Probability

If the surface area of a certain cube is 2400, what is its volume? 8000(√6)3 1000(√6)3 8000 6400 4000

C Correct. Remember, the surface area of a cube is 6(Side²). Also, the volume of a cube is Side³. Given that the surface area of the cube is 6(Side²) = 2400, Side = √(2400/6) = √400 = 20. The volume of the cube = 203 = 23·103 = 8000. Hence, this is the correct answer.

John tossed a fair coin 3 times. What is the probability that the outcome was "tails" exactly twice? 1/8 1/4 3/8 1/2 9/10

C You overestimated the time this question took you. You actually solved it in 1 minutes and 16 seconds. Correct. Probability of getting a result of "heads" = Probability of getting a result of "tails" = 1/2. This question presents several possible scenarios to reach the desired outcome. Identify these scenarios, calculate each one using the following method, then add the probabilities of all scenarios, since they are connected by an "OR" relationship. Within each scenario: Break down the triplets of tosses into separate, single events (first toss, second toss, third toss); calculate the probability for each event One-at-a-time; then MULTIPLY the probabilities, as the events have an AND relationship between them. There are three "good" scenarios here: 1) "Tails" on toss 1 and 2, and "Heads" on toss 3: 1/2 × 1/2 × 1/2 = 1/8 OR 2) "Tails" on toss 1 and 3, and "Heads" on toss 2: 1/2 × 1/2 × 1/2 = 1/8 OR 3) "Tails" on toss 3 and 2, and "Heads" on toss 1: 1/2 × 1/2 × 1/2 = 1/8 That's a total probability of 1/8 + 1/8 + 1/8 = 3/8

Consider the following question: What is the probability of getting a 7 when rolling a six sided die, numbered 1-6? 0 1 7/6

A Correct. Out of all possible outcomes, the numbers 1-6 on the die, there is no 7, therefore you have no wanted outcomes within the possible 1-6. Use the probability formula: 0/6 = 0. The dry algebraic facts of this case are supported by the use of common sense. There is no way you can get a 7 out of a six sided die. This means it is never going to happen - a probability of zero.

To sum up:

To find the number of all possible outcomes: Event A AND Event B = (Number of outcomes in A)×(Number of outcomes in B) To find the probability of event A and event B happening together: Probability (A AND B) = Probability (A) × Probability (B)

We will talk about the actual calculations later. For now, just remember this:

An outcome is one thing that may happen in the setup of the problem. In the GMAT, probability problems ask about the chances or likelihood that events may happen. An event is any single choice in which there are a number of wanted outcomes out of a total number of outcomes.

Probability of zero means something can never happen. Probability of one means something MUST happen. Probability value is always between zero and one, inclusive. In other words, it is a fraction, and as such may be also expressed as a decimal or percent. In the GMAT, probability problems are just another way for GMAC to test part-to-whole relationships. You are likely to see 1-2 probability problems in your GMAT.

Now try this one: What are the chances of getting a number within the range of 1-6 inclusive, when rolling a six-faced die? 0 1/6 1 C Correct. There are six wanted outcomes (the numbers 1-6) out of six possible outcomes (the numbers 1-6). Use the probability formula: 6/6 = 1. You MUST get a number that is in the range 1-6 inclusive when rolling a six-faced die.

To sum up: Harder Prob Qs

Probability questions require careful definition of the scenarios presented by the question before attempting to calculate them. While the preferred method of choice remains breaking down the problem into a series of events and calculating their seperate probabilities before combining them, harder probability questions sometimes lend themselves to calculating the denominator (total number of outcomes) and numerator (the number of wanted outcomes) of the probability formula using Combinations counting methods. Keep an open mind, and if one approach starts leading you down a darker path with no visible end, try a different one.

To Sum up: 1

Many GMAT probability questions will present several scenarios, or several ways (which may consist of a series of events) of reaching the desired result: 1) Calculate the probability of each scenario separately. 2) Since the scenarios are different ways of reaching the same result with an OR relationship between them - ADD the probabilities together.

If k is a positive integer and the greatest common divisor of k and 45 is 15, then the greatest common divisor of k and 900 may be any of the following EXCEPT 15 45 60 150 300

B Correct. Remember that the GCD (Greatest Common Divisor) of two numbers is the greatest number by which both numbers are divisible. According to the question stem, the GCD (greatest common divisor) of k and 45 is 15. That means that the greatest number both k and 45 are divisible by is 15. If k would have been divisible by 45, the GCD of k and 45 would have been 45 itself. Hence, k is not divisible by 45. If k is not divisible by 45, then the GCD of k and any other number (in this case 900) cannot be 45.

To sum up:2

Geometric probability Qs are the same as regular probability questions, with a slight twist: Prob = Area of wanted figure / Total available area

The overall approach is therefore:

1) Break down the "at least" into the distinct scenarios covered by it. In our case, at least 1 white oyster allows any number of white oysters between 1 and the maximum of 3. Use the list of "good" scenarios" to define the list of "Forbidden" scenarios. In the above case, our "forbidden" scenario was 0 white oysters, 3 black oysters. 2) Calculate the probability for the "Forbidden" scenario using the relevant method, one-at-a-time. 3) Subtract the number of "Forbidden" combinations from 1 to find the probability of "Good" scenarios.

If n is an integer between 1 and 96 (inclusive), what is the probability that n is divisible by 8? 1/8 1/4 3/8 1/2 5/8

A Correct. The probability here is the number of multiples of 8 out of the total numbers in the range. The probability here is the number of multiples of 8 out of the total numbers in the range. The denominator, the number of items in the range is the difference between the extremes plus 1: 96-1+1 = 96 numbers. Now to the numerator, the number of multiples of 8 in the range: When counting the multiples of x within a given range (in this case x = 8): 1) Find the relevant extremes - the nearest multiples of 8 within the specified range: The effective range is actually 8 to 96 inclusive. 2) Subtract the relevant extremes, and divide by 8: 96 - 8 = 88. 88/8 = 11 3) Add one: 11 + 1 =12 Thus the probability that n is a multiple of 8 is 12/96 = 1/8.

A jar contains 40 marbles, of which 15 are orange and 25 are beige. if 18 of the marbles were randomly removed, what is the probability that the next marble to be removed at random is beige? (1) Of the marbles removed, the number of beige marbles is twice that of orange marbles. (2) Of the first 16 marbles removed, 10 are beige. statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked; statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked; BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked; EACH statement ALONE is sufficient to answer the question asked; statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

A Correct. This is a "what is the value of..." DS question. In this type of question, a statement will be sufficient only if it leads to a single value of the variable (or expression) that you're asked about. The issue here is the probability of drawing a beige marble after some of the marbles were taken out. In order to answer this question you have to know how many beige marbles remained in the jar after removing 18 marbles. Note that you already know the total number of marbles after the exclusion. According to Stat. (1), If a total of 18 marbles were removed in which x are orange are and 2x are beige, then: --> x+2x = 18 --> 3x = 18 --> x = 6 Thus the number of beige marbles removed, is 2x = 12, and after the removal there are 25-12 = 13 beige marbles in the jar and 15-6 = 9 orange marbles in the jar. The probability of drawing a beige marble in this situation is thus 13/22 (the number of beige marbles out of the total number of marbles remaining). Stat.(1)->S->AD. According to Stat. (2), There is no telling what is the color of the marbles drawn after the 16th marble: The two additional marbles could be either beige or orange. Therefore, you cannot know how many beige marbles remained after the exclusion, and thus you cannot calculate that number out of the remaining 22 marbles. Stat.(2)->IS->A.

In a basketball contest, players must make 10 free throws. Assuming that Shmill O'Real has a 90% chance of making each of his shots, how likely is it that he will make all of his 10 shots in the contest? (9/10)^10 (9/10)^2 9/10 (8/9)^10 (9/10)^9

A Correct. This is a multiple event probability question - ten events of throwing a free throw. Break down the question into separate, single events; calculate the probability for each event One-at-a-time; then MULTIPLY the probabilities, as the events have an AND relationship between them. The probability of O'Real making a single shot is 9/10, and this is the probability that he makes each of the 10 free throws. Probability of making all 10 shots = P(making first shot) AND P(making 2nd shot) AND etc. Since there's an "AND" relationship between shots, multiply 9/10 by itself 10 times to get (9/10)10.

Bill's compact disc player randomly plays a song, so that no song is repeated before the entire album is played. If Bill plays a disc with 14 songs, what are the chances that the third song he hears will be his favorite? 1/14 1/12 1/11 3/14 1/3

A Correct. This is a multiple event probability question - three different playings of a track from the disc. Break down the question into separate, single events; calculate the probability for each event One-at-a-time; then MULTIPLY the probabilities, as the events have an AND relationship between them. When considering each event one-at-a-time, assume that the previous events yielded a wanted response, and consider each event under the new conditions created by this assumption. For questions without repetition, this means that the conditions (i.e. total number of outcomes and number of wanted outcomes) change for each successive event. For the third song to be Bill's favorite, songs 1,2 and 4 through 14 should not be Bill's favorite song. There are 13 unfavorable songs. The probability of that happening is: 13/14 × 12/13 × 1/12 × 11/11 × 10/10 × 9/9 × ...... × 1/1 = 1/14 Note that from the 4th pick onwards the probability of an unfavorable song being played is always 1, since the only favorite has already been played. Reduce the first three fractions: 13/14 × 12/13 × 1/12 = 1/14 Alternative method: Note that there is an equal probability of 1/14 for any of the 14 songs on the CD to be the third that is played randomly, thus the probability of Bill's fav song being the third played is also 1/14.

