the percent change in his total score. Since the figures are already calculated for you in terms of average, you can calculate the percent change just given the averages: (300 - 240)/240 = 1⁄4, so John would have needed to increase his score by 25%.

MEAN MEDIAN MODE RANGE STANDARD DEVIATION

Mean = average = sum of the terms in a set number of terms in a set Median = the middle value = the middle term, in ascending order, of a set with an odd number of terms OR the average of the two middle values in a set with an even number of terms Mode = most frequently occurring value(s) = the term or terms that appear the greatest number of times in a set Range = the difference between the largest value and the smallest value Standard Deviation = a measure of the dispersion of a set of data from the mean (We'll cover the calculation later in this section.)

PROBABILITY OF ONE EVENT AND ANOTHER EVENT Simultaneous Events

Simultaneous Events One very confusing type of GMAT problem for students is that which asks you to pick two balls from a jar at the same time or two cards from the deck at the same time. Consider this example question: A jar contains 4 black and 3 white balls. If you reach into the jar and pick two balls at the same time, what is the probability that one ball is black and one is white? Any question in probability that asks you to calculate the probability of two things happening simultaneously is really asking you to do it one at a time, without replacement. It may help to think about it this way - even if the selections were completely simultaneous (which is nearly impossible at the nanosecond level), you'd still visually identify one ball at a time when seeing which ones were chosen. "Simultaneous" really doesn't happen - you can pick them at the same time but you'd still notice them in a sequence. And the probability doesn't change, so think of "simultaneous" as "in rapid succession, but still one at a time without replacement." In this question, rephrase it as the following: what is the probability of picking a white ball and a black ball on two consecutive draws from a jar that contains 4 black and 3 white balls? Consider the probability of picking black and then white: 4/7 x 3/6 = 2/7 . This probability will always be the same as picking white and then black: 3/7 x 4/6 = 2/7 . And since each sequence (black, then white; or white, then black) takes you to the desired outcome of one of each, you'll add those sequences together to determine that there is a 4/7 chance of selecting one of each when you draw at the same time. Any question that asks for the probability of two events simultaneously can and should be thought of as "one at a time, without replacement" If the problem involves dependent events, make the calculations accordingly. If the problem involves independent events, just multiply the fixed probability.

STANDARD DEVIATION

Standard Applications of the Standard Deviation In order to calculate the standard deviation of a set, you would need to perform the following five steps: 1. Find the mean. 2. Compute the differences between the mean and each number in a set. 3. Square these differences and add them together. 4. Divide the sum of the squared differences by the number of terms. 5. Take the square root of the result. Now, notice this: That's a ton of labor without a calculator, and it's also memorization- intensive. Neither of those items matches the typical GMAT modus operandi, so you will not need to calculate the standard deviation on the GMAT. But it is important to recognize what the standard deviation represents: • It's an averaging (step 4 is "divide the sum of the squared differences by the number of terms"). • It hinges on the distance between each point and the mean. Knowing this will help you gain a deep conceptual understanding of GMAT standard deviation. Consider the following drill questions:

For the set {1, 2, 3, 4, 5}: 1. What is the mean? 2. What is the median? 3. What is the range?

The mean of the first set is 3. But note that, because the set is evenly spaced, you didn't need to do the full calculation (sum divided by five terms). Visually, you can just see that the outer terms (1 and 5) average to 3, the next terms (2 and 4) average to 3, and 3 is just 3. On mean-based problems, there is often a more efficient and logical way to see or anticipate the average without needing to perform tedious calculations. Know that, for evenly spaced sets, the mean will equal the median. 2. The median of the first set is 3—and note that regardless of how the set is presented (even if it were in the order 2, 1, 5, 4, 3), the median would be 3. To find the median, you must arrange the numbers in ascending (or descending) order. 3. The range is 4: 5 (the highest value) - 1 (the lowest value) = 4.

Mutually Exclusive Events

Understanding the concept of mutually exclusive events is very important for working with the general case formula for probability, to which you will be introduced shortly. Two events are called "mutually exclusive" if they can never occur together—that is, the occurrence of one event completely eliminates the probability of occurrence of the other. Examples of mutually exclusive events: • Getting a 750 and a 730 on the same GMAT test • The Yankees and the Red Sox winning the World Series in the same year • Getting a head and a tail on one flip of a coin • Winning and coming in second in the Boston Marathon in the same year If events A and B are mutually exclusive, P(A and B) = 0.

PROBABILITY OF ONE EVENT AND ANOTHER EVENT Independent Events

When determining the probability of multiple independent events, simply multiply the individual probabilities. Consider the following example: On a given day, if the probability of winning the lottery is one in a million, and the probability of getting struck by lightning is one in a million (assume these two events are independent), what is the probability that Bill wins the lottery and gets struck by lightning on that day? When determining the probability of one event and another event, multiply the two probabilities together. In this example, the probability of one event has nothing to do with the other event, so they are independent events. This simplifies the problem and allows you to simply multiply the individual probabilities: 1/(10^6)*1/(10^6)=1/(10^12) Therefore, the probability of both events taking place is one in a trillion. For independent events A and B, the probability of both events occurring is the product of the probabilities of the two events: P(A and B) = P(A) x P(B)

PERMUTATIONS: Multiple Problems in One Question -LOOK FOR IMPORTANT TRIGGER WORDS Often Combinatorics questions will involve wording such as "at least,""at most," and "or." In such counting problems, you must often make multiple calculations and then sum the results. Consider the following two Combinatorics questions: It is important to remember that you never have to do tedious calculations on the GMAT. Obviously, if you cannot find a shortcut to avoid a tedious calculation, then just do it and get the problem correct. However, the time that you waste doing such calculations will eventually catch up with you in the form of missed problems at the end of the test. When the math gets messy, you should always look to back-solve and use the answer choices to shortcut calculations. In counting problems, that shortcut usually involves leveraging your understanding of the units digit.

11. Mike, a DJ at a high school radio station, needs to play two or three more songs before the end of the school dance. If each composition must be selected from a list of the 10 most popular songs of the year, how many unique song schedules are available for the remainder of the dance, if the songs cannot be repeated? (A) 6 (B) 90 (C) 120 (D) 720 (E) 810 12. A company assigns product codes consisting of all the letters in the alphabet. How many product codes are possible if the company uses at most three let- ters in its codes, and all letters can be repeated in any one code? (A) 15,600 (B) 16,226 (C) 17,576 (D) 18,278 (E) 28,572 The words "at least," "at most," and "or" indicate that there will be multiple counting problems in one. For the first example (#11), you must calculate two different permutations: 1. How many three-song schedules can you make from 10 songs? 10 x 9 x 8 = 720 2. How many two-song schedules can you make from 10 songs? 10 x 9 = 90 Since any three-song schedule is inherently different from any two-song schedule, you must sum the results to get the total number of possible two- or three-song schedules: 720 + 90 = 810. Answer choice E is correct. The second problem (#12) is trickier, because you need to remember that you can repeat the letters and because the calculations seem very tedious. Since the problem asks for how many codes can be made from "at most three letters," it is three separate problems: How many three-letter codes are possible if you can repeat letters? 26 X 26 x 26 = 263 How many two-letter codes are possible if you can repeat letters? 26 x 26 = 262 How many one-letter codes are possible? 26 The key to this problem is to realize that you would never need to actually calculate 263 on the GMAT. As you learned from the previous problem, the answer to this question is the sum of the number of possible three-letter codes, possible two-letter codes, and possible one-letter codes. Your understanding of the units digit plays a key role in shortcutting this problem, as you can figure out quite quickly which answer choice it must be: SKILLS MEET STRATEGY Use Answer Choices and Don't Get Baited into Tedious Calculations It is important to remember that you never have to do tedious calculations on the GMAT. Obviously, if you cannot find a shortcut to avoid a tedious calculation, then just do it and get the problem correct. However, the time that you waste doing such calculations will eventually catch up with you in the form of missed problems at the end of the test. When the math gets messy, you should always look to back-solve and use the answer choices to shortcut calculations. In counting problems, that shortcut usually involves leveraging your understanding of the units digit. 263 is some large number ending in 6: +262 a3digit number ending in6: + 26: =some number ending in8 The correct answer choice must be D. _ _ _ _ _ 6 + _ _ 6 + 26 _ _ _ _ _ 8

NOTES

for evenly spaced sets, the mean will equal the median. Conceptually, standard deviation is pretty true to its name: It involves the average (standard) amount of dispersion (deviation) from the mean. So adding an additional value to the set that provides no variance from the mean will reduce the average variance. Even without the calculation itself, you should recognize that a set in which the terms are closer to the average will have a smaller standard deviation. % Change in Average = % Change in Sum (if the number of terms does not change) when all of the terms in a set are multiplied by a value with an absolute value greater than 1, the standard deviation will increase. The logical reason for that is that the multiplier spreads the terms out farther from the mean. Remember: Standard deviation only relies on distance from the mean. With the positive integers, the mean is 4 and the distances are 3, 2, 1, 0, 1, 2, and 3. When the terms are all negative, the mean is -4 and the distances stay the same. Because standard deviation involves squaring the differences, then taking the square root, the standard deviation will never be negative, so answer choice D is a trap choice here. Standard deviations essentially measure distances. You wouldn't say that the next town over is "negative 20 miles away," so you'll never say that a standard deviation is -1.5. Standard deviations are either positive or zero, as you will see in the next explanation. NOTE: If you are confused about whether a counting problem is a combination or a permutation, ask yourself the following question: If I change the order, would the new arrangement be considered distinct? NOTE: There are some permutation and combination problems that cannot be solved with this approach and that require specialized formulas and strategies (which we will address later), but most are best solved with this method. On some problems you will be summing the number of counts from various cases or dividing to remove repeated sequences, but to determine any individual count you must multiply. Almost all counting problems can be thought of in terms of the number of spots that are available and the number of possibilities for each spot. REMEMBER: Todeterminethenumberofpossiblearrangementsinmostcounting problems, simply multiply the number of choices for each available spot in the arrangement.

