Probability

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.

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.

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

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.

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.

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.

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.

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

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.

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

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

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.

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

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.

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.

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.