How many 6 digits phone numbers are there if the first digit is 6, the second is 0 and all the other digits are different from each other and can include 6 and 0? 5040 5240 5420 6346 7880

A Correct. This problem presents a case of Single source - DIGITS. Break the problem in a step-by-step method using SeBoxes. There are 6 digits, so you need 6 SeBoxes. Since the digits after the second digit have to be different, there's no repetition and the SeBoxes' size changes from one digit to the next. The first digit is 6, therefore it only has 1 choice. The second digit is 0, therefore it only has 1 choice. The third digit can be any digit (including 6 and 0) - 10 options. Since the fourth digit must be different from the third digit, it has only 9 possible choices, and so on: 1×1×10×9×8×7 = 5040

A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red ball and a white ball (not necessarily in that order) in two successive draws, each ball being put back after it is drawn? 2/27 1/9 1/3 4/27 2/9

A Incorrect. This is a multiple event probability question - two events of drawing a ball. Break down the question into separate, single events; calculate the probability for each event One-at-a-time; then MULTIPLY the probabilities, as the events have an AND relationship between them. When considering each event one-at-a-time, assume that the previous events yielded a wanted response, and consider each event under the new conditions created by this assumption. When the balls are put back, each event starts with all items, and the conditions do not change. Note: This is the probability of drawing a red ball first and a white ball second, but what about drawing a white ball first and a red ball second? D Correct. There are 2 good scenarios in this question: 1) Drawing a red ball, putting it back and drawing a white ball: There are 3 red balls and 3+4+2 = 9 balls in total, thus the probability of drawing a red ball is 3/9 = 1/3. Now put back the red ball. There are 2 white balls out of 9, thus the probability of drawing a white ball is 2/9. Total probability = (1/3)×(2/9) = 2/27 OR 2) Drawing a white ball, putting it back and drawing a red ball: There are 2 white balls and 3+4+2 = 9 balls in total, thus the probability of drawing a red ball is 2/9 Now put back the white ball. There are 3 red balls out of 9, thus the probability of drawing a red ball is 3/9 = 1/3. Total probability = (1/3)×(2/9) = 2/27. Note that the order of drawing the balls does not change the probability. Since There's an OR relationship between scenarios, add: 2/27 + 2/27 = 4/27

Miguel rolls a six-sided die, and records the number shown on the upper face. What is the probability that the recorded number is "1" or "5"?

Recall the probability formula: Probability = No of wanted outcomes / Total no of possible Finding the total number of possible outcomes on a single roll of a die is easy: the possible outcomes are {1, 2, 3, 4, 5, 6}, or 6 possible outcomes. The number of wanted outcomes here is also easy: 1 and 5 are both wanted outcomes, so there are 2 wanted outcomes out of 6 possible ones. The probability of the described event is 2 / 6, or 1/3.

2 people are to be selected from Abraham, Benjamin, Chris, and Dave. What is the probability that both Abraham and Benjamin will be selected? 1/12 1/6 1/3 2/3 5/6

A Incorrect. This question presents two possible scenarios to reach the desired outcome. Identify these scenarios, calculate each one using the following method, then add the probabilities of all scenarios, since they are connected by an "OR" relationship. Within each scenario: Break down the pair of chosen persons into separate, single events (first choice, second choice); calculate the probability for each event One-at-a-time; then MULTIPLY the probabilities, as the events have an AND relationship between them. When considering each event one-at-a-time, assume that the previous events yielded a wanted response, and consider each event under the new conditions created by this assumption. B Correct. The two wanted scenarios are: Abraham is chosen first - 1/4, and then Benjamin is chosen - 1/3. a total probability of 1/4×1/3 = 1/12 OR Benjamin is chosen first - 1/4, and then Abraham is chosen - 1/3. a total probability of 1/4×1/3 = 1/12 Since this is an "OR" relationship, add the probabilities: 1/12+1/12 = 2/12 = 1/6 Alternative method using combinations: When calculating complex probabilities, start from the denominator. It is the number of all pairs. This is equal to picking 2 out of 4, not ordered, with no repetition, since you cannot pick the same person twice. C(4,2) = 6 Abraham and Benjamin are 1 pair, so the probability that they will be chosen is 1/6

Let's look at the scenario of choosing Private Benjamin second: For the first choice, we want anyone BUT private Benjamin - we're saving him for second. Therefore, there are only 9 wanted outcomes of "Not Benjamin" out of 10 possible outcomes, or 1st choice 9/10 × 2nd choice × 3rd choice × 4th choice For the second choice, we want private Benjamin - so there are 9 remaining possible outcomes, and only 1 wanted outcome: 1st choice 9/10 × 2nd choice 1/9× 3rd choice × 4th choice

As before, once Benjamin is chosen, we don't care who gets chosen for 3rd and 4th places. So the remaining probabilities are 8/8 and 7/7 for 3rd and 4th choice, respectively. We end up with: 1st choice 9/10 × 2nd choice 1/9× 3rd choice 8/8 × 4th choice 7/7 = 1/10 It should come as no surprise that the probability of Private Benjamin being chosen second is equal to the probability of private Benjamin being chosen first - there's no real logical reason why one should be greater than the other. The remaining scenarios (private Benjamin being chosen 3rd and 4th) will yield the same probability (work it out yourself if you're not sure). Since these scenarios all have a relationship of OR between them (private Benjamin can be chosen first OR second OR third OR fourth), add all probabilities together to get: 1/10 + 1/10 + 1/10 + 1/10 = 4/10 = 2/5 which was our original answer.

Now try this one, What is the probability of getting a "2" on a single roll of a regular six sided die? 0 1/6 2/6 4/6 1

B Correct. There is one wanted outcome (getting the number "2") out of 6 possible outcomes (the numbers 1-6).

If John rolls a plain die twice, what is the probability that both rolls will give the same result? 1/36 1/6 1/3 1/2 2/3

B Correct. This is a multiple event probability question - two events of rolling a single die. Break down the question into separate, single events; calculate the probability for each event One-at-a-time; then MULTIPLY the probabilities, as the events have an AND relationship between them. Note that the desired outcome is "same result" - not any single particular result. Therefore, there are several scenarios that lead to a desired result: in this case, the results (1,1), (2,2), (3,3), (4,4), (5,5) or (6,6) are all wanted outcomes. When there are multiple scenarios think of the different events that are included in the "wanted" outcomes, calculate the probability of one of those events and ADD the probabilities for each of the different scenarios, as they present an "OR" relationship between them. Calculate the probability of one of those scenarios: for example, the chance to receive 1,1 - The chance to receive "1" on a die toss is 1/6. Since there are 2 tosses of "1", multiply 1/6×1/6 to receive a probability of 1/36. The probability for each of the 6 possible "good" scenarios (e.g. 2,2) is the same, and since there is an "OR" relationship between them, add 6 times 1/36 = 6×(1/36) = 1/6 Alternative Method: Look at each event as a whole: First toss: It does not matter to John what he rolls in the first toss. Thus there are six possible outcomes, all of which are wanted outcomes, and the probability is 6/6 = 1. Second toss: whatever number came up on the first toss, John needs the same particular number on the second toss. The odds of getting a single particular result on a die toss = 1/6. Multiply the probabilities of the two events to get a total probability of 1×(1/6) = 1/6

A number is selected at random from the first 30 natural numbers. What is the probability that the number is a multiple of either 3 or 13? 17/30 2/5 7/15 4/15 11/30

B Correct. This question presents several possible scenarios to reach the desired outcome. Identify these scenarios, calculate each one using the following method, then add the probabilities of all scenarios, since they are connected by an "OR" relationship. The probability here is the number of multiples of 3 OR the number of multiples of 13 out of the total numbers in the range. Find the total number of numbers in the range, that's your denominator (the total number of possible outcomes). Find the number of multiples of 3 and add to the number of multiples of 13, that's your numerator (the number of "good" results). The denominator, the number of items in the range is the difference between the extremes plus 1: 30-1+1 = 30 numbers. Now to the numerator. The number of multiples of 3 in the range: When counting the multiples of x within a given range (in this case x = 3): 1) Find the relevant extremes - the nearest multiples of 3 within the specified range: The effective range is actually 3 to 30 inclusive. 2) Subtract the relevant extremes, and divide by 3: 30-3 = 27. 27/3 = 9 3) Add one: 9 + 1 = 10. So there are 10 multiples of 3 in the range. Now, do the same for multiples of 13: When counting the multiples of x within a given range (in this case x = 13): 1) Find the relevant extremes - the nearest multiples of 13 within the specified range: The effective range is actually 13 to 26 inclusive (the other steps are a bit redundant in the case of 13, since there are only 2 multiples --> 13 and 26, the "extremes"). 2) Subtract the relevant extremes, and divide by 13: 26-13 = 13. 13/13 = 1. 3) Add one: 1 + 1 = 2. So there are 2 multiples of 13 in the range. Before you rush and answer, stop for a moment and think - are there any multiples of BOTH 3 and 13 in the range? Because if there are such multiples, we have just counted them twice (once for 3, and once for 13), and thus you need to subtract their number once from the total. Ok, there aren't any multiples of 3 AND 13 in the given range, the first multiple of 3 and 13 is their product, 39, which is not in the range. Phew. There are 10 + 2 = 12 multiples of 3 OR multiples of 13 in the first 30 natural numbers, Thus the probability that the number is a multiple of 3 or 13 is 12/30 = 2/5.

What is the remainder when dividing 2^21 by 3? 4 3 2 1 0

B You slightly underestimated the time this question took you. You actually solved it in 3 minutes and 13 seconds. Incorrect. When dividing a number by 3, the remainder cannot be 3. Eliminate this answer choice. This question is about the remainder of a division. There is no straightforward way for calculating the remainder, and 221 is an unbelievably large number. There must be a pattern of some sort. To understand what is going on, take a few steps using simpler versions of the question: What is the remainder when dividing 2^1=2 by 3? 2 divided by 3 is zero with a remainder of 2. remainder of 2^2=4 by 3? 4 divided by 3 is one with a remainder of 1. Keep going until you see a repeating pattern. C Correct. remainder of 2^3 / 3? 8 / 3 is two with a remainder of 2. remainder of 2^4 / 3? 16 / 3 is five with a remainder of 1. remainder of 2^5 / 3? 32 / 3 is ten with a remainder of 2. It seems that when dividing 2n by 3, the remainder is 2 when n is odd, and 1 when n is even. Since 21 is odd, it follows that the remainder of 221 divided by 3 is 2.