BIONOMIAL PROBABILITY

sequential probability where there Are exactly the same two choices of each events. E.g:- coin Q: probability 2heads and a tail in 3 flips of a coin 1st way : 1/2 x 1/2 X1/2=1/8. (H H H , H T H , T H H ) 2nd way: 3 positions - - - , 2x 2x 2=8 outcomes only one is favourable so 1/8 BIONOMIAL PROBABILITY WITH MANY EVENTS: PROBABILTIY of getting 3 heads on 5 flips of a fair coin? 32 outcomes =( 2x2x2x2x2) 3heads and 2 tails = favourable outcomes = 2 things are repaying one of them is reappearing 3 times the other 2 times so (5!)/(3!2!). =10 So 10/32=fav /total =5/16

1. What is the median of the set {-8, -5, 3, 5, -6, 1}?

1. -2 . To find the median you must first put the set in ascending order: {-8, -6, -5, 1, 3, 5} If the set has an odd number of terms, the middle number is the median. If the set has an even number of terms, as in this example, take the average of the two middle terms to determine the median: (-5 + 1) = -4 = -2 22 2. 26 . 14 is the largest value, and -12 is the smallest value. 14 - (-12) = 26 3. 10 . Because this set is evenly spaced, you do not need to calculate the sum and divide by the number of terms. Just note that the middle number (the median) will equal the mean. This means that the fifth of nine terms will be the median/ mean, so you can stop at the fifth term in the series: 2, 4, 6, 8, 10. 10 is the mean. 4. A, C, E, and F . Note that if J is greater than 5, then 5 will be the middle number. If J is less than 0, then 0 will be the middle number. If J is between 0 and 5, then J will be the middle number. And if J equals 0 or 5, then the mode will include the middle number. So the median could be anything between (and including) 0 and 5; answer choices A, C, E, and F are all possibilities. 5. 2 . Arrange the set in order to better determine the median: {-9, -5, 2, 5, 12, and n}. Because n is the median, it must be the average of the two middle terms, giving you these possibilities: • n is less than -5: n would be the average of 2 and 5 (which is not possible). • n is greater than 5: n is the average of 2 and 5 (which is not possible). • n is between -5 and 5: n is the average of 2 and n. This means that n must be 2. 5a. 2 . Because n is 2, then 2 is the most commonly occurring value and therefore the mode. 6. C . The range of any six consecutive integers is 5: (n + 5) - n = 5. Even with negative integers, the largest minus the smallest value will be positive. For example, consider: -1 - (-6) = 5. 27 skillbuilder 6a. A . The range of this set is (positive) 5, but the mean and median will be a negative number because all values in this set are negative. For example, consider the set {-101, -102, -103, -104, -105, -106}. Both the mean and median are -103.5, but the range (-101 - (-106) = 5) is 5, so the range will be the largest value.

Face cards

4 in each suite a,j,k,q Total 16

combinatorics: Basic Counting Principles: Summary from Skillbuilder -UNDERSTAND THE BASIC COUNTING PRICIPLE

Below is a summary of the important skills for basic permutation problems with several drill questions. For any permutation problem consider the following: • Read carefully. Make sure that it is a permutation (in which order matters), not a combination (in which order does not matter). Also make sure you always ask the following questions: Can I repeat elements or not? Are any of the elements identical? To see the importance of this consider two quick drill questions: 1. If a company needs to assign two letter codes to all of its partner companies, how many such codes are possible? 2. In how many ways can the letters ABB be arranged in a row? • Multiply. Map out the number of spots in which items will be arranged and then determine the number of choices for each spot. To find the total number of possibilities, always multiply the number of choices together. Generally the difficulty comes from determining correctly how many choices there are for each spot. Consider the following example: 3. How many license plates are possible if the plate must contain exactly five digits, and the plate cannot start with 0 or repeat any digits? (The digits must be in a row and the plate contains no other characters.) • Use logic. If you are confused, use logic and consider writing out the possibilities. Except for problems involving combinations and permutations with repeating elements, you will never need to use a formula to solve GMAT Combinatorics problems. Solutions 1. Since it does not say that you cannot repeat letters, there are 26 x 26 or 676 such unique codes. 2. Because there are two identical elements, the answer is not six but three: BBA, BAB, and ABB. 3. There are nine choices for the first spot (can't use 0), nine choices for the second spot (can't repeat the first number, but now you can use 0), eight choices for the third spot, seven for the fourth spot, and six for the fifth spot: 9 x 9 x 8 x 7 x 6 = 27,216. NOTE: You would not have to actually calculate this on the GMAT. Only one answer choice would end in 6.

Dependent Events

Events A and B are dependent if the occurrence of one event affects the probability of another. In calculating the probability of multiple events that are dependent on each other, the probability will change with each event. Examples of dependent events: • Picking balls out of a jar without replacement • Picking cards out of a deck without replacement With probability involving dependent events, the individual probabilities are affected by what has occurred previously.

STATISTICS : AVERAGE - BE CAREFUL WITH MULTIPLICATION /DIVISION IN SETS. The most commonly tested statistic on the GMAT is the average. While people tend to understand the concept well, they do not always attack average problems in the most efficient way. Also, there are several situations in which these problems can be quite confusing and counterintuitive. The following two problems address both efficiency in calculation and common difficulties in average problems:

KILLS MEET STRATEGY Look at the Answer Choices Before Doing Work Using answer choices is an essential problem-solving strategy. Good problem solvers always follow a regimented procedure: 1.) Digest the problem carefully and understand what is being asked; 2.) Look at answer choices and decide how much (and what kind of) work is required to pick the correct answer; 3.) Execute the necessary calculations (if any). On this problem, if you look at the answer choices before starting your work, then you can simply select answer choice A and move to the next problem! 2. The average of five numbers is 6.8. If one of the numbers is multiplied by 3, the average of the numbers increases to 9.2. Which of the five numbers is multiplied by 3? LEARNING BY DOING Be Careful with Multiplication/Division in Sets On a problem like this, it would be tempting to shortcut it as follows: The sum of the set before the change is 6.8 x 5 (Sum = average X number of terms) = 34. The sum of the set after one term is multiplied by 3 is 9.2 x 5 (Sum = average X number of terms) = 46. Since the net gain after the change is 12, it seems logical to ask: What number multiplied by 3 will give me 12? The answer, of course, is 4, but that is the sucker choice on this problem. Why? Because the original number is still in the set. The correct question to ask yourself is this: What number when multiplied by 3 creates a net change of 12 after the original number is subtracted? The answer to that question is 6. 6 x 3 = 18 - 6 = desired increase of 12. If you worked it out algebraically, this is clear: Beforechange: a+b+c+d+e =6.8soa+b+c+d+e/5=34 Afteronenumberismultipliedby3: 3a+b+c+d+e /5=9.2so 3a+b+c+d+e=46 Subtracting the first equation from the second, you see that 2a = 12 and a = 6. Answer choice E is correct. The most important takeaway from this problem (which is true for many hard problems on the GMAT) is this: If you want to solve things conceptually or with logic, make sure that you really understand the concept or you are better off doing the math. If you did take the time to do the math on this problem, then you would almost surely get it right. This problem is a classic (and very clever) example of how testmakers create problems to exploit particular conceptual mistakes. It is almost impossible to resist picking 4 unless you have seen a similar problem before or you realize conceptually what is going on (that you need a net change of 12). These types of problems are particularly dangerous on the test, because you have no idea that you are missing them. Your only hope the first time you see this problem is to understand the concept well and/or play devil's advocate (4 seems to fall in your lap a little too easily!). Remember: Every time you encounter one of these commonly exploited mistakes, commit it to memory and avoid it on future problems.

RESTRICTION PROBLEMS: COMBINATION WITH SEVERAL GROUPS:-

Q: from a group of 10 B and 7 G , how many different hockey teams of 6 players can be formed if the team must consists of 3G and 3B? ------ No of groups of boys that can be created (10)!/(3!(10-3)!) = 120 No of groups of girls that can be created= (7!)/(3!4!) = 35 total = 120 x 35 = 4200 teams.

TO INCREASE DIFFICULTY

TO INCREASE DIFFICULTY , THE TESTMAKERS CAN TURN A SIMPLE PERMUTATIONS PROBLEM INTO A COMPLEX PERMUTATION PROBLEM , OR CAN ADD EXTRA RESTRICTIONS. HOWEVER , NO MATTER THE QUESTIONS DIFFICULTY LEVEL , IT WILL ALWAYS RELY ON THE SAME PRINCIPLES YOU HAVE ALREADY LEARNED. SOMETIMES, permuataion with restrcition Q's can be answered by finding the no of permutations that would exist without the restrictions - which is almost always one of the answer choices in itself ( usually D or E) and assesing the relative impact of the restrictions. The remaining answer choices will typically vary enough that an estimate will often suffice. To make probability Q difficult , GMAT . A dependan probability elements is introduced into a mostly independant probability Q. They want you to think that a dependent event is actually independant. The product of any 3 consecutive positive integers= multiple of 2, 3, 6.

RESTRICTION PROBLEMS:- TREAT EACH COUPLE OR GROUP AS A SINGLE ENTITY OR UNIT AND THEN DEAL WITH THE WAYS TO ARRANGE THE PEOPLE WITHIN THE COUPLE OR GROUP AT THE END.

TREAT EACH COUPLE OR GROUP AS A SINGLE ENTITY OR UNIT AND THEN DEAL WITH THE WAYS TO ARRANGE THE PEOPLE WITHIN THE COUPLE OR GROUP AT THE END. Q: A group of 8 students take a full row .there are 2 couples and in each couple 2 people must sit together . in how many different arrangements can the students sit? A: 8 students-2 couples = 8-4 = 4 remaining 4 remaining +2 couples = 6! ways. now couples can be switched = each couple = 2! there fore =2! x 2! ans: 6! x 2! x 2!=2880 Q: How many ways can 3 groups of 3 people be seated in 9 seats, if groups must sit together? 9people- 9sets-3groups x 3people in each group now 3! x 3! x 3! x 3! = 6 x 6 x 6 x 6 =1296 ways

If K is the median of the set {2, 5, K, 8, 10}, is K > 5?