Mike's pencil box contains 30 pens. 15 of the pens are red, and all of the others are either blue or black. If Mike were to choose a random pen from the pencil box, what is the probability that it would be blue? (1) The probability that the pen will be red minus the probability that the pen will be black equals 0.3 (2) The probability that the pen will be red or blue is 0.8 statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked; statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked; BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked; EACH statement ALONE is sufficient to answer the question asked; statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

C According to Stat.(1), The probability that the pen will be red minus the probability that the pen will be black equals 0.3. This would allow us to calculate the probability of the pen being black. (since P(red)-P(black)=0.3.). If you know the probability of black and red, you know the probability of blue: P(red)+P(black)+P(blue) = 1 Therefore this statement is sufficient. Stat.(1)->S->AD. D Correct. This is a "what is the value of..." DS question. In this type of question, a statement will be sufficient only if it leads to a single value of the variable (or expression) that you're asked about. The issue in this question is probability, so keep in mind the probability formula: In order to find the probability of choosing a blue pen, you need to know the number of blue pens out of the total 30 pens. Note that 15 red pens out of a total of 30 pens means that probability of red = 15/30 = 0.5. According to Stat.(1), The probability that the pen will be red minus the probability that the pen will be black equals 0.3. This would allow us to calculate the probability of the pen being black. (since P(red)-P(black)=0.3.). If you know the probability of black and red, you know the probability of blue: P(red)+P(black)+P(blue) = 1 Therefore this statement is sufficient. Stat.(1)->S->AD. According to Stat.(2), The probability that the pen will be red or blue is 0.8. Since the probability of a single event with two possible outcomes with an OR relation equals to the sum of their probabilities, this means that: P(red)+P(blue)= 0.8 By plugging in P(red) in the equation above, it is possible to calculate P(blue). Therefore, this statement is sufficient. Stat.(2)->S->D.

Next scenario: What is the probability of getting a "5" and a "2"? 1/36 1/6 × 1/6 The same as the probability of getting a "6" and a "1"

C Correct! The probability of getting a "5" on a die 1 is 1 / 6. The probability of getting a "2" on die 2 is 1/6 as well. Since we want a "5" AND a "2", multiply the probabilities to get 1/6 × 1/6 = 1/36. The same goes for all other scenarios - each of them has a probability of 1/36. Our Table looks as follows: Add the probabilities together to get 1/36+1/36+1/36+1/36+1/36+1/36 = 6/36 = 1/6.

If a^2 = a, which of the following must be true? |a| = -a a = -a a = |a| -a^2 = |a| a^3 = -a^2

C Correct. Manipulate the equation into the form of a2-a=0. Factorize the equation by extracting the common factor a: a2-a = a(a-1) = 0. Based on this, the roots of the equation (and the only possible values of a which satisfy the equation) are a=0 and a=1. Plug these values into the answer choices, and POE any answer choice which does not hold true for one of the values. This answer fits all possible values of a: 1 is indeed equal to |1| 0 is indeed equal to |0|. All of the other answer choices do not hold for a=1, and are thus eliminated. Thus, C remains the only correct answer choice.

If a is a non-negative integer less than 10, for how many different values of a is a^2=2^a? 0 1 2 3 4

C Correct. Plug in 0,1,2,4,8 for a, to eliminate answer choices. All other integers less than 10 (3, 5, 7, 9) have no chance as they include factors other than 2. Plug in a=0. 0^2 (0) is not equal to 2^0 (1) Plug in a=1. 1^2 (1) is not equal to 2^1 (2) Plug in a=2. 2^2 (4) is equal to 2^2 (4) Plug in a=4. 4^2 (16) is equal to 2^4 (16) Plug in a=8. 8^2 (64) is not equal to 2^8 (256) Since only 2 and 4 satisfy the equation, this is the correct answer.

(Goal Sept Excel) In a recent street fair students were challenged to hit one of the shaded triangular regions on the large equilateral triangular board below with a ping pong ball. Each of the triangular regions is an equilateral triangle whose side is a third of the length of the large triangle board. If the ping pong ball hits the large triangular region, what is the probability of hitting a shaded triangle? 1/5 1/4 1/3 1/2 2/3

C Correct. This is a geometric probability question. Remember the formula for geometric probability: In this case, the issue is area of the three black triangles. You are asked to find the area of the small black triangles out of the area of the big triangle (the entire shape). In order to avoid unnecessary calculations, try to divide the hexagon into equilateral triangles - you only need the ratio between the shaded triangles and the unshaded ones. Remember - GMAT figures are drawn to scale unless otherwise stated. Divide the unshaded hexagon into 6 identical equilateral triangles, by drawing the main diagonals:The black triangles are identical to the 6 triangles of the hexagon - their angles must also be 60º (in order to supplement the hexagon's angles, 120º, and they each share a side with the hexagon's triangle. They even look the same area.) Thus, the entire shape is made of 9 identical triangles, 3 of them shaded. Therefore, the chance of hitting them in a random throw of a ball is 3/9 = 1/3. Avoid unnecessary calculations - ballpark!

The Full House Casino is running a new promotion. Each person visiting the casino has the opportunity to play the Trip Aces game. In Trip Aces, a player is randomly dealt three cards, without replacement, from a deck of 8 cards. If a player receives 3 aces, he or she wins the game. If the deck of 8 cards contains 3 aces, what is the probability that a player will win the game? 1/336 1/120 1/56 1/720 1/1440

C Correct. This is a multiple event probability question - three events of picking a card. Break down the question into separate, single events; calculate the probability for each event One-at-a-time; then MULTIPLY the probabilities, as the events have an AND relationship between them. When considering each event one-at-a-time, assume that the previous events yielded a wanted response, and consider each event under the new conditions created by this assumption. There is only one scenario that results in a win: receiving three aces. The probability of picking out the first ace, when there are 3 aces in a deck of 8 cards, is 3/8. The probability of picking out the second ace, when there are 2 aces in a deck of 7 cards, is 2/7. The probability of picking out the first ace, when there is 1 ace in a deck of 6 cards, is 1/6. You need the first ace AND the second ace AND the third ace in order to win, so multiply the discrete probabilities to receive the probability of winning: (3/8) × (2/7) ×(1/6) = 1/56.

Jason flips a coin three times. What is the probability that the coin will land on the same side in all three tosses? 1/16 1/8 1/4 1/3 1/2

C Correct. This is a multiple event probability question. Break down the question into separate, single events; calculate the probability for each event One-at-a-time; then MULTIPLY the probabilities, as the events have an AND relationship between them. It does not matter on which side the coin falls on the first toss, since both sides are "wanted outcomes" for the purpose of 3 tosses of the same kind, thus the probability that the coin will fall on either side in the first toss is 2/2. Once Jason tossed the coin, it must fall on the same side on the next two tosses as well. Thus, regardless of the outcome on the first toss, there is only 1 "wanted" outcome out of two for each of tosses 2 and 3, and the probability that the coin will fall on a specific side is 1/2. The total probability is thus: 2/2 × 1/2 × 1/2 = 1/4 Alternative method: Same side = 3 "heads" OR 3 "tails". Consider each scenario seperately: The probability of 3 "heads" is: 1/2×1/2×1/2 = 1/8 and this is the same for 3 "tails". Since you need 3 "heads" OR 3 "tails", add the probabilities to get: 1/8+1/8 = 2/8 =1/4

A car rental service facility has 10 foreign cars and 15 domestic cars waiting to be serviced on a particular Saturday morning. Because there are just a few mechanics, only 6 cars can be serviced. If the 6 cars are chosen at random, what is the probability that first 3 cars selected are domestic and the other 3 are foreign? 15×14×13×10×9×8 / 25! 15/25 × 14/24 × 13/23 × 12/22 × 11/21 × 10/20 15/25 × 14/24 × 13/23 × 10/22 × 9/21 × 8/20 15×14×13×12×11×10 / 25! 1 - 15/25 × 14/24 × 13/23 × 10/22 × 9/21 × 8/20

C Correct. This is a multiple event probability question. Break down the question into separate, single events; calculate the probability for each event One-at-a-time; then MULTIPLY the probabilities, as the events have an AND relationship between them. When considering each event one-at-a-time, assume that the previous events yielded a wanted response, and consider each event under the new conditions created by this assumption. There are 15 domestic cars in 25 cars total, so the probability that the first car chosen is domestic is 15/25. The second, 14/24, the third, 13/23. The probability that the fourth car is foreign (there are 10 foreign cars in the 22 remaining cars) is 10/22, the fifth - 9/21, the sixth 8/20. Multiply, because multiplication is FUN! (and also because there's an "AND" relationship between car picks):

The probability of rain showers in Barcelona on any given day is 0.4. What is the probability that it will rain on exactly one out of three straight days in Barcelona? 0.144 0.072 0.432 0.72 0.288

C Correct. This question presents several possible scenarios to reach the desired outcome. Identify these scenarios, calculate each one using the following method, then add the probabilities of all scenarios, since they are connected by an "OR" relationship. Within each scenario: Break down the three days into single events (first day, second day, third day); calculate the probability for each event One-at-a-time; then MULTIPLY the probabilities, as the events have an AND relationship between them. Note that if the probability of rain is 0.4 then the probability of no rain on any given day is 1 - 0.4 = 0.6, because: The probability of x not happening = 1 - the probability of x happening. There are three "good" scenarios here: 1) Rain on the first day and no rain on the other days: 4/10 × 6/10 × 6/10 = 144/1000 = 0.144 OR 2) Rain on the second day and no rain on the other days: 4/10 × 6/10 × 6/10 = 144/1000 = 0.144 OR 3) Rain on the third day and no rain on the other days: 4/10 × 6/10 × 6/10 = 144/1000 = 0.144 That's a total probability of 0.144+0.144+0.144 = 0.432

What is the probability of randomly selecting one of the shortest diagonals from all the diagonals of a regular hexagon? 1/4 1/3 1/2 2/3 7/9

C Incorrect. This is a probability question involving geometry. The issue is the number of diagonals in a hexagon. Figure out the total number of outcomes first: the number of diagonals of a polygon is given in the formula:n(n-3) / 2 , where n is the number of sides. In the case of the hexagon, the n=6. As to the number of wanted outcomes - draw a regular hexagon and its diagonals, and see how many of the diagonals are shorter. D Correct. A regular hexagon has 9 diagonals. Since we're talking about a hexagon here, it's quite simple to draw the diagonals: and see that there the 9 diagonals are divided into two groups: 3 diagonals are longer (the main diagonals, connecting opposite vertices) and the other 6 are shorter. That's a 6/9 = 2/3 chance of randomly picking a shorter diagonal.