Your first inclination is likely "yes," as you'll probably think that K must be between 5 and 8. But how would you answer this question? What is the median of the set: {2, 5, 5, 8, 10}? Well, those values are in ascending order, and 5 is the third of the five values. So the median here is 5. The answer to the first question is "we don't know." K might be greater than 5, or it just might be 5 itself. It is possible for the median to be a repeat term, and you will find that this concept makes for a tricky Data Sufficiency setup.

Statistics tricks

a few examples: For average, you need to be careful about net change versus total change in multiplication/division problems; for median, you must remember to put the terms in ascending order first; for mode, you must remember that you can have multi-modal sets; for standard deviation, you need to always consider the number of terms when comparing sets. If you are aware of these traps ahead of time, it is punlikely that you will fall for them, so make sure you keep mental notes of common traps associated with each statistic.

% Change in Average = % Change in sum (if the number of terms does not change) Over a five-game stretch, Rasheed scored 22, 16, 12, 14, and 21 points. His actual average was what percent less than his goal of averaging 20 points per game?

if the number of terms in a set does not change, then the percent change in the average will equal the percent change in the sum of values. Note that here you're given data in two forms: Rasheed's actual scoring comes in the form of a sum, and his goal comes in the form of an average. Because the number of values doesn't change—we're talking about a five-game set in both cases—you do not have to average the five scores that he posted. Because his goal was 20 points per game for five games (which equals 100 total points), it's easier to calculate the percentage in terms of sum to sum. He scored 85 points, which is 15 less than his goal of 100. And 15/100 is 15%, so you can conclude that he fell 15% short of his targeted average.

6. In how many ways can the letters AABB be arranged uniquely in a row?

. While you might be tempted to answer 4! (24) to this problem, it is actually 6. Because some of the elements are identical, there will not be as many unique arrangements. Clearly, if the question asked for the number of ways to arrange ABCD in a row, then the answer would be 24. However, in a problem like this you must either write them out or apply a formula that you will learn in detail in the full lesson portion of this book. The number of unique ways is: AABB, ABAB, ABBA, BBAA, BABA, and BAAB. (The formula is given in the summary section that follows, but you will learn how to apply it in the full lesson.) The answer applying the formula is 4! /(2!*2!)= 6. One important takeaway from this problem is that you can use logic and simply write out many counting problems on the GMAT.

3. What is the probability that Bill gets a total of two or a total of 12 on two rolls of a fair six-sided die?

3. This problem is more difficult to interpret than the previous two examples. When you roll a die two times, there is only one way to get a total of 2 (1 and then another 1), and the probability of that is 1 /6x 1/6 = 1/36 . The same is true for getting a total of 12: There is only one way (6 and then 6), and the probability ofthatisalso 1/6 x 1/6 = 1/36 .Itisimpossibleto get both a 2 and a 12 on two rolls of a die, so these events are mutually exclusive. To get the probability of one or the other happening, simply add the two probabilities: 1/36 + 1/36 = 1/18 . The answeris 1/18 .

4. The standard deviation of set B is 0. If set B consists of {3, x, y}, which of the following must be true? (Select all that apply.) (A) x=y (B) The average of set B is 3. (C) The mode of set B is 3.

4. A, B, and C. When a standard deviation is 0, that means that "the typical value does not deviate at all from the mean." This means that there is no deviation. The only way for this to be true is if all of the values are the same, so in this set, x and y must each equal 3.

6. Identify which of the following pairs of events are dependent. (A) Rolling a die and flipping a coin (B) Picking one ball and then another out of a bowl with replacement (C) Picking one person and then another from a group of people

6. Coin flips and rolls of a die are classic examples of independent events, so it cannot be answer choice A. With answer choice B the balls are being replaced each time, so each pick will be independent of whatever has happened previously. With answer choice C, these events are dependent. Consider this example: If there are 10 people in a group, of which six are men and four are women, what is the probability of getting two women when you randomly select two? The answer is (4/10) x (3/9) = (2/15) . As you can see, the probability of the second pick is dependent on what happened in the first pick. The correct answer choice is C.

For Statistics, Combinatorics, and Probability—three fairly difficult math content areas that make up roughly 15% of GMAT quantitative questions—the underlying material is actually very relevant to business

Even while the content of this lesson clearly matters for business school, the same rules apply on these questions as on questions testing Geometry, a content area that clearly does not matter for business school: You must understand the core basic concepts well, but the problems are made difficult not by testing obscure content, but rather asking the questions in challenging ways. This is good news, as Statistics and Combinatorics concepts get very difficult when you move beyond the basic high school level. Remember: These content areas are still a vehicle for testing logic and problem solving, so if you understand the basic content and learn how to apply it with GMAT problems, you will succeed on the five or six questions from these content areas.

Binomial Probability (Heads/Tails Probability): Write Out Possibilities When Possible On any binomial probability question with three or fewer events, simply write out the possibilities and use logic to solve the question. Make sure to read the problem carefully and clarify exactly which outcomes are favorable. To this point you have seen several difficult probability concepts: 1. Problems with "or" in which you have to subtract overlapping probability 2. Tricky dependent probability problems in which you need to calculate the conditional probability 3. Problems with the phrase "at least one" in which you are better off calculating the complementary probability 4. Special dependent probability, such as "pairs probability" The remaining content area that is heavily tested on GMAT Probability questions is called binomial probability (probability in which there are only two possible outcomes). The most common binomial probability problems on the GMAT are those involving coin flips. Consider the following drill problem: What is the probability of getting two heads and a tail (in any order) on three flips of a coin?

Given what you learned in the Skillbuilder, the temptation would be to say 1/8 , but that is not correct because there are multiple ways to get two heads and one tail on three flips of a coin. On many binomial probability problems, it is helpful to think of the problem more from a counting perspective. Instead of thinking about multiplying the probabilities as shown in earlier heads/tails problems, think about favorable outcomes over total outcomes. In three flips of a coin (or any three binomial events), you know there are eight outcomes, because if you have three spots to fill with two choices for each, there must be 2 x 2 x 2 = 8 possible outcomes. On binomial probability problems with three or fewer events, it is always easy to write out the possible outcomes and answer any question about them. To illustrate this, let's write out the eight possible outcomes that exist with three flips of a coin: TTT HHH TTH HTH THT HHT THH HTT Each of these outcomes is equally probable, and if you were asked for one such as "What is the probability of getting a head, then a head, and then a tail?" the answer would be 1/8 . However, as you can see, the answer to the drill question is 3/8 , as three of the eight outcomes have two heads and one tail.

RESTRICTION PROBLEMS: PERMUTATIONS WITH ALTERNATING SEATING:-

In how many ways can 3M and 3W be seated in 6 seats if they must alternate. MWMWMW 3M places x 3W places = 3! X 3!=36 36 X 2 = 72 (WMWMWM)

PERMUTATIONS: Permutations with Restrictions -FIX ELEMENTS AND ARRANGE AROUND THEM Permutation problems get very difficult very quickly when restrictions are introduced. What does a restriction mean? It is some condition that limits somehow the total number of possibilities. Consider the question stem from a few example problems so you understand what these will look like: 1. Three couples go to the movies and sit in six consecutive seats. If the couples must sit together, in how many ways can they be seated? 2. There are three girls and three boys who need to be arranged in six consecutive seats. If the boys and girls must be alternated, in how many ways can they be seated? 3. Steve goes to the movies with four friends but refuses to sit next to one of them. In how many ways can they be arranged? It is important that you prepare for restriction problems broadly and that you understand how to contend with several of the most common restrictions covered in Section 4 (couples, circular arrangements, etc.). However, because these questions are difficult, many people looking to score highly often spend way too much time trying to learn obscure restrictions that would never be on the test. Use your study time wisely and spend it improving your abilities on the hard problems that you are much more likely to see: factor/ number line, Geometry, hard Data Sufficiency, Word Problems, etc.

14. John and four friends go to a Lakers game. In how many ways can they be seated in five consecutive seats, if John has to sit between any two of his friends? (A) 144 (B) 120 (C) 96 (D) 72 (E) 48 In approaching restriction problems, there are always two ways to attack them: 1. Add up all the ways that are allowed by the restriction. or 2. Get the total without any restrictions and then subtract the number of ways that break the restriction. On certain problems adding up the allowed arrangements is faster, and on others subtracting out the prohibited arrangements will be faster. As a result, you should be comfortable with both approaches. In this example it does not matter which approach you use, so let's do it first by adding up all the ways that are allowed by the restriction. If John must sit between two friends, then he can sit in the second, third, or fourth seat. To think about it properly, fix John in the second seat and then consider how many ways everyone can be seated around him in that case: Alex John Bill Cindy Dan With John fixed, you can see that the four people (A, B, C, D) can be arranged 4 factorial ways around him. So with John in the second seat, there are 4! or 24 possible seating arrangements. Taking this to the next logical step, you see that there are 24 seating arrangements for each of the five seats that John could sit in. Because he is allowed to sit in three of the five seats, you can simply sum the allowed arrangements for each case. In this case it is 24 + 24 + 24 or 3 x 24 = 72 ways. Answer choice D is correct. The other approach would be to consider all the ways the friends could be seated if there were no restrictions. In this case that is 5! or 120 ways. John is not allowed to sit on either of the end seats, so if you subtract out the number of ways that break the restriction from 120, then you will also get the answer. Of the 120 possible ways, 24 of them have John in the first seat and 24 of them have him in the last seat. Subtracting 48 from 120, you come up with the same answer of 72 ways. An even quicker way to utilize this approach would be to realize that because John is barred from two out of five seats, two-fifths of the permutations are not allowed. Thus the number of allowed permutations is three-fifths the number of permutations without restrictions, or three-fifths of 120 = 72. (In Section 4, you will see more of the other restrictions and learn how to attack them.)