There are two lines in the neighbourhood supermarket. In the first line there are 2 people, one with 10 items in her cart and the other with 14 items. In the second line there are 3 people, one with 11 items, one with 12 items and one with 13 items. What is the probability of randomly picking 2 people (it doesn't matter from which line), whose combined number of items is 23? 1/10 1/5 2/5 6/25 16/25

C Incorrect. This question presents several possible scenarios to reach the desired outcome of 23 on two shoppers. Identify these scenarios, calculate each one using the following method, then add the probabilities of all scenarios, since they are connected by an "OR" relationship. Within each scenario: Break down the pair of shoppers into separate, single events (first choice, second choice); calculate the probability for each event One-at-a-time; then MULTIPLY the probabilities, as the events have an AND relationship between them. When considering each event one-at-a-time, assume that the previous events yielded a wanted response, and consider each event under the new conditions created by this assumption. A Incorrect. B Correct. When picking the first person there are only 4 "wanted" shoppers (which we will name after the number of items in their carts) - 10,11,12 and 13. 14 is not a "wanted" choice, since it does not have someone to complete it to 23 (9 is not in the supermarket that day). Count the number of "good" pairs - pairs whose sum reaches 23: (10,13), (13,10), (11,12), (12,11) Now, calculate the probability of getting a "good" pair (for example, 10 and 13): The probability of picking "10" is one out of five: 1/5. Once you've picked "10", the probability of picking its wanted partner "13" is one out of the remaining four: 1/4. That's a total probability of: 1/5 × 1/4 = 1/20 Since there are four such pairs with the same probability, and you need pair 1 OR 2 OR 3 OR 4, ADD the probabilities of each pairs to get: 1/20 + 1/20 + 1/20 + 1/20 = 4 × 1/20 = 4/20 = 1/5 When picking the first person there are only 4 "wanted" shoppers - 10,11,12 and 13. So the probability of picking a good customer on the first choice is 4 "good" customers out of 5 total customers = 4/5. When picking the second person there are only 4 people to choose from, because you cannot pick the same person twice. Now, once you already have the first number, there is only 1 person that will complete it to the desired 23 - If the first one is 10 - you need only 13; if the first one is 11, you need only 12; and so on. Thus the probability of that happening is 1/4. Multiply to get the total probability: There are four "good" pairs.

What happens when the two separate events happen simultaneously, such as rolling two dice together to get a 6-6?

What happens indeed? Look at the following version of the above problem: If John rolls two six-sided dice, what is the probability that both dice will give a result of "6"? Although the two rolls happen simultaneously, they are still two separate events. We still need a "6" on one die AND a "6" on the other die. Therefore, the method remains the same: break down the question into separate, single events; calculate the probability for each event One-at-a-time; then MULTIPLY the probabilities, as the events have an AND relationship between them. The result will still be 1/6 × 1/6 = 1/36.

A basic model of a slot machine randomly shuffles and presents different combinations of the letter of the word MATH. What is the probability that the position of the "A" remains unchanged when the letters of the word MATH are re-arranged? 1/4 1/6 1/3 1/24 1/12

C Incorrect. Use the probability formula: Calculate the denominator first - the total number of ways in which the word MATH can be re-arranged, then calculate the good choices - those in which the "a" doesn't change its place. A Correct. Your denominator is the total number of ways in which the word MATH can be re-arranged. It has 4 different letters, thus the number of arrangements = 4! = 4×3×2×1 = 24 (number of arrangements in a row for n different items = n!) Now, if the position of the "A" does not change, the first, third and the fourth positions are reserved for consonants and the vowel A remains at the second position. The consonants M, T and H can be re-arranged in the first, third and fourth positions in 3! = 6 ways. Therefore, the required probability = 3!/4! = 1/4

What is the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains at least 1 woman? 11/102 77/204 77/102 91/102 31/34

C Very good! You took 1 minutes and 24 seconds to answer this question. Incorrect. In AT LEAST ONE questions use the following method: 1) Break down the "at least" into the distinct scenarios covered by it. Use the list of "Good" scenarios" to define the list of "Forbidden" scenarios. 2) Calculate the probability for the "Forbidden" scenarios using the relevant method, one-at-a-time. 3) Subtract the number of "Forbidden" combinations from 1 to find the probability of "Good" scenarios. D Correct. At least one woman allows 1, 2, 3 or 4 women on the committee. Therefore, the "Forbidden" scenarios in this question are only those in which no woman is chosen among the four people. Calculate the probability of that happening. There are 7 women in 18 people, thus 11 are not women: 11/18 × 10/17 × 9/16 × 8/15 This is the probability of NOT PICKING ANY WOMEN, thus the probability of picking AT LEAST ONE WOMAN is: 1 - 11/18 × 10/17 × 9/16 × 8/15 = (after reduction) 91/102

David and Rachel are getting married. The extended family wants to have its picture taken, but David's father is unwilling to stand next to Rachel's mother. How many options does the photographer have to arrange the 10 family members in a row for the picture? 9! 9×8! 8×9! 10!/2! 10!

C You grossly underestimated the time this question took you. You actually solved it in 3 minutes and 43 seconds. Correct. This question presents a "Forbidden scenario". Use the "Forbidden choices" formula: [Total combinations - "Forbidden" combinations] = Good combinations In this case the forbidden choices are those in which the two enemies stand next to each other. There are 10! options to arrange 10 people in a row. Now, calculate the forbidden options. To do that, consider the two enemies as a single "person". There are 9! ways to arrange 9 people in a row. We're not done: since the two enemies have 2! (=2) internal arrangements between them (Father on the left, then mother, and vice versa), the number of forbidden arrangements is: 9!×2! Subtract the forbidden choice from the total number of choices to get only the good options: 10! - (9!×2) = ? --> 10! = 10×(9×8×7×6×5×4×3×2×1) can be written as simply 10×9!: --> 10! - (9!×2) = 10×9! - 2×9! It is then possible to extract 9! as a common factor: --> 10×9! - 2×9! = 9! × (10-2) = 9! ×8

What is the probability of randomly selecting an arrangement of the letters of the word "MEDITERRANEAN" in which the first letter is E and the last letter is R? 1/13 1/20 1/26 1/50 1/100

C You slightly overestimated the time this question took you. You actually solved it in 1 minutes and 21 seconds. Correct. Use the probability formula: Calculate the denominator first - the total number of ways in which the word "MEDITERRANEAN" can be re-arranged, then calculate the good choices - those in which the first letter is E and the last letter is R. Pay attention to the identical letters - these indicate an "internal ordering" type combinations problem, where the total number of arrangements must be divided by the number of internal arrangements of each group of identical letters. The total number of arrangements: There are 13 letters, of which 3 are E, 2 are N, 2 are A and 2 are R. This is an "internal ordering" problem: There are 13! arrangements for 13 different letters, but you have to divide by the number of internal arrangements of the 4 groups of identical letters: 13! / 3!×2!×2!×2! Now for the good choices, in which the first letter is E and the last one is R. There is only 1 choice for the first and last letter. Of the other 11 letters, there are 2 Ns, 2Es and 2As. Again, an "internal ordering" question: divide the number of ways of arranging 11 different letters by (11!) by the number of internal arrangements of each group of identical letters: 11! / 2!×2!×2! Thus the probability is: (11! / 2!×2!×2!) / (13! / 3!×2!×2!×2!) = 1/26

Some Probability problems involve the concept of geometric probability: A dart board is composed of two concentric circles. The small circle has a radius of 2, and the large circle has a radius of 7. If a dart lands somewhere on the board, what is the probability that it lands within the smaller circle? Recall the probability formula: Prob = No of Wanted Outcomes / Total no of possible The term "number of wanted outcomes" doesn't make much sense when discussing areas of circles (or other geometric shapes), but the concept of probability as representing a part-to-whole relation between a wanted outcome and the total possible outcomes remains the same. In simple words - the probability that a dart lands within a specific wanted area is simply: Prob = Area of wanted figure / Total available area

In our case: The area of the wanted figure is the area of the small circle: πr^2 = π2^2 = 4π. The total available area is the area of the large circle: πr^2 = π7^2 = 49π. Therefore, the probability of a dart landing within the small circle is simply 4π / 49π = 4 / 49.

If John rolls two six-sided dice, what is the probability of that the sum of the two rolls is 7? An interesting question. Clearly, there are two separate events presented by this Question: roll of die 1, and roll of die 2. For example, a result of 7 can be reached by rolling a "6" on die 1 and a "1" on die 2. However, this is not the only scenario leading to the desired result: A "5" on die 1 and a "2" on die 2 will also result in a sum of 7. List all the possible scenarios of the two events in your notebook. How many possible scenarios of the two events will lead to a sum of 7? 4 scenarios 5 scenarios 6 scenarios 7 scenarios

Correct. There are in fact 6 different scenarios of reaching a sum of seven from the results of two six sided dice: Die 1 Die 2 6 1 5 2 4 3 3 4 2 5 1 6 From the table we learn that reaching a result of 7 means: getting a "6" and a "1" OR a "5" and a "2" OR a "4" and a "3" OR etc. Once we've established these relationships, the way forward is as follows: 1) Calculate the probability of each scenario separately. 2) Since the scenarios are different ways of reaching the same result with an OR relationship between them, treat them the same way we've learned about different outcomes of the same EVENT with a OR relationship - ADD the probabilities together. Let's do this. Begin with the scenario of getting a "6" and a "1": calculate the probability if each event separately, one-at-a time: The probability of getting a "6" on die 1 is 1 / 6. The probability of getting a "1" on die 2 is 1/6 as well. Since we want a "6" AND a "1", multiply the probabilities to get 1/6 × 1/6 = 1/36.

Jerry has 3 flavors of ice cream in his parlor. How many options are there for George to pick a one-flavor, two-flavor or three-flavor order? 2 3 4 7 8

D Correct. Calculate the different groups (1,2,3) seperately and then add the results. 1 flavor: 3 choice 2 flavors: use the combinations formula for picking k=2 out of n=3 ice cream flavors. Order doesn't matter, because we only need to know which flavors are chosen - not which is chosen first, then second. C(3,2) = 3 3 flavors: Only one way - choose all three flavors. Since George can order 1 flavor OR 2 flavors OR 3 flavors, add the results to receive 3+3+1 = 7 option

A student is entered in a college housing lottery for 2 consecutive years. what is the probability that the student receives housing through the lottery for at least 1 of these years? (1) 80% of the students in the lottery do not receive housing through the lottery in any given year. (2) Each year 1 in 5 students receives housing through the lottery. statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked; statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked; BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked; EACH statement ALONE is sufficient to answer the question asked; statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

D Correct. This is a "what is the value of..." DS question. In this type of question, a statement will be sufficient only if it leads to a single value of the variable (or expression) that you're asked about. The issue here is finding the probability of getting housing in AT LEAST ONE year out of two. In at least questions it's best to work with the probability of the "bad" cases = not getting housing at all, because: The probability of x happening + the probability of x not happening = 1 According to Stat. (1), 80% of the students do not get housing in any given year, thus 20% do get housing. The probability of winning the housing lottery in a certain year is thus 1/5 (=20%). You're asked to find the probability that the student will receive housing in AT LEAST one of the two years. For At Least questions, It's easier to find the probability of this not happening, and subtract it from 1. The probability of the student not getting housing in any of the two years: The probability of NOT GETTING housing in a single year = 4/5, so in two years = (4/5)*(4/5) = 16/25. The probability of getting housing in at least one of the years = 1 - the probability of not getting housing in any of the years = 1 - 16/25 = 9/25. Stat.(1)->S->AD. According to Stat. (2), You have the same information as you are given in statement 1, thus 2 is also sufficient. Stat.(2)->S->D.

A game at the state fair has a circular target with a radius of 10 cm on a square board measuring 30 cm on a side. Players win prizes if they throw two darts and hit only the circular area on both throws. If Jim hit the board on both throws, what is the probability that Jim won the game? 1/3 1/9 π/9 π2/81 π2/9

D Correct. This is a geometric probability question. In this game you have to hit twice to win. Break down the question into two events (first throw, second throw), calculate the probability for success in each one, then multiply the probabilities to find the probability of event A and event B happening together. Remember the formula for geometric probability: Prob = Area of wanted figure / Total available area In this case, the issue is the area of the circle. You are asked to find the area of the circle, and its relation to the area of the big square it's situated on (the entire shape). The area of the circle is all the "wanted" results, while the area of the square (including the circle) contains the total available area possible when throwing a dart. The area of the square = side2 = 30^2 = 900 The area of the circle = πr2 = 100π Thus the chance of hitting the circle on a single throw = 100π/900 = π/9 Since in order to win Jim has to hit the target on the first try AND on the second try, multiply: π2/81

A group of pictures of butterflies contains 10 pictures. Jim had bought 3 of them. If 2 pictures are to be picked out from the 10 pictures, what is the probability that both pictures are not those that were already bought by Jim? 14/15 3/5 6/13 7/15 7/10

D Correct. This is a multiple event probability question - two events of choosing a butterfly picture. Break down the question into separate, single events; calculate the probability for each event One-at-a-time; then MULTIPLY the probabilities, as the events have an AND relationship between them. When considering each event one-at-a-time, assume that the previous events yielded a wanted response, and consider each event under the new conditions created by this assumption. Since you cannot pick the same picture twice, there is no repetition, and this means that the conditions (i.e. total number of outcomes and number of wanted outcomes) change for each successive event. Since 3 pictures were bought, then the probability that the first picture was not bought is 7/10. After picking the first picture, there are 9 pictures to choose from, of which only 6 were not yet bought, so that's 6/9 chance that the second picture was also not yet bought. Multiply because of the "AND" relationship to get: (7/10)×(6/9) = 7/15

There are 12 balls in a jar: 6 red, 2 blue and 4 green. If a single ball is drawn from the jar, what is the probability of that it is either blue or red? 1/12 1/4 1/2 2/3 3/4

D Correct. This probability question presents a case of a single event - one draw of a single ball. If the question presents several wanted outcomes of the same event with an OR relationship between them, ADD the probabilities. Recall the probability formula: The probability of drawing a blue ball = the number of blue balls out of the total number of balls = 2/12 The probability of drawing a red ball = the number of red balls out of the total number of balls = 6/12 Since you need a blue ball OR a red ball, add the above results to get: 2/12 + 6/12 = 2/3

A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher? 1/4 1/3 2/3 6/7 8/9

D Very good! You took 1 minutes and 36 seconds to answer this question. Correct. For probability questions using "at least", use the following approach: Probability of "Good" Scenarios = 1 - Probability of "Forbidden" Scenarios This is because the "Good" Scenarios may contain many events, and thus may take up time to calculate directly. Break down the "at least" into the distinct scenarios covered by it. Use the list of "Good" scenarios" to define the list of "Forbidden" scenarios. Then, calculate the probability for the "Forbidden" scenarios using the relevant method, one-at-a-time. Finally, subtract the number of "Forbidden" combinations from 1 to find the probability of "Good" scenarios. In this question, the good scenarios are committees with at least one preacher, allowing committees with 1 preacher and one teacher, or committees with two preachers. Thus, the "Forbidden" scenarios are committees with no preachers, or in other words, 2 teachers. The probability of picking 2 teachers: For the first teacher, you have 3 teachers (the numerator) out of 7 people total (the denominator) = 3/7 After picking one teacher, there are 2 teachers to pick from out of 6 people = 2/6 = 1/3 Thus the total probability of a 2-teacher committee is (3/7)×(1/3) = 1/7 Remember, The probability of getting an "at least one preacher" committee = 1 - the probability of getting a "no preachers" committee. Therefore: The probability of getting an "at least one preacher" committee = 1 - (1/7) = 6/7

What is the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains exactly 1 woman? 77/204 77/832 11/77 308/1411 22/832

D You grossly underestimated the time this question took you. You actually solved it in 3 minutes and 10 seconds. Incorrect. This question presents several possible scenarios to reach the desired outcome of just one woman out of 4 people: a woman chosen first, second, third or fourth. Calculate each one using the following method, then add the probabilities of all scenarios, since they are connected by an "OR" relationship. Within each scenario: Break down the foursome into separate, single events (first choice, second choice, etc.); calculate the probability for each event one-at-a-time; then MULTIPLY the probabilities, as the events have an AND relationship between them. When considering each event one-at-a-time, assume that the previous events yielded a wanted response, and consider each event under the new conditions created by this assumption. B Incorrect. A Correct. A woman chosen first (and no other women chosen): There are 7 women in 18 people, thus 11 are not-women. Break it down into 4 events - choosing a woman, followed by 3 events of choosing non-women: (7/18)×(11/17)×(10/16)×(9/15) Don't calculate yet, think about the other scenarios. What about the probability of choosing a woman only in the second pick? non-woman × woman × non-woman × non-woman (11/18) × (7/17) × (10/16) × (9/15) Indeed. That's exactly the same as the first scenario, since the probability of a woman being picked anywhere is identical. Thus the total probability of picking just one woman is 4 times the probability of each of the scenarios: 4 × (11/18) × (7/17) × (10/16) × (9/15) Reduce: 4 with 16, 9 with 18: 11/2 × 7/17 × 10/4 × 1/15 Reduce the 10 with the 2 and 15: 11/1 × 7/17 × 1/4 × 1/3 = 77/204

What is the probability that out of the combinations that can be made using all the letters of the word EXCESS, Jerome will randomly pick a combination in which the first letter is a vowel and the last letter is a consonant? 96/320 24/180 33/100 48/180 96/180