probability

all probability values lie between 0 and 1, or between 0% and 100%, inclusive. Probability of a Single Event The probability of a single event is determined by the ratio of the outcomes when this event occurs to the total number of possible outcomes. This is commonly described as favorable outcomes over total outcomes. For example, when you make a blind guess on a GMAT question, you have a total of five outcomes (answer choices A through E) but only one favorable outcome. Thus, the likelihood of answering any GMAT question correctly based on a blind guess is one out of five ( 1/5 ) or 0.2 or 20%. The probability of an event A is calculated as follows: P(A) =(# of outcomes when A occurs)/( # of total possible outcomes) Questions dealing with the probability of a single event are generally straightforward from a probability standpoint, but questions involving multiple events become much more complicated. Before you can attack those types of questions you must first understand several important concepts and definitions that are presented on the next few pages. PROBABILITY OF ONE EVENT AND ANOTHER : 1. INDEPENDENT EVENTS :- P(A and B)=P(A).P(B) 2. DEPENDENT EVENTS:- P(A and B)=P(A).Pa(B) 3. SIMULTANEOUS EVENTS:- one at a time without replacement Confusing PROBABILTIY OF ONE EVENT OR ANOTHER EVENT: P(A or B)= P(A) + P(B)-P(A and B) OR=add or subtract sometimes 1. MUTUALLY RXCLUSIVE EVENTS : where P(A and B)=0 2. NOT MUTUALLY EXCLUSIVE EVENTS: P(A or B)=P(A)+P(B) -P(A and B) COMMON PROBABILTIY QUESTION TYPES: -ATLEAST ONE -PAIRS PROBABILITY

"Or" Probability and Dependent Probability: DEPENDENT PROBABILITY IS EASY TO OVER LOOK When assessing any probability, always make sure it is not dependent (conditional) on some other outcome. If it is, calculate that dependent probability accordingly. Tricky dependent probability is a mainstay on harder GMAT probability questions, and it is easy to miss on problems like #18 above (and #17, if you use the general case formula to solve for it) WRITE OUT POSSIBLE OUTCOMES WHENEVER POSSIBLE: There is a lot to learn from problem #17. If you simply consider what happens when you roll one die and write out those possibilities, the question is very easy. However, applying the formula is incredibly difficult, because people forget that the second probability is necessarily dependent on the first. Here's the takeaway: If you can solve a probability problem by using logic to write out total outcomes and write out favorable outcomes, that is always preferable and less error- prone than using a formula.

"Or" Probability and Dependent Probability 17. If you roll one fair six-sided die, what is the probability that the number is even or less than 4? (A) 1/6 (B) 1/3 (C) 2/3 (D) 3/4 (E) 5/6 For the first problem (#17), the best way to solve it is simply to use logic. (As you will see, applying formulas to this is tricky and error-prone.) When you roll a die there are six possible outcomes: 1, 2, 3, 4, 5, or 6. The problem is asking for the probability that it is even or less than 4. Since 1, 2, and 3 are all less than 4, they are all favorable. 4 and 6 are both even, so they are also favorable. The only outcome of the six possibilities that is not favorable is 5. Therefore, five numbers are favorable and one number is not; the answer is 5/6 (answer choice E). Of course, you could also use the general case formula (and it is a good opportunity to review it): P(A or B) = P(A) + P(B) - P(A and B). Here the P(<4) = 3/6 and the P(even) is also = 3/6 . It is the probability of both—i.e., P (<4 and even)—that is very tricky dependent probability. The probability that the roll is <4 is 3/6 , but then that affects the probability that it is even, and we must adjust this dependent probability. Of the numbers less than 4, only one is even, so the conditional probability that the die is even given that the die is <4 is not 3/6 as many people will think but rather 1/3 . You then multiply 3/6 x 1/3 to see that the probability of both is 16 . Plugging back into the formula, you see that P(<4 or even)= 3/6 + 3/6 - 1/6 = 5/6 . Again, answer choice E is correct. 18. The acceptance rate at a certain business school is 15% for first-time applicants and 20% for all re-applicants. If David is applying for admission for the first time this year and will attend if he is accepted, what is the probability that he will have to apply no more than twice before he is accepted? (A) 20% (B) 30% (C) 32% (D) 35% (E) 40% The second example (#18) is another "or" problem in which you need to add probabilities. The question is asking for the probability that he is accepted on the first attempt or on the second attempt. Since the two events are mutually exclusive, you only have to add the two probabilities àP(A and B), as shown in the general case formula above, is equal to zero. The first probability is relatively simple P(accepted on first attempt) = 15%. However, the probability on the second attempt is much trickier dependent probability. It is not 20%, because to be accepted the second time, he must not have been accepted the first time. Therefore the probability that he is accepted the second time given that we was not accepted the first = 85% chance not accepted first year X 20% chance accepted second year = (.85) (.20) = .17 or 17%. So the probability he is accepted this year or next year = 15% + 17% = 32%—answer choice C.

1. Set Y consists of unique integers a, b, c, d, and e. If each of the inte- gers in set Y is multiplied by 10, the standard deviation of the set will _____________________. (A) increase (B) decrease (C) remain unchanged

1. A. Regardless of which values a, b, c, d, and e represent, multiplying each of them by 10 will make their distances from the mean increase. Say that the set was {-2, 0, 2, 4, 6}. The average of this set is 2, and the distances from the mean are 4, 2, 0, 2, and 4. Multiply each term by 10 and the new set is {-20, 0, 20, 40, and 60}. That average is 20, and the distances from the mean are 40, 20, 0, 20, and 40. Because the distances from the mean are far greater, and the number of terms is the same, the standard deviation will increase. As a general takeaway, when all of the terms in a set are multiplied by a value with an absolute value greater than 1, the standard deviation will increase. The logical reason for that is that the multiplier spreads the terms out farther from the mean.

1. On a probability midterm, John has an 82% chance of getting at least one problem correct. What is the probability that he gets none of the problems correct?

1. Whenever you see the words "at least one" in a probability problem, you should be focusing on the concept of complementary events. As long as you are dealing with whole-number events, 0 will always be complementary to at least 1. For instance, if you are asked for the probability of getting at least one head on four flips of a coin, the easy way to solve that problem is to find the probability of getting no heads ( 1/16 ) and subtract that from 1 to get the 15/ 16 On this problem, if John has an 82% chance of getting at least one problem correct, then he must have a 1 - .82 or 18% chance of getting none correct.

COMBINATIONS: DONT CONFUSE MULTIPLICATION WITH ADDITION Up until this point you have dealt exclusively with problems in which order matters in determining whether an arrangement is unique (permutations). But a large portion of combinatorics involves problems in which the order does not matter. To further clarify this difference, consider the following two example problems: 1. In how many ways can four people be placed in two seats? From what you learned previously, you know the answer is 4 x 3 = 12. To make a point, let's write out the possibilities if the four people are Andy, Bill, Charlie, and Dan: AB, AC, AD, BA, BC, BD, CA, CB, CD, DA, DB, DC 2. How many groups of two can you make from four people? In this problem, the order does not matter so the answer is not 12. Look carefully at the list above and you will understand why. When being arranged in seats in a row, AB is necessarily different from BA. But as a group of two people, AB is identical to BA. With some logic you can realize that each pair in the permutation calculation above has been double-counted when considering them as combinations. There are only six groups of two you can make from four people: 1.AB=BA 2. AC=CA 3.AD=DA 4.BC=CB 5.BD=DB 6.CD=DC If you are not sure about whether to multiply or add at the end of this problem, you should not just guess and/or go with your best instinct. Use a simple example and prove it! If you had three possible groups of women and two possible groups of men, could you make five teams or six teams? The answer is 6, so you must multiply. Call the female groups A, B, and C and the male groups 1 and 2. Here are the six possible teams: A1, A2, B1, B2, C1, C2. When you are confused about a concept on the GMAT, try to prove it to yourself with a simple example instead of just guessing conceptually. Clearly this question is designed to exploit one common mistake: using addition at the end instead of multiplication. If you are confused about whether to add or multiply, try to determine if you are still creating new arrangements (as in this case) or whether you are simply adding together the different results from several problems with different values for K (as in the high school dance song sequence problem). Remember: The only time you add in Combinatorics problems is when you are summing together the results from several different problems (with language such as "at least," or in restriction problems in which you sum up what is or is not allowed).

15. At a high school track tryout, there are eight women and five men competing for the three male and three female spots on the decathlon team. How many different combinations of decathletes are possible on the final team for the six spots? (A) 112,896 (B) 3,136 (C) 560 (D) 66 (E) 18 This problem starts with the application of the combination formula on two separate groups. You must first determine how many groups of three women you can pick from eight women. This is a simple combination of 8 pick 3, which looks like this: 8!/3!5! =56 Next, determine the number of groups of men that are possible: 5! /3!2!=10 At this point, you are still putting together two elements to create a final set of possible arrangements, so multiplication is required. For instance, if you were told that there are 56 burger choices and 10 drink choices, then there are 560 meals you could make (remember the counting tree!). Clearly, this question is designed to make you pick 56 + 10, or 66, but that is not logical in this case (you are not summing up separate results as in the "at least" problems). The correct answer choice is C.

4. John flips a coin and rolls a die at the same time. What is the probability that he gets a head or a 6 on his flip and roll?

4. This example is perhaps trickier than the last because you might miss that these events are not mutually exclusive; you could get both a head and a 6 when you flip a coin and roll a die. Clearly the probability of getting a head is 1/2 and the probability of getting a 1 is 1/6 . The probability of getting a head and a 6 is 1/2 x 1/6 = 1/12 . Apply the general case formula to get the probability ofaheadora6: 1/2 + 1/6 − 1/ 12= 7 /12.Theanswer is 7 /12.

COMBINATIONS: Combinations with Restrictions -SUBTRACT WHAT IS NOT ALLOWED Combinations with restrictions are approached in the same manner as permutations with restrictions. For most students, however, they are conceptually more difficult and account for some of the hardest counting questions possible on the GMAT. For most combinations with restrictions, it is faster to find the arrangements that break the restriction and subtract them from the number of possibilities without any restrictions. Consider the following restriction problem: Combination restriction problems are a common type of difficult counting problem on the GMAT. There is good reason for this: They are more abstract than permutations (because order doesn't matter), they are confusing, and they require good reasoning skills. Given that, it is important that you understand well how to approach them. Almost without exception, combinations with restrictions are best solved by subtracting what is not allowed from the total. While you might be able to add what is allowed, it is almost always more time- consuming and error-prone.