D You slightly overestimated the time this question took you. You actually solved it in 1 minutes and 24 seconds. Correct. This problem presents a case of Single source. Break the problem in a step-by-step method using SeBoxes. There are 6 items in the required combinations, so you need 6 SeBoxes. Since you can use each letter only once, there's no repetition and the SeBoxes' size changes. Finally, since changing the order of the chosen letters yields a different word, the order of choice matters. After calculating the good choices, calculate the total different arrangements possible to figure out the probability to receive a good choice. Let's start with the good choices: Start filling the SeBoxes according to the possible choices - The first letter can only be a vowel - there are two vowels (the two Es), so there are 2 choices. The sixth letter can only be a consonant - 4 choices (X, C, S, S). Then move to the other terms: The second through fifth letters can be anything, but since we already used two letters, there are only 4 choices left for the second letter (2 out of 6 have already been used) and then 3, 2 and 1 choices respectively for the third , fourth, and fifth letters (the number decreases with each consecutive letter, because there is no repetition). The SeBoxes should therefore look like this: 1st Digit(2) × 2nd(4) × 3rd(3) × 4th(2) × 5th(1) × 6th(4) But! Since we have 2 "E"s and 2 "S"s, this number of arrangements includes some redundant choices displacing similar letters. These double letters turn this question into an internal ordering one. The internal ordering between the two Es and the two Ss doesn't matter, so we need to divide twice by 2! so as to get rid of the redundant arrangements. Thus the good arrangements are actually: 2×4×3×2×1×4 / 2!×2! = 48 good choices Out of how many combinations? There are 6! arrangements for 6 letters, but since there are 2 "E"s and 2 "S"s, divide twice by 2! to receive: 6! / 2!×2! = 180 total arrangements Thus the percentage of good choices is 48/180. Alternative Explanation: Treat the problem as a series of events with no repetition, and figure out the probability for each event. Since the first and last digit have limitations, start there: The Probability of the first letter being a vowel is 2/6 (two vowels - E and E - out of six letters). The probability of the last letter being a consonant is now 4/5 (four consonants - X, C, S, S - out of five letters, as one was already chosen for the first). Once these conditions have been set, you don't really care what happens to the four letters between the first and last: for each of those, all of the remaining letters are wanted outcomes. Thus the probability of a wanted outcome for the second letter is 4/4 (4 letters available, and all 4 are wanted outcomes), 3/3 for the third letter, 2/2 for the fourth, 1/1 for the fifth. In effect, the probability of a wanted outcome for the four middle letters is 1 - since we don't care what happens there, the probability of a wanted outcome is guaranteed. Thus, the final probability is 2/6×1×1×1×1×4/5 = 8/30 This does not appear directly in any of the answer choices, but a quick expansion of the fraction by 6 in both the numerator and the denominator leads back to 8×6 / 30×6 = 48/180.

There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 Portuguese books. What is the probability of choosing 2 books in different languages? 31/90 3/10 1/3 31/45 28/90

D You underestimated the time this question took you. You actually solved it in 2 minutes and 57 seconds. Correct. Break the problem into a series of events - pulling the first book, then the second book. The probability of choosing 2 books in different language then depends on the language of the first book: 1) Picking an English book AND picking a not-English Book. OR 2) Picking an Spanish book AND picking a not-Spanish Book. OR 3) Picking a Portuguese book AND picking a not-Portuguese Book. In questions that present several scenarios, or several ways (which may consist of a series of events) of reaching the desired result: 1) Calculate the probability of each scenario separately. 2) Since the scenarios are different ways of reaching the same result with an OR relationship between them - ADD the probabilities. 1) There are 5 English books out of 10 total books, so the odds of picking an English book are 5/10. There are 5 books not in English, so the odds of pulling a second book not in English are 5/9 (only 9 books in total, because one is already out after the first pick). So the probability of this scenario is: 5/10 × 5/9 = 25/90 2) There are 3 Spanish books out of 10 total books, so the odds of picking a Spanish book are 3/10. There are 7 books not in Spanish, so the odds of pulling a second book not in Spanish are 7/9 (only 9 books in total, because one is already out after the first pick). So the probability of this scenario is: 3/10 × 7/9 = 21/90 3) There are 2 Portuguese books out of 10 total books, so the odds of picking a Portuguese book are 2/10. There are 8 books not in Portuguese , so the odds of pulling a second book not in Portuguese are 8/9 (only 9 books in total, because one is already out after the first pick). So the probability of this scenario is: 2/10 × 8/9 = 16/90 Since there's an "OR" relationship between scenarios, add them to get the total probability: 25/90 + 21/90 + 16/90 = 62/90 = 31/45 Alternative Method: Probability is the number of wanted results out of the number of total results possible. The number of total options to pick 2 books out of 10 (not ordered) is 10!/[(10-2)!2! = 45 To easily calculate the number of wanted results, subtract the unwanted results from this total. The number of unwanted results, is the number of ways of to select 2 books in the same language. this depends on what that language is: 2 English books OR 2 Spanish books OR 2 Portuguese books: 2 English books - pick 2 out of 5 not ordered = 5!/[{5-2!)2! = 10 2 Spanish books = 3 options (it's like picking the unwanted book out of 3 possibilities) 2 Portuguese books - 1 option (there are only 2 books in that language) So the unwanted options are 10+3+1 = 14, which makes the good options 45-14 = 31, and the probability for a wanted option 31/45.

A game at the state fair has a circular target with a radius of 10 cm on a square board measuring 30 cm on a side. Players win prizes if they throw two darts and hit only the circular area on at least one of the two attempts. What is the probability that Jim won the game? 1- (9-∏)/9 1- (18∏-∏²/18) 9-∏ / 9 (9-∏)² / 81 18∏-∏² / 81

D You underestimated the time this question took you. You actually solved it in 3 minutes and 55 seconds. Incorrect. Note: This is the probability of NOT HITTING the target on both attempts. This is a geometric probability question. In this game you have to hit at least once to win. For "At least questions, it is easier to calculate the odds of NOT HITTING the target on both shots, and then subtracting the probability of this happening from 1: Probability of "Good" Scenarios = 1 - Probability of "Forbidden" Scenarios The probability of NOT hitting twice is equal to the probability of not Hitting on the first throw AND not hitting on the second throw. Remember the formula for geometric probability: In this case, the issue is the area of the circle. You are asked to find the area of the circle, and its relation to the area of the big square it's situated on (the entire shape). The area of the circle is all the ""wanted" results, while the area of the square (including the circle) contains the total available area possible when throwing a dart. E Correct. The area of the square = side2 = 302 = 900 The area of the circle = πr2 = 100π Thus the chance of hitting the circle on a single throw = 100π/900 = π/9 The chance of not hitting on a single throw is thus 1-the chance of hitting = 1 - π/9 = 9/9 - π/9 = (9-π)/9 The chance of missing the first shot AND the second shot is: (9-π)/9 × (9-π)/9 = (9-π)²/81 And thus the chance of NOT missing both shots, or in other words hitting the target AT LEAST ONCE is: 1-(9-π)²/81 Using Recycled Quad II you get: 18π-π²/81

If a is an integer and (a^2)/(12^3) is odd, which of the following must be an odd integer? a/4 a/12 a/27 a/36 a/72

E You slightly underestimated the time this question took you. You actually solved it in 3 minutes and 40 seconds. Correct. Think about what value a could be. There are two things we know about a2, which we can use to infer about a: 1) a2 is divisible by 123, so it must contain all of the prime factors of 123 (at least) in order to be able to reduce the top of the fraction with the bottom and come up with an odd integer result. 2) a2 must be a perfect square, i.e. must have an integer square root. Break down 123 : (4⋅3)3 = 43⋅33 = (22)3⋅33 = 26⋅33. So a2 should at least equal 26⋅33. Can we take the square root of 26⋅33 and get an integer result? Taking the square root is the equivalent of multiplying the powers by 1/2, so if we take the square root of a2 = 26⋅33 we get a=23⋅33/2 - which is not an integer. square root of 26 is √26 = 26/2 = 23. That's fine - the problem is with the 33 - the odd power doesn't allow an integer square root. This means that a2 must include at least an extra 3, to make a2 = 26⋅34, leaving a as √(26⋅34) = 23⋅32= 8⋅9 = 72. Which works, since in this case (a2)/(123) = 26⋅34 / 26⋅33 = 3, which is indeed an odd integer, as the question requires. All this leads us to the conclusion that a COULD equal 72 - it's not the only value of a, but it's a valid plug in. According to the question stem, one of the answer choices must always result in an odd integer for every value of a. Work with what you have: plug in a=72 into the answer choices and see if you get an odd result. If you don't, then eliminate the answer choice, as we've found a single counter example of a that does not give an odd result for that answer choice. A 72/4 = 18 - not odd B 72/12 = 6 - not odd. C 72/27 = that's not even an integer. D 72/36 = 2 - not odd. E 72/72 = 1 - the only odd answer. Since this is enough to eliminate answer choice A, B, C and D, E remains as the only right answer.

If n is an integer between 1 and 96 (inclusive), what is the probability that n×(n+1)×(n+2) is divisible by 8? 1/4 1/2 5/8 3/4 7/8

E You underestimated the time this question took you. You actually solved it in 3 minutes and 3 seconds. Incorrect. Plug in a few numbers for n to see what's going on: n=1 --> n×(n+1)×(n+2) = 1·2·3 = 6 --> not a multiple of 8. n=2 --> n×(n+1)×(n+2) = 2·3·4 --> a multiple of 8, since it has a 2 and a 4. n=3 --> n×(n+1)×(n+2) = 3·4·5 --> not multiple of 8. n=4 --> n×(n+1)×(n+2) = 4·5·6 --> a multiple of 8, since it contains one multiple of 4 and one more even number. Some experimentation reveals that there are two scenarios in which n×(n+1)×(n+2) is divisible by 8: 1) When n AND n+2 are even (thus one is a multiple of 2 and the other a multiple of 4). OR 2) When n and n +2 are odd, but n+1 is a multiple of 8. Calculate the probability of each scenario using the probability formula, then add the probabilities of all scenarios, since they are connected by an "OR" relationship. C Correct. Between 1 and 96 half of the numbers are even and half are odd, thus the probability that n is even (which will make n+1 odd and n+2 even) is 1/2. This is the probability of the first scenario. As for the second scenario: The probability that a number in a range is a multiple of 8 is the number of multiples of 8 in the range divided by the total number of items in the range. The denominator, the number of items in the range is the difference between the extremes plus 1: 96-1+1 = 96 numbers. Now to the numerator, the number of multiples of 8 in the range: When counting the multiples of x within a given range (in this case x = 8): 1) Find the relevant extremes - the nearest multiples of 8 within the specified range: The effective range is actually 8 to 96 inclusive. 2) Subtract the relevant extremes, and divide by 8: 96 - 8 = 88. 88/8 = 11 3) Add one: 11 + 1 =12 Thus the probability that n+1 is a multiple of 8 is 12/96 = 1/8. Finally, the total probability is 1/2 + 1/8 = 4/8 + 1/8 = 5/8

Most of the GMAT probability problems are more complex than single outcome events. This, in turn, means you have to deal with series of events rather than with a single event.