16. A committee of four is to be chosen from seven employees for a special project at ACME Corporation. Two of the seven employees are unwilling to work with each other. How many committees are possible if the two employees do not work together? (A) 15 (B) 20 (C) 25 (D) 35 (E) 50 If there were no restrictions, then it would be a relatively easy combination problem. N = 7 and K = 4, so the number of arrangements is: 7!/4!3! = 35. However, some of those groups are not allowed, because they would contain the two employees that won't work together. To figure out exactly how many of these 35 are not allowed, imagine the two people together as a single unit, and then see how many ways you can fill the remaining two spots. Since the two "enemy" employees are already grouped together as one, there are five people left to fill the two remaining spots. This is a basic combinationwhenN=5andK=2,whichis: 5! /3!2!=10.SO OF THE 35 WAYS to make a committee of four from seven people, 10 of them have the two enemy employees together and must be eliminated. That leaves 25 possible committees with the restriction given in the problem. The correct answer choice is C. Remember: To solve a problem like this, group the "enemies" together as a single unit, see how many ways they can be arranged with the other people, and subtract that from the total. While this problem can be solved by adding all the ways that are allowed, it is much more tedious and prone to error.

"At Least One" Probability Questions: Misdirection Find the Easier Way, but Answer the Proper Question Oneofthemostimportantsetsoftriggerwordsinprobabilityis"atleastone."Whenever you see that language, you can be sure that finding the complementary probability will be easier. (REMEMBER: None is always complementary to "at least one.") Consider this technique on the following problem: "At least one" problems are some of the best examples of misdirection. By asking for the probability of "at least one," testmakers are baiting you into incredibly time- consuming (and sometimes almost-impossible) calculations, when in reality the problem is really quite simple if you understand and recognize complementary probability. Remember: Calculating the probability of none is generally one simple calculation, while calculating the probability of "at least one" could involve two, three, four, or even more difficult calculations. Don't let the testmakers lead you astray! When it is easier to find complementary probabilities, you should always do that and then subtract the value from 1. Importantly, however, you then must make sure you are answering the proper question. In the problem you just did, remember that 5/17 is not the probability of what the question is asking for but rather the complement. To get the answer for "at least one" yellow, you must then subtract 5 /17from 1 to get the correct answer: 12/17. A common mistake using this method is to forget to subtract the complementary probability from 1 (and of course that answer choice will be there!).

19. In a bowl of marbles, eight are yellow, six are blue, and four are black. If John picks two marbles out of the bowl at random and at the same time, what is the probability that at least one of the marbles will be yellow? (A) 5/ 17 (B) 12 /17 (C) 25/ 81 (D) 56/ 81 (E) 4 /9 Whenever you see the phrase "at least one" in a probability question, it is probably easier to find the complementary probability of "none." In the Skillbuilder, you learned that the sum of the probabilities of complementary events is always 1. Since "none" is always complementary to "at least one," on this problem you should find the probability of getting no yellow marbles and subtract that from 1. (It is much easier to find the probability of no yellow than to add up the three ways to get at least one yellow.) Since 10 of the 18 marble are not yellow, the probability on the first pick that you will not get a yellow marble is 10/18, and on the second pick the probability of no yellow is 9/17. To get the probability of no yellow and then no yellow again, you need to multiply those two probabilities: 10 /18x 9/17 = 5 /17. For the final answer, subtract that value from 1 tO get the probability of getting at least one: 1− 5/17 = 12/17. Answer choice B is correct. For those who want to see the calculations for adding up the probabilities of the different ways to get at least one yellow, here they are: P(yellow then not yellow) = 8/18 x 10 /17= 40 /153P(not yellow then yellow) = necessarily the same 10/18 x 8 /17= 40/153 P(yellow then yellow) = 8/18 x 7/17 = 28/153 Summing those probabilities together (you want the probability of one or the other or the other and they are all mutually exclusive) = 40/153 + 40/153 + 28/153 = 108/153 = 12/ 17 If you needed any motivation to use the complementary approach on problems such as this, those tedious calculations should provide it!

2. Set Z consists of five consecutive even integers. If each term in set Z is divided by 2, the standard deviation of the set will ______________________. (A) increase (B) decrease (C) remain unchanged

2. B. For the same reason as above, when the terms are all divided by the same number (which has an absolute value greater than 1), the terms will become closer together, reducing their differences from the mean and creating a smaller standard deviation. Say the set was {10, 12, 14, 16, and 18}. The mean is 14, and the differences from the mean are 4, 2, 0, 2, and 4. Divide each by 2, and the new set of {5, 6, 7, 8, and 9} has a mean of 7 and differences from the mean of 2, 1, 0, 1, and 2. Smaller distances from the mean and the same number of terms yield a smaller standard deviation.

Special Dependent Probability: Probability questions can be made difficult when certain conditions must be met within a problem. Consider this classic example of "pairs" probability that has been used in many finance job interviews:

20. What is the probability that you get a pair when picking the top two cards off of a randomly shuffled deck of cards (52 total cards, with 13 sets of four matching suits)? (A) 12/ 2,652 (B) 16/ 2,652 (C) 1 /17 (D) 1 /13 (E) 1/2 This question is asking for the probability of getting any pair, not one specific pair (say a pair of 2s or 3s). Thinking about the problem one pick at a time (which is how you should always think about events at the same time), you realize that the first pick does not matter. In other words, the probability of a favorable outcome on the first card is 52/52 After you pick a card, you must now match that card with the second pick. For whatever card you have chosen, there will be three favorable outcomes left in the deck out of 51 cards (remember you used one in the first pick). Therefore the probability to get anypairpickingthetoptwocardsoffthedeckis52/52× 3 /51= 3 /51or 1/17.AnswerchoiceCis correct. Problems like this are very logic-based, so they are favorites for tricky Probability questions. Once you learn how to think about "pairs probability," it is not that difficult, but the first time students see this problem they tend to have a low success rate. Consider one more quick drill to reinforce this important concept: The individual socks that form exactly five matching pairs are randomly assorted in the bottom of a laundry bag. If you reach in at random and grab two socks, what is the probability that you are holding a pair?

What Happens When You Can't Write Out the Possibilities? When the number of events is greater than three, it becomes impractical to write out the number of outcomes. While these types of questions are uncommon and only faced by students facing the upper-bin questions, they do appear on the GMAT. Consider the following example:

21. What is the probability of getting exactly three heads on five flips of a fair coin? (A) 1 /32 (B) 3 /32 (C) 1/4 (D) 5 /16 (E) 1/2 The easiest place to start on this problem is to determine the number of total outcomes when you flip a coin five times. Since there are five spots with two choices for each, there are always 32 possible outcomes in any binomial problem with five events (2 x 2 x 2 x 2 x 2). It is not practical to write out that many outcomes in a reasonable amount of time, so you must use your understanding of counting to figure out how many of these 32 outcomes contain exactly three heads. What you do know is that each outcome has the same probability, so you must only calculate the number of outcomes that contain exactly three heads and two tails, and put that over 32 for the overall probability (favorable outcomes over total outcomes). Going back to your knowledge of counting, the question is really asking for the number of unique ways to write out HHHTT. This can be done with your formula for permutations with repeating elements (remember the Mississippi problem). In this case N = 5 and there are two things repeating, one of them two times and the other three times. Therefore the formula says there are 5!/3!2! , or 10 ways to uniquely write out HHHTT. Therefore, of the 32 outcomes ranging from HHHHH to TTTTT, exactly 10 of them have the favorable combination of three Hs and two Ts. The answer then is 10/32 = 5/16 . Answer choice D is correct.

PROBABILITY: FIND SIMPLE APPROACH Remember: On any GMAT problem that seems extremely complicated, there must be a simple approach. There simply is not enough time to solve a problem like this by considering all the different ways you could get at least one pair. On any Probability problem that seems overly complex, slow down and consider the wording carefully, looking for logical shortcuts based on the concepts presented in this lesson. PUTTING IT ALL TOGETHER Many of the harder Probability questions simply require you to leverage several individual concepts at once. Consider the following example, which draws on two of the more important concepts from this lesson:

22. If four fair dice are thrown simultaneously, what is the probability of getting at least one pair? (A) 1/6 (B) 5/18 (C) 1/2 (D) 2/3 (E) 13 /18 The only way to solve this question in a reasonable amount of time is to determine the probability of getting no pairs, and then subtract that probability from 1. Again, the trigger words on this question to prompt that approach are "at least one." To find the probability of getting no pairs, take each roll individually, and consider the approach discussed in the previous pairs problems. For the first roll, it does not matter what you get, so all numbers are favorable, and the probability is 1. On the subsequent rolls, you must determine the probability that you do not get a pair. On the second roll, there are five favorable outcomes out of six, because five of the numbers will not match the original roll. On the third roll, there are four favorable outcomes out of six, because four of the numbers will not match the two numbers that have already been rolled. On the fourth roll, there are there are three favorable outcomes out of six, because only three of the numbers will not match the first three rolls. As a result, the probability ofgettingnopairsis 6/6 x 5 /6x 4 /6x 3/6 = 5 /18andtheprobabilityofgettingat leastonepairis1− 5/18 =13/18 Answer choice E is correct.

3. Identify which of the following pairs of events are mutually exclusive but not complementary. (A) Getting heads/getting tails on one flip of a coin (B) The Red Sox winning the 2012 World Series/the Yankees winning the 2012 World Series (C) Getting at least one head/getting no heads on three flips of a coin

3. All complementary events are mutually exclusive, but there are many mutually exclusive events that are not complementary. For answer choice A, the two events are both complementary and mutually exclusive; the probability of getting heads or tails is 1, and the probability getting heads and tails is 0. The same is true for answer choice C: The probability of getting no heads or at least one head is 1, but the probability of getting both outcomes is 0. For answer choice B, the two events cannot happen together, so they are mutually exclusive. While people living in those two areas might think that the Red Sox and the Yankees are the only teams that matter, other major league teams can win the 2012 World Series. Therefore they are not complementary. The probability of the Red Sox or the Yankees winning the 2012 World Series is not 1, so the correct answer choice is B.

3. Set A consists of {1, 2, 3, 4, 5, 6, 7}. If each of the integers in set A is multiplied by -1, the standard deviation of set A will ______________________. (A) increase (B) decrease (C) remain unchanged (D) be multiplied by -1

3. C . Remember: Standard deviation only relies on distance from the mean. With the positive integers, the mean is 4 and the distances are 3, 2, 1, 0, 1, 2, and 3. When the terms are all negative, the mean is -4 and the distances stay the same. Because standard deviation involves squaring the differences, then taking the square root, the standard deviation will never be negative, so answer choice D is a trap choice here. Standard deviations essentially measure distances. You wouldn't say that the next town over is "negative 20 miles away," so you'll never say that a standard deviation is -1.5. Standard deviations are either positive or zero, as you will see in the next explanation.

STATISTICS: Number Properties and Statistics -PROVE IT IF YOU ARE NOT SURE Many of the trickiest Statistics problems on the GMAT are logically parallel to the Arithmetic number property questions that you saw in that lesson. The problems are more about how a particular statistic works (and what that statistic measures) than about actually doing calculations. Often these are presented in Data Sufficiency form, and they can be quite difficult, particularly those involving standard deviation. To see one example, consider the following problem, which tests your conceptual understanding of the median:

3. Set D is a new set created by combining all the terms of sets A, B, and C. No other terms are added to set D other than those in sets A, B, and C. What is the median of set D? (1) Sets A, B, and C each have a median of 125. (2) Sets A, B, and C each have the same number of terms. Remember the biggest takeaway from the previous average problem: Make sure that you do some math if you're not sure conceptually. The same holds true on this commonly missed problem. It seems logical to students that the number of terms in each set should matter, so many people pick answer choice C. But note that this question has all the makings of a classic "C trap" Data Sufficiency problem: The two statements together definitely give you sufficiency, so that should raise the possibility that statement (1) could be sufficient on its own. (Clearly the second one is not.) You can solve this conceptually, but you better be sure if you use this approach (and if not, you should test some sets as shown below). In a situation like this, the number of terms does not matter. If each set has the same median, then the combined set will have that same median. By definition, the median means that there are an equal number of terms on each side of the median. Regardless of whether set A has three terms or 1,000 terms, half of its terms are to the left of the median and half are to the right. And as the same is true for sets B and C, when you combine the sets, half of all terms will fall to the left of the median and the other half will fall to the right. Answer choice A is correct. NOTE: Ifthereisanoddnumberofterms,then"half"isn'ttechnicallytrue,asthe middle term is the median itself. But of the remaining terms, half will be above and half will be below, so the demonstration still holds. One of the most important strategies for attacking number property questions seems quite simple: Do something! Since number property questions can be quite abstract and confusing, doing some simple calculations or using parallel examples often gets the ball rolling. If you are unsure about the answer to this problem, then you shouldn't keep thinking about it. Rather, you should make some sets (which can be done in less than two minutes) and either prove or disprove the answer that you think is correct conceptually.

3. In seven nights as a waitress, Jenna's tips totaled $202, $195, $194, $206, $190, $187, and $212. If her goal was to average $200 per night, by what percent was her income lower than her goal?

3. This question adds a bit of a twist on the "minimize calculations" theme. Notice that her tip totals all fall within a few dollars of her goal of $200, but also that the actual numbers are a bit messy to calculate in your head or on paper. In a question such as this, you can simply calculate the differences between each total and the average: $202 is +2; $195 is -5; $194 is -6; $206 is +6; $190 is -10; $187 is -13; and $212 is +12. Keeping track of the +/- from the mean, you have + 2-5-6+6-10-13+12=-14. And-14 spread over seventerms is an average of -2, so she averaged $198. That's 1% below her goal.

In seven nights as a waitress, Jenna's tips totaled $202, $195, $194, $206, $190, $187, and $212. If her goal was to average $200 per night, by what percent was her income lower than her goal?

3. This question adds a bit of a twist on the "minimize calculations" theme. Notice that her tip totals all fall within a few dollars of her goal of $200, but also that the actual numbers are a bit messy to calculate in your head or on paper. In a question such as this, you can simply calculate the differences between each total and the average: $202 is +2; $195 is -5; $194 is -6; $206 is +6; $190 is -10; $187 is -13; and $212 is +12. Keeping track of the +/- from the mean, you have + 2-5-6+6-10-13+12=-14. And-14spreadoverseventermsisanaverage of -2, so she averaged $198. That's 1% below her goal.

For the set {3, 1, 2, 3, 4, 5}: 4. What is the mean? 5. What is the median? 6. What is the mode? 7. What is the range? 8. Is the standard deviation higher or lower than it was for the previous set?

4. Note the change here: Because the only number added to the old set was the average itself, the average stays the same at 3. 5. Now that you have six terms (an even number of terms), the median is the average of the two middle terms: 3 and 3, so the median is 3. 6. Now that there is a duplicate term, 3, this set has a defined mode: 3. 7. Because the highest and lowest terms were unaffected from the previous set, the range stays the same at 4. 8. This question is a bit tricky, but should demonstrate to you the way in which the GMAT will expect you to understand standard deviation. Conceptually, standard deviation is pretty true to its name: It involves the average (standard) amount of dispersion (deviation) from the mean. So adding an additional value to the set that provides no variance from the mean will reduce the average variance. Even without the calculation itself, you should recognize that a set in which the terms are closer to the average will have a smaller standard deviation. Therefore, the second set will have a smaller standard deviation than the first.

4. What is the probability of getting four heads in a row on four flips of a coin? 5. What is the probability of getting three heads in a row and then a tail on four flips of a coin? 6. What is the probability of getting two 3s in a row followed by a different number on three rolls of fair six-sided die?

4. To calculate the probability of multiple independent events, simply multiply the different probabilities together. Here that is 1/2 x 1/2 x 1/2 x 1 /2= 1/16 . 5. This trick question has the same answer as the previous one. The probability of headsandtailsarethesame,sotheanswerwillstillbe 1/2 x 1/2 x 1 /2x 1 /2= 1/ 16 6. This is independent probability with specific conditions put on the favorable outcomes. The probability of getting a 3 on the first role is 1/6 and on the second role is also 1/6 . On the next role anything but 3 is favorable, so there are five favorable outcomes, for a probability of 5/6 . You then multiply those probabilitiestogettheanswer: 1/6 x 1/6 x 5 /6= 5/216 .

STATISTICS: Evenly Spaced Sets, Median, and Mean -LEVERAGING STATISTICAL RULES One of the most important concepts from the Skillbuilder lesson is the relationship between the mean and median in evenly spaced sets. In evenly spaced sets, the mean and the median are the same. Remember to leverage this fact on any Statistics problem:

4. What is the sum of all even integers from 650 to 750, inclusive? (A) 3,500 (B) 35,000 (C) 35,700 (D) 70,000 (E) 70,700 In this problem, you are dealing with a textbook example of an evenly spaced set (all even integers, evenly spaced apart by 2). Knowing that, you can employ the following two rules to make the calculation efficient: In evenly spaced sets, the median equals the mean. Mean = sum of the terms , so sum of the terms=# of terms × mean With the first rule, you can determine that the mean is just the middle number between 650 and 750, which is 700. Using the second rule, you can see that the mean (700) equals the sum of the terms (that's what the question asks about) divided by the number of terms. To solve for the sum, you'll multiply the mean times number of terms. At this point, all you have to do is determine the number of terms in the set, but that is the tricky part of the problem. In the Arithmetic lesson, you were exposed to the inclusive set rule, which applies here. To review, consider a sample problem: If John is going to build 50 feet of fence and put a post every 10 feet, how many posts does he need? To solve, divide 50 by 10 to get 5, but then you must add 1 to account for both ends. Visually you can see this clearly: To determine the number of terms in any inclusive set, first take the difference between the terms and divide by the frequency of the terms. Then add 1 to that result, and you have the number of terms. In the sample problem above, the difference is 50 (50 - 0) and the frequency is every 10. Divide 50 by 10 and add 1: 5 + 1 =6. On the actual problem, the difference between 750 and 650 is 100. Since you only want even numbers, the frequency is every 2. Divide 100 by 2 and add 1: 50 + 1 = 51. 51 ∙ 700 is 35,700, so the correct answer choice is C. Another important takeaway from this problem is that you can use statistical definitions to your advantage even when a question doesn't technically ask about stats. Any time that a questions asks you for a seemingly incalculable sum, you find it by using your knowledge of the mean. Since the mean = sum of the terms/# of terms , any unusual sum question requires only that you find the average and the number of terms and multiply them together. The tricky component in these problems is usually finding the number of terms. The common mistake on this problem is either to forget to divide by 2 or forget to add 1 (or both!) when determining the number of terms. Of course, all of those incorrect answers are accounted for in the answer choices. If you get confused and forget the inclusive set rule, you can easily prove it to yourself with a simple example such as the fencepost problem above.

5. If you are drawing one card from a regular deck of cards (52 cards, with 13 sets of four), what is the probability that you draw a 4 or a spade?

5. Again you should note that these two events are not mutually exclusive; you could get both a 4 and a spade (the 4 of spades). So you must apply the general case formula and remember to subtract the overlapping probability. P(4)= 4/52 = 1/13 , P(Spades)= 13/52 = 1 /4,andP(4andSpades)= 1/52 P(4orSpades)= 4/52 + 13/52 − 1 /52= 16/52 = 4 /13.The answer is 4/13

5. Fernando purchased a university meal plan that allows him to have a total of three lunches and three dinners per week. If the cafeteria is closed on weekends and Fernando always goes home for dinner on Friday nights, in how many ways can he allocate his meals during the week?

5. This Combination problem is a bit more complicated than the first four examples (and more like you will see in the lesson portion of this book). First, you must realize that order is unimportant. Having lunch on Wednesday, Thursday, and Friday is no different from having lunch on Thursday, Friday, and Wednesday. Next, you must understand that it is really two separate problems. First figure out how many different ways the three lunches could be allocated, and then figure out how many ways the dinners could be allocated. Since there are five days in which to allocate the three lunches, the number of lunch allocations is a combination 5 pick 3 = 5!/(3!2!) = 5 x 4/2 = 10. For the dinners you must read carefully, as you only have four days in which to allocate the three dinners (because he always eats dinner at home on Fridays). The number of dinner allocations is a combination 4 pick 3 = 4! /(3!1!)= 4. Now you must understand the counting tree and realize that putting these together creates the final set of possibilities, so you must multiply. The answer is 10 x 4 = 40 possible allocations for lunch and dinner.