If John rolls a six-sided die twice, what is the probability that both rolls will give a result of "6"? Remember the probability formula: One way of dealing with this question will therefore be: 1) list the total number of possible outcomes when a single dies twice: 1-1, 1-2, 1-3,...5-6, 6-6 2) Calculate the number of wanted outcomes: in this case, a single outcome of 6-6. For this simple question, listing the total number of possible outcomes is not that difficult. It can even be done using a simple matrix: Each square in the above matrix represents a single possible combination of the results of two rolls. For example, the square marked as "x" represents getting a "1" on the first roll AND a "2" on the second roll. The total number of squares in the above matrix represents the total number of possible outcomes when rolling two six-sided dice: 6×6 = 36 possible outcomes. The square marked as ✔ represents the single wanted outcome according to the question: an outcome of 6-6. Therefore, the probability of getting a six on both rolls is 1 / 36. Clearly, this method of counting outcomes, via visual or other counting methods. is limited to simple problems of the sort presented above. Extending the above problem to three dice would already render drawing a matrix of the total combinations unfeasible. A more general approach is needed.

To sum up, remember this important fact for more advanced questions:

If the question presents several wanted outcomes of the same event with an OR relationship between them, ADD the probabilities.

You may have heard the term probability in many instances. There is a 50% likelihood of rain tomorrow. The chance of winning the lottery is 1 in a 1,000,000,000. There is a ¼ probability of Superman beating Spiderman.

Probability means the chance or likelihood that something may happen. For instance: What is the probability of getting a result of "1" when rolling a regular six-sided die? Use the probability formula probability = No of wanted outcomes / Total no of possible The question asks for a 1, so there is only one wanted outcome out of the total six possible outcomes (the numbers 1-6). Hence, the probability of getting a 1 is 1/6.

Back to George and his white oysters, but with a little twist: George has a bucket of oysters, containing only 3 white oysters and 7 black oysters. If George takes a handful of three oysters at random, without replacing them, what is the probability that at least one white oyster will be removed?

Pay extra attention to the phrase "at least", as it is the main cause of difficulty in this version of the question. Let's break it down and see what it means to pull at least one white oyster out of three oysters removed. Pulling one white oyster and two black oysters is a possible scenario, but "at least" one means we can't stop there. Our possible scenarios are: One white oyster, two black oysters OR Two white oysters, one black oyster OR Three white oysters, no black oysters. Each of these is a different scenario. The obvious way to deal with this question is the same method we've seen for dealing with multiple scenarios: Work out the number of probability for each one, then add the results, as these scenarios have an OR relationship. However, the obvious choice is not the best one here. It will take too long, and leave much room for careless errors. Notice that our scenario list is missing a single instance: 0 white oysters, 3 black oysters. This is the only "forbidden" scenario according the terms of the question, requiring at least one white oyster. Together, these scenarios present the entire body of possible results - George will pull either 0, 1, 2, or 3 white oysters out of his basket. We've already seen that if several scenarios map out the entire body of possible results, the sum of their probabilities equals 1 (remember the chicken that can either cross safely or get run over?). In other words: Probability of "Good" Scenarios + Probability of "Forbidden" Scenarios = 1. Since the question asks for the probability of the "Good" scenarios, isolate that to get Probability of "Good" Scenarios = 1 -Probability of "Forbidden" Scenarios Therefore, instead of calculating the different probabilities of all the "good" scenarios, calculate the probability of the single "Forbidden" scenario, and subtract it from 1.

To sum up: Atleast Probability Qs

Probability "At least" questions are considered one of the tougher question types in this already tough subject, but the correct approach will quickly unravel them. The major first step is identifying them by the use of "at least" in the question. "At least" questions are usually constructed in a way that the "Forbidden" scenarios will be fewer in number and easier to calculate than the "good" scenarios. Therefore, The presence of the phrase "at least" is a surefire trigger to breaking down the question using the "1 minus" approach: Probability of "Good" Scenarios = 1 -Probability of "Forbidden" Scenarios

Let's review the basic concepts of probability:

Probability means the chance or likelihood that something may happen. Probability is given in the formula probability = No of wanted outcomes / Total no of possible Probability value is always between zero and one; Probability of zero means something can never happen. Probability of one means something must happen. Probability can be expressed as fraction, decimal, or percent. You are likely to see 1-2 probability problems in your GMAT.

Most probability questions can be solved by the usual method of breaking down the problem into a series of events and dealing with them one-at-a-time. However, some harder Probability questions require the use of combination counting methods we've learned under Combinations and Permutations. Take a look at the following example: Private Benjamin is a member of a squad of 10 soldiers, which must volunteer 4 of its members for latrine duty. If the members of the latrine patrol are chosen randomly, what is the probability that private Benjamin will be chosen for latrine duty?

Remember the probability formula: The question presents a case of choosing a small group out of a larger one, which is the basic premise of most Combinations and Permutations questions. We can therefore use what we've learned of Combinations to work out the numerator and denominator of the probability formula. Start by working out the denominator - the total number of possible outcomes. In this case, the total number of outcomes are all possible ways of choosing the latrine patrol - 4 people out of 10. Work out the problem according to our combinations method: 1) Consider: Single source / multiple sources - This is clearly a case of choosing from a single source - privates. 2) Find the size of the smaller group (k) - in this case, k=4 - the four poor members of our latrine Patrol. 3) Find the size of the Larger group (n) - in this case, n=10 - the 10 possible members of the squad. 4) Consider: With / Without repetition? - we can't choose the same member twice, since he will then take up two slots and the latrine patrol will number less than 4 members. Therefore, this problem is a case without repetition - n × (n-1) × (n-2)... 5) Consider: Order matters / doesn't matter? - The latrine patrol has no defined "titles" - we're just concerned with who is on the patrol, regardless of the order they're chosen for it. Therefore, this is a case where the order doesn't matter: use the Combinations formula C(n,k) = n! / (n-k)! k!, or calculate the SeBoxes and divide by k! = 4!. Bottom line: the number of ways of choosing 4 people out of 10 (without repetition, order doesn't matter) is: C(10,4) = 10! / 6!×4! = 210

Here's the general approach for probability questions: Whenever possible, divide the problem into single events, calculate them one a-at-a-time, then combine the results. In our case, treat each roll as a separate event. The desired result of getting a "6" on both rolls is broken down into getting a "6" on the first roll AND a "6" on the second roll. Calculate the probability of getting a result of "6" on the first roll: The total number of possible outcomes on a single roll is is 6: {1, 2, 3, 4, 5, 6}. Out of these possible outcomes, a single outcome of "6" is the wanted outcome. Therefore, the probability of getting a result of "6" on an event of a single die roll is 1/6.

Repeat this process for the second roll: the probability of reaching a "6" on the second roll is still 1 out of six, or 1/6. Now, combine the two events: Since the two events have a relationship of AND between them, MULTIPLY the two probabilities. The probability of getting a "6" on the first roll AND "6" on the second roll = 1/6 × 1/6 = 1/36. Notice that we reach the same result as the we did using the matrix to count the possible outcomes. Splitting the problem into separate events allows for dealing with more complex questions with 3 or more separate events.

Now for the interesting part - figure out the numerator of the probability formula - the number of wanted outcomes. Since the question requests the probability that private Benjamin will be chosen for latrine duty, the number of wanted outcomes is the number of ways of choosing 4 people for a latrine patrol that must include Private Benjamin.

Take the next logical step for defining this wanted scenario: our latrine patrol of 4 must include poor Benjamin and 3 more people. Therefore, the number of wanted outcomes is simply the number of ways of choosing the other 3 people out of the remaining 9 members. Other than defining the new k=3 and n=9, all other parameters of this scenario are the same as the denominator's: single source, without repetition, order doesn't matter. Therefore, use the Combinations formula: C(9,3) = 9! / (9-3)!3! = 84 Alright, we have the numerator and the denominator of the probability formula. The probability that private Benjamin is one of the 4 privates chosen for latrine duty is therefore Probability = No of wanted outcomes / Total no of possible = No of ways of choosing Benjamin n 3 others / Total no of ways of choosing 4 people = 84/210 = 12/30 = 2/5 A) Got it B) I tried breaking the problem into separate events and working it one at a time and got 1/10×9/9×8/8×7/7 = 1/10 and not 2/5. What am I doing wrong? This is a common mistake. Let's label the calculation above to see what's really going on: 1st choice 1/10 × 2nd choice 9/9 × 3rd choice 8/8 × 4th choice 7/7 The first fraction is the probability that Private Benjamin is chosen for latrine duty out of the entire squad of 10. Once we know that Private Benjamin is chosen, we really don't care who are the other three. Therefore, we get 9 wanted outcomes out of 9 possible outcomes for the second choice, 8 out of 8 for the third choice, and 7 out of 7 for the fourth, resulting in a final probability of 1 out of 10. However, this is not the answer to the original question - this is just the probability of the limited case that Private Benjamin is chosen first. The question asked for the probability that private Benjamin is chosen in general, which means we also need to consider (and add) the scenarios where Benjamin is chosen second, third OR fourth.

Cracking probability problems relies on breaking down their structure. Consider the following example: Miguel rolls a six-sided die, and records the number shown on the upper face. What is the probability that the recorded number is even?

The die has numbers 1-6, each of which is an outcome. Thus, the total number of possible outcomes is six. Probability problems ask about an event. We define an event as a single instance requiring the calculation of a probability fraction of In other words, an event is any single choice in which there are a number of wanted outcomes out of a total number of outcomes. Examples are A single roll of a die with a requested outcome, a single draw of card, etc. An event may be a single outcome or any combination of series of outcomes. In this case the event (an even result on a roll of a six-sided die) is composed of the outcomes 2,4,6 (the even numbers in the range 1-6). And so, the wanted event is made up of a series of three wanted outcomes. Therefore, according to the probability formula, the probability of the event of an even outcome on a six-sided die is 3/6=1/2.

Notice what actually happens here: There are two wanted outcomes: "1" and "5". The probability of getting each of these outcomes separately is 1 / 6. Since the question presents either "1" OR "5" as the wanted outcome, we simply ADD the probabilities of "1" and the probability of "5" to get 1/6 + 1/6 = 2/6.

The general rule from this example is that whenever the question presents several wanted outcomes of the same EVENT with an OR relationship between them, calculate the probability of each outcome separately, then ADD the probabilities. In other words: Outcome A OR Outcome B (of the same event) Means Add Probabilities. Notice the important phrase "...outcomes of the same EVENT.." in the above rule. An OR relationship between separate events does not automatically mean "ADD". For example, if John flips a coin twice, the probability of getting a tails on the first OR the second throw is NOT 1/2 + 1/2 = 1. A probability of 1 means a certain event: intuitively, we know that John is NOT certain to get tails on either one of the two throws. This dissonance is due to the fact that flipping a coin twice constitutes two separate events, while our rule of "OR means ADD" only applies to different outcomes of the same event.

The probability that a chicken crosses the street safely without getting run over is 1/3. What is the probability that a chicken gets run over when crossing the street? One thing is certain: our poor chicken will either cross the street safely, or get run over. Together, these two outcomes form the entire body of possible outcomes, and thus form a certain event, which, as you recall, has a probability of 1. Translating this to math language, we come to following conclusion: [Probability of chicken crossing safely] + [probability of chicken NOT crossing safely] = 1.

The practical outcome of this equation, which will be useful for solving much more complicated GMAT probability questions, is this: [Probability of chicken NOT crossing safely] = 1 - [probability of chicken crossing safely] Therefore, our poor chicken has a probability of 1 - 1/3 = 2/3 of getting run over when crossing the road. In general terms, we say that: Probability (event happens) + Probability (event does NOT happen) = 1 Or, in other words: Probability (event does NOT happen) = 1 - Probability (event happens)

Breaking down probability questions into a series of events and dealing with each event separately is the key to handling most probability questions. Read the following question and think about the correct breakdown of the question into events, according to our above definition of an event: George has a bucket of oysters, containing only 3 white oysters and 7 black oysters. If George takes a handful of three oysters at random, without replacing them, what is the probability that all three oysters removed are white?

The question asks for the probability of getting three white oysters. Technically, we could treat this problem as a single event , calculated as: Probability = no of ways of choosing 3 white oysters out of 10 / no of ways of choosing 3 oysters out of 10 However, remember that we defined an event as a single choice requiring a probability fraction. This definition facilitates an easier and more intuitive way of dealing with the problem: break it down into 3 separate events, each consisting of a single draw of one oyster; calculate the probability of a wanted outcome for each successive event; then combine the three events.

Remember George and his multi-colored oysters? George has a bucket of oysters, containing only 3 white oysters and 7 black oysters. If George takes a handful of three oysters at random, without replacing them, what is the probability that all three oysters removed are white? We've seen that the way to treat this sort of problem is to break down the question into a series of single events, calculate the probability for each event one-at-a time, them multiply the probabilities. In other words, the probability of pulling a handful of three white oysters is equal to: the probability of pulling a white Oyster on the FIRST draw AND the probability of drawing a white Oyster on the SECOND draw AND the probability of drawing a white Oyster on the THIRD draw.

This is an example of a probability question involving a series of events without repetition. Notice the keywords "without replacing", which add an additional twist. The fact that the oysters are not returned to the basket indicates that this is an example of a probability question involving a series of events without repetition. The practical outcome of this is that the situation (number of possible outcomes and number of wanted outcomes) changes between draws. Let's work the problem. Calculate the probability of drawing a white Oyster on the FIRST draw: The Total number of possible outcomes is the total number of Oysters: 3 white + 7 black = 10 oysters. Since George wants a white oyster on the FIRST draw, the number of wanted outcomes is the number of white oysters = 3 white oysters. Therefore, the probability of drawing a white oyster on the FIRST draw is 3 / 10. Now, calculate the probability of drawing a white oyster on the SECOND draw. Important note: when calculating the probability for the second draw, assume that the first draw was successful in reaching a wanted outcome. In other words, assume that George has already pulled a white oyster on the first draw and didn't return it; now consider the new conditions in the basket. What's the total number of possible outcomes now? 9 Correct! Since George has already drawn a white oyster and did not replace it, the basket now contains only 2 white oysters + 7 black oysters = 9 oysters total. Okay. Now what's the new number of wanted outcomes? 2 Correct. Since George has already drawn a white oyster and did not replace it, the basket now contains only 2 white oysters. Our probabilities so far are therefore: 3/10 for getting a white oyster on the FIRST draw AND 2/9 for getting a white oyster on the SECOND draw. What is the probability of drawing a white oyster on the THIRD draw? Go ahead and calculate this on your own. Type in the probability in the text box in the form of "numerator/denominator", without spaces. e.g. "1/2" 1/8 Good! After successfully drawing two white oysters on the FIRST and SECOND draws, there are 1 white oyster + 7 black oysters = 8 oysters in the basket. Out of these, only 1 outcome is a wanted outcome - the single white oyster left in the basket. Therefore, under these new conditions, the probability of drawing a white oyster on the THIRD draw is 1/8. The probabilities of our three events are: 3/10 for getting a white oyster on the FIRST draw AND 2/9 for getting a white oyster on the SECOND draw AND 1/8 for getting a white oyster on the THIRD draw. Since the three events have an AND relationship between them (the desired result is drawing three white oysters out of the basket), multiply the three fractions to get 3/10 × 2/9 × 1/8 = 1/120

OK, so now we ask "what is the probability of pulling 0 white oysters and 3 black oysters from the bucket?" Break down the question into a series of single events: FIRST draw, SECOND draw, THIRD draw. For the desired result of 3 black Oysters, George needs to pull a black oyster on the FIRST draw AND a black oyster on the SECOND draw AND a black oyster on the THIRD draw

What is the probability of drawing a black oyster on the FIRST draw? Go ahead and calculate this on your own. Type in the probability in the text box in the form of "numerator/denominator", without spaces. e.g. "1/2" 7/10 Correct! The Total number of possible outcomes is the total number of Oysters: 3 white + 7 black = 10 oysters. Since we want a black oyster on the FIRST draw, the number of wanted outcomes is the number of black oysters = 7 white oysters. Therefore, the probability of drawing a black oyster on the FIRST draw is 7/10. Now, calculate the probability of drawing a black oyster on the SECOND draw. Type in the probability in the text box in the form of "numerator/denominator", without spaces. e.g. "1/2" Remember: assume that the first draw was successful in reaching a wanted outcome. In other words, assume that George has already pulled a black oyster on the FIRST draw and didn't return it; now consider the new conditions in the basket. 6/9 Good! Since George has already drawn a black oyster and did not replace it, the basket now contains only 3 white oysters + 6 black oysters = 9 oysters total. We want a black oyster for the SECOND draw, so number of wanted outcomes is the number of black oysters currently in the basket. Since George has already drawn a black oyster and did not replace it, the basket now contains only 6 black oysters. Our probabilities so far are therefore: 7/10 for getting a black oyster on the FIRST draw AND 6/9 for getting a black oyster on the SECOND draw. What is the probability of drawing a black oyster on the THIRD draw? Go ahead and calculate this on your own. Type in the probability in the text box in the form of "numerator/denominator", without spaces. e.g. "1/2" 5/8 Excellent. After successfully drawing two black oysters on the FIRST and SECOND draws, there are 3 white oysters + 5 black oysters = 8 oysters in the basket. Out of these, only 5 outcomes are wanted outcomes - the 5 black oysters left in the basket. Therefore, under these new conditions, the probability of drawing a black oyster on the THIRD draw is 5/8. The probabilities of our three events are: 7/10 for getting a black oyster on the FIRST draw AND 6/9 for getting a black oyster on the SECOND draw AND 5/8 for getting a black oyster on the THIRD draw. Since the three events have an AND relationship between them (the desired result of our "Forbidden scenario" is drawing black oysters out of the basket), multiply the three fractions to get 7/10×6/9×5/8 Reduce to get 7/1×1/3×1/8 = 7/24 Important note: by the time you're done calculating this scenario, it is very easy to forget WHY you're doing it. 7/24 is the probability if the "Forbidden scenario" - not the answer to the question!!! The question asked for the probability that at least one white oyster will be removed. The probability of the "Forbidden" scenario of pulling three black oysters is only useful for the formula we've established: Probability of "Good" Scenarios = 1 -Probability of "Forbidden" Scenarios Therefore, as a last step: The probability of pulling at least one white oyster = 1 - probability of pulling three black oysters = 1 - 7/24 = 17/24. GMAC is happy to capitalize on this sort of careless mistake. You just know that 7/24 is going to appear in one of the answer choices as a powerful distractor. Always re-read the question after a long period of calculation - remember what you're looking for. It's also a good idea to begin your work by writing the above formula in your notebook as a reminder of the final step.

The main take home message from this lesson is this:

When considering each event one-at-a-time, assume that the previous events yielded a wanted response, and consider each event under the new conditions created by this assumption. For questions without repetition, this means that the conditions (total number of outcomes and number of wanted outcomes) change for each successive event.