5. If there are 20 people in a room and each person shakes everyone else's hand exactly once, how many handshakes take place?

5. This trickier problem requires that you think logically. First, realize that you have 20 choices for the first spot and only 19 for the second spot. (You cannot shake your own hand.) That would give you 20 x 19, or 380 handshakes. That is not the answer, however, because you are double-counting all the handshakes. For instance, when Ailene shakes Bill's hand, that is the same handshake as when Bill shakes Ailene's hand. As you will learn later, this is really a combination problem, but it is included here because it can be solved with simple counting principles and some basic (but tricky!) logic. To get the correct answer, simply divide 380 by 2 to eliminate all the double-counting. The correct answer is 190.

STATISTICS: The Lesser Statistics: Mode and Range - "COULD BE " QUESTIONS IN STATISTICS While not tested as commonly, mode and range will still show up on the GMAT. For mode, you must remember that sets can be multi-modal (have more than one mode) or they can have no mode (if no number appears more than once). The concept of range is straightforward, but it can be made confusing with abstract "could be" questions:

5. Which of the following could be the range of the following set: {-21, 23, x, y, z}? I. -54 II. 40 III. 63 The range is defined as the largest value minus the smallest value, so you should know that if x, y, and z are between -21 and 23, the range will be 23 - (-21) = 44. What that also means is this: The range of 44 cannot be decreased! The only way it can be changed is if one (or more) of the variables is less than -21 or greater than 23, in which case the range would increase. Any other values inside of -21 < x < 23 will have no effect on the range. Accordingly, the range cannot be any less than 44, so from the options given it can only be 63, making answer choice C correct. Beware of abstraction in statistics-based questions. Your natural inclination is to want to see all of the terms in a set before you make a determination on it, but your knowledge of statistics definitions and rules can allow you to make decisions without a complete picture. You will encounter many abstract "could be" Statistics questions on the GMAT.

STATISTICS : Standard Deviation -DONT FORGET ABOUT NO OF TERMS In the Skillbuilder, you learned how to actually calculate standard deviation for a set, but on the GMAT you will never be asked to do that calculation. You will be asked many number property questions about how that calculation works and what standard deviation is measuring. Remember: Standard deviation is a measure of the dispersion of terms around the mean. It is calculated by determining the average squared difference (variance) of each term from the mean and then taking the square root of the variance. Here are two examples of how standard deviation is tested on the GMAT:

6. Sets A, B, and C are shown below. If the number 100 is added to each of these sets, which of the following represents the correct ordering of the sets in terms of the absolute increase in their standard deviation, from largest to smallest? A {30, 50, 70, 90, 110} B {-20, -10, 0, 10, 20} C {30, 35, 40, 45, 50} (A) A,C,B (B) A,B,C (C) C,A,B (D) B,A,C (E) B,C,A 7. If M is a negative integer and K is a positive integer, which of the following could be the standard deviation of set {-7, -5, -3, M, 0, 1, 3, K, 7}? I. -1.5 II. -2 III. 0 (A) I only (B) II only (C) III only (D) I and III only (E) None of the above The first problem (#6) shows that if you cannot calculate standard deviation, you can recognize what it means. It relates to the dispersion of data points from the mean. So in adding any one term and trying to gauge its impact, you need to consider two things: 1.) How does it change the mean; and 2.) How far from the mean will it fall relative to the current dispersion? In this case, you should see that set B will be the set most affected. The current mean is 0, so adding 100 can only pull it up so far. (The current terms net to 0, so the new sum will be 100, which divided by 6 terms leaves an average of a little over 16.) This not only increases the deviation from -20 and -10 to a good bit farther from the mean, but 100 remains quite far itself. Set C will also be significantly affected by the change, as 100 falls substantially out of its range. You should see immediately, however, that because the current mean is closer to 100 already, the shift won't be as dramatic (the mean will rise from 40 to 50), and the terms will still fall relatively close to the mean, with even 100 significantly closer than it was in set B. Set A will be the least affected. Note that 100 won't extend its range at all (the set already includes a higher value, 110) and so, while 100 will pull the mean farther from the lower terms 30 and 50, it itself will fall quite close to the new mean, and it will bring the mean closer to 90 and 110. So the order is B, C, A; answer choice E is correct. More importantly, you should recognize that this is an eyeball question; you shouldn't have to do the math to see the relative impacts of the additions of 100. In the second problem (#7), remember from the Skillbuilder that the standard deviation is never a negative number. It represents the typical distance of a data point from the median, and distances are expressed in the terms of > 0. That "or equal to 0" caveat applies when a set has no deviation from the mean—when all the values are the same. As this set clearly has multiple values, and therefore some deviation, 0 is not a possibility. The answer must be "none of the above"—answer choice E. If there was a "close call" in your analysis of the sets in the first problem, it would be between sets B and C, as you could argue that set C is already such a small standard deviation (its distances from the mean are half those of set B) that perhaps a shift in both the mean and an added term would wreak more havoc. But C, B, A is not a choice! Set A should clearly be the least affected set, so C, A, B is clearly not correct. When in doubt, check the answer choices to see if you really have to work harder. Here, some may be tempted to prove their choice by actually calculating standard deviations, but the answer choices show that it's not necessary. In the second problem, if you don't look at answer choices then you will waste all kinds of time. It is really a trick question, as none of them can be the standard deviation! Testmakers will often bait you into doing more work on standard deviation questions than is really necessary. Since you never have to actually calculate a standard deviation on the GMAT, these questions are always conceptual in nature. Think carefully about how standard deviation works and you will unlock the correct answer. 8. Is the standard deviation of set A > standard deviation of set B? (1) Set A consists of consecutive multiples of 10. (2) Set B consists of consecutive multiples of 2. (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked. (B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked. (C) BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient. (D) EACH statement ALONE is sufficient to answer the question asked. (E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed. A common mistake with standard deviation is focusing only on the spacing of the terms and forgetting about the number of terms. In this example, each statement alone is clearly insufficient, as you need information about each set to be able to compare standard deviations and answer this question. The answer, then, must be answer choice C or E. In assessing the statements together, be careful about any assumptions you might make. If you have two sets with the same number of terms, then the one that is more spaced (multiples of 10) would indeed have a greater standard deviation than the one that is more narrowly spaced (multiples of 2). However, no information is given here about how many terms are in each set. Since standard deviation is a measure of dispersion around the mean, the number of terms is a hugely important piece of information. If set B had a million consecutive multiples of 2, it would clearly have a greater standard deviation than set A if it only contained 10 consecutive multiples of 10. Therefore the relationship between the two standard deviations cannot be determined, and the correct answer choice is E. Now that you have completed all of the questions in this section and the Skillbuilder, you should notice that for each statistic there are certain traps that are exploited. Here are

combinations

Because rearranging the elements in questions where the order is irrelevant will not create a new Combination, combination problems deal with selecting a smaller subset from a larger pool. Otherwise, if all elements from the pool have to be included in the selection and the order of items is unimportant, there is only one way to make this choice. For instance, there is only one way to select a group of five new hires from five interviewees, or to choose a team of eight participants from eight nominees. No additional combinations can be created. Combinations Formula Unlike permutations, combination problems are best solved using a formula. While you could use your understanding of basic counting principles and logic to solve these, it would be more cumbersome and time-consuming. Combinations Formula: N !/( K!(N-K)!)

Probability Summary

Below is a summary of the important concepts, formulas, and strategies that are required for success on GMAT Probability questions: Probability—describes the likelihood of a certain event; all probability values fall between 0% and 100%, or between 0 and 1, inclusive. Probability of a single event: P(A) = # of outcomes when A occurs /total # of possible outcomes Mutually exclusive events—events that can never occur together (i.e., the occurrence of one event completely eliminates the probability of occurrence of the other). Examples: Getting an A, a B, and a C as the final grade on one class Answering 17 questions right and answering 21 questions right on a particular exam Important Properties of Mutually Exclusive Events 1. Mutually exclusive events never occur together: P(A and B) = 0. 2. If A and B are mutually exclusive, then P(A or B) = P(A) + P(B). Two events are complementary if one and only one of them must occur. All complementary events are mutually exclusive (because only one of them can occur), but not all mutually exclusive events are complementary. Examples: Failing or passing a certain course (contrast with getting a particular grade) Getting at least one head on three flips of a coin or getting no heads

PROBABILITY OF ONE EVENT AND ANOTHER EVENT Dependent Events

Dependent Events When determining the probability of multiple dependent events, multiply the probability of the first event times the probability of the second event, assuming the first event has occurred. Consider the following example: There are four black balls and three white balls in a jar. What is the probability of picking one black ball and then one white ball, without replacement? As with independent events, you need to multiply the probability of the two events. However, in this case the probability of the second event is dependent on the first event. For the first event, the probability is 4/7 . For the second event, the probability has changed based upon the first event. Since one ball has been taken out, there are now six total outcomes and, assuming that the ball picked was black, three of the remaining balls are white and thus favorable. Therefore, the probability of picking a white ball on the second pick is 3/6 . The probability ofthetwoeventsisthus 4/7 x 3/6 = 2/7 . The probability of dependent events occurring together is calculated using the general-case AND formula: P(A and B) = P(A) • Pa (B) Here P(A and B) is the probability that both these events will occur; P(A) is the probability of event A; and Pa (B) is the probability that event B will occur assuming that A has already occurred. Pa (B) is called "conditional probability."

Important Properties of Complementary Events

Important Properties of Complementary Events 1. Complementary events never occur together: P(A and B) = 0. 2. Because one of the complementary events must occur, their probabilities sum up to 1: P(A or B) = 1; P(A) = 1 - P(B); P(B) = 1 - P(A). Events A and B are dependent if the occurrence of one event affects the probability of another. Examples: Drawing balls without replacement Allocating a limited number of prizes among the audience (Each subsequent allocation of a prize reduces the probability of the remaining participants winning.) Important Properties of Dependent Events 1. P(A and B) = P(A) • PA(B) Where P(A and B) is the probability that both events will occur; P(A) is the probability of event A; PA(B) is the probability that event B will occur assuming that A has already occurred; PA(B) is called conditional probability. Events A and B are independent if the occurrence of one event does not affect the probability of the occurrence of another. In case of sequential events, the initial situation is restored before each subsequent experiment. Examples: Tossing a coin Selecting balls from a jar with replacement

Important Properties of Independent Events

Important Properties of Independent Events • P(A and B) = P(A) • P(B) OR formulas (probability of A or B): • General case: P(A or B) = P(A) + P(B) - P(A and B) • Mutually exclusive events: P(A or B) = P(A) + P(B) • Complementary events: P(A or B) = 1 AND formulas (probability of A and B): • • • Other • • • • Mutually exclusive and complementary events: P(A and B) = 0 Dependent events: P(A and B) = P(A) • PA(B) Independent events: P(A and B) = P(A) • P(B) Important Strategies: Always consider if you are dealing with dependent probability. It is easy to miss dependent probability on questions. Often it is easier to find the complementary probability, particularly when you see the words "at least one." If a question seems too complicated, it may be because you can easily solve for the complementary probability. Read carefully, particularly in binomial probability questions. There is a big difference, for instance, between asking the probability of getting "two heads in a row and then a tail" and "two heads and a tail in any order." Many probability questions are best solved using combinatorics or writing out the possibilities. Probability is not always multiplying or adding probabilities; sometimes it is just figuring out the number of favorable outcomes and the number of total outcomes.

2,3,5,7,11,13,and17 For set J, which of the following values would reduce the mean? 1, 7, 9, 15, 18 For set J, which of the following values would reduce the median? For set J, which of the following values would reduce the range?

Knowing that the mean is between 8 and 9, you don't need to do any calculation here. You know that any value lower than the current mean would reduce the new mean, so 1 and 7 would both reduce the mean, and the others would all increase it. Because the current median is 7, only numbers to the left of 7 would reduce the median. Only 1 does so. 7 itself would keep the median the same, as the new set of {2, 3, 5, 7, 7, 11, 13, 17} would have as its two middle terms 7 and 7. Their average, 7, keeps the median the same. None of these would reduce the range. Even though 1 would shift the set leftward on the number line, by doing so it would increase the range even while reducing the mean and median. Numbers that are inside the current high/low values of 2 and 17 (that means 7, 9, and 15) wouldn't affect the range, and 18 would increase the range by extending the set higher.

RESTRICTION PROBLEMS:- PERMUTATIONS WITH COUPLES OR GROUPS THAT CANNOT BE CHOSEN TOGETHER:

PERMUTATIONS WITH COUPLES OR GROUPS THAT CANNOT BE CHOSEN TOGETHER: Q: Jill brought 6 glasses - white, red, black , grey , yellow, and blue and would like to display 3 of them on the shelf next to each other , if she decides that red and a blue glass cannot be displayed at the same time , in how many different ways can jill arrange the glasses? A: 1. Find total permutation, unconstrained = 6!/(6-3)! = 120 2. To find where constraint would be violates , assume red and blue on the shelf -only 1 place is left with 4 choice of colours so combination = (4!)/(1!(4-1)!) = 4!/3!=4 no of colour groupings= 4 no of ways in which we can arrange each of these voilated constraint groups = 4 x ( no of permutations of 3 items displayed in shelf for each combination of 3 constraint voilating colours) = 4! x 3!=24 3. Now 120-24 = 96

STATISTICS :SUMMARY

Statistics Summary The summary below outlines the important concepts, formulas, and strategy points for Statistics: Arithmetic mean (or simply the mean)—the average value in a set. Arithmeticmean= sumoftheterms/# of terms • On problems involving averages, be ready to calculate both the mean and the sum. If the problem provides an average, the solution typically requires computing the sum. • If a problem asks you to calculate an unusual sum, find the average and the number of terms, and multiply them together. • If all terms of a set are multiplied by a constant, the new average can be derived by multiplying the initial mean by the same constant. • Be careful with problems in which you are multiplying or dividing a term that is already in the set. In those cases, it is a good idea to do the math. Median—the middle value of a set ordered in ascending order. If a set contains an even number of terms, the median is the average of the two middle values. • To find the median, always rearrange the terms in ascending or descending order before finding the middle value. If a set contains an even number of terms, average the two middle values. • There is no fixed relationship between the median and mean of a set unless the sets are evenly spaced. • If a set is evenly spaced, the median equals the mean. If that same set is in ascending order, then you can determine the median and mean by averaging the first and last terms in the set. Mode—the most frequently occurring value in a set. If several values occur with the same frequency, each of these values represents a mode (i.e., the sequence has several modes). • To find the mode, identify the most frequently occurring value(s) in a set. Remember that there can be multiple modes, and if no number appears more than once then there is no mode. Range—the difference between the largest and the smallest value in a set. • To find the range, subtract the smallest value from the largest value in the set. Remember that range cannot be negative, and it is only 0 if there is one term in the set or all terms are the same. Standard deviation—measures how closely the terms in the set are spread around its mean. • Remember that standard deviation can never be negative and is only 0 if the range of the set is 0 (one term or all terms are the same). • Don't forget about the number of terms in determining standard deviation. Use the "Map Strategy" below to deduce the effect on standard deviation for certain set manipulations: Map Strategy for Standard Deviation • Adding or subtracting a constant from each element in the set has no effect on the standard deviation. Shifting a map does not change the distances. • Multiplying each term of a set by a number with an absolute value greater than 1 increases the standard deviation, while multiplying by a number with an absolute value less than 1 decreases the standard deviation. Increasing the scale of the map increases distances, while reducing the scale shrinks them. • Dividing each element in a set by a number with an absolute value greater than 1 decreases the standard deviation, while dividing by a number with an absolute value less than 1 increases the standard deviation. Increasing the scale of the map increases distances, while reducing the scale shrinks them. Changing the sign of all elements in the set or multiplying by -1 has no effect on the standard deviation.

Combinatorics Summary

The following is a summary of the different types of counting problems as well as strategies for attacking them. Remember: Most counting problems can be solved logically with the basic counting principles. Basic Counting Principle—to determine the total number of possible arrangements always multiply the number of possible choices for each available spot. Factorial—a product of all positive consecutive integers less than or equal to the number under the factorial. Examples:N!=N•(N-1)...•2•1 6!=6•5•4•3•2•1=720 General Permutations—ordered arrangements of elements without repetition. Use basic counting principles. Examples: Letters in a password (without repetition), Arrangements of five students in five seats, Number of ways to arrange N distinct items in N spots = N! Permutations with Repeating Elements—permutations of elements, some of which are indistinguishable from others. Use permutations with repeating elements formula. Examples: Arrangements of different vases, two of which are identical Arrangements of letters, some of which are repeated Number of arrangements of N items, of which A,B... are the same = N! /A! B!... Combinations—unordered arrangements of elements. Use combinations formula. Examples: Members of a committee or students in a study group General Combinations Formula Number of unordered K element sets created from N elements = N!/K! (N − K)! Counting Problems with Restrictions—place items where they are or are not allowed to be and arrange around them. In most cases, it is better to subtract what is not allowed than to add what is allowed. Other Important Strategies for Combinatorics: • Read carefully. It is essential that you interpret the wording properly and check to see if you can repeat or not, whether words like "at least" are present, etc. • Use shortcut calculations. Many of these problems can be shortcut with your understanding of units digit properties. Don't get baited into messy math. • Only use formulas for certain problems. You only need to use formulas on two types of problems (and they are actually one in the same): permutations with repeating elements and combinations.

REPEATED VALUES AND MEDIAN Another interesting setup in Statistics-based questions involves the presence—or, in more tricky situations, the possibility—of a repeat value at the median. Consider the question: If K is the median of the set {2, 5, K, 8, 10}, is K > 5? Your first inclination is likely "yes," as you'll probably think that K must be between 5 and 8. But how would you answer this question? What is the median of the set: {2, 5, 5, 8, 10}?

Well, those values are in ascending order, and 5 is the third of the five values. So the median here is 5. The answer to the first question is "we don't know." K might be greater than 5, or it just might be 5 itself. It is possible for the median to be a repeat term, and you will find that this concept makes for a tricky Data Sufficiency setup.

COMBINATORICS

When Does the Order Matter? Combinatorics problems can be divided into two broad categories: those for which the order of arrangements is important (permutations and basic counting) and those for which the order does not matter (combinations). To provide some insight into these concepts, let's consider a few examples demonstrating the differences between ordered and unordered arrangements. Permutations Permutations, or scenarios in which the sequence of elements is important, include the arrangement of books on a shelf, the finishing order of a race, or numbers in a password when the numbers cannot be repeated. Each variation in the sequence (e.g., switching places of letters, numbers, etc.) will generate a different permutation. If items can be repeated, it is not a true permutation, but the order still matters and these can be solved with your understanding of the basic counting principles to be discussed shortly. Another distinguishing feature of permutations lies in the fact that the number of elements often corresponds to the number of available slots for these elements. Typical problems involving permutations include arranging people in a line (every person needs one slot) or sitting an audience in a movie theatre (each person must be seated). Combinations By contrast, questions dealing with combinations (unordered arrangements) refer to arrangements in which the order does not matter and in which a smaller group of elements is drawn from a larger pool. Examples of such problems include selecting a three-person case competition team from a class of 250 MBA students, choosing several items from a menu in a student lounge, or purchasing three suits at a department store to wear to your second-round interviews. In each of these cases, the sequence of choices is unimportant and does not create a new combination. For instance, if students John, Jenny, and Lisa are selected to represent the school at a case competition, changing the order in which they were chosen does not yield a new combination. For instance, a team consisting of Lisa, John, and Jenny selected in that order is the same as the original team where John was selected first, followed by Jenny and Lisa. Similarly, if you order a turkey sandwich, then salad, then a glass of orange juice, you will end up with the same meal, regardless of which item you ordered first. If the order is not important in combinations, then what is? New combinations are created by selecting a different set of items from the pool of available possibilities. For instance, in our menu example, ordering the same turkey sandwich and salad with apple juice rather than orange juice would constitute a new combination of a meal. Similarly, replacing any (or all) of the three students with other candidates on the case competition team would also result in a new combination.