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GMAT Data Sufficiency Practice Questions on Probability

First off, some of my previous blogs that may be relevant:

1) Data Sufficiency Tips

2) Probability Rules

3) The probability of the “at least” scenario

4) Probability problems involving counting

Here are eight practice questions.  Solutions will appear at the end of the article.

1) In nine independent trials, what is the probability that Outcome A happens at least once?

Statement #1: The probability that Outcome A does not happen even once in any of the nine trials is 0.026

Statement #2: the probability of Outcome A resulting in a single trial is 1/3.

2) A class contains boys and girls.  What is the probability of selecting a boy from a class?

Statement #1: there are 35 students in the class

Statement #2: the ratio of boys to girls is 3:4

3) From a group of M employees, N will be selected, at random, to sit in a line of N chairs.  There are absolutely no restrictions, either in the selection process nor in the order of seating — both are entirely random.  What is the probability that the employee Andrew is seated somewhere to the right of employee Georgia?

Statement #1: N = 15

Statement #2: N = M

4) From a group of J employees, K will be selected, at random, to sit in a line of K chairs.  There are absolutely no restrictions, either in the selection process nor in the order of seating — both are entirely random.  What is the probability that the employee Lisa is seated exactly next to employee Phillip?

Statement #1: K = 15

Statement #2: K = J

5) Bert has $1.37 of loose change in his pocket  —- pennies ($0.01), nickels ($0.05), dimes ($0.10), and quarters ($0.25).  He reaches into his pocket and pulls out one coin at random.  What is the probability that the coin is a nickel?

Statement #1: There are exactly seven pennies in his pocket

Statement #2: There are exactly three quarters in his pocket

6) In a single Epsilon trial, the probability of Outcome T is 1/4.   Suppose a researcher conducts a series of n independent Epsilon trials.  Let P = the probability that Outcome T occurs at least once in n trials.  Is P > 1/2?

Statement #1: n > 3

Statement #2: n < 6

7) Suppose A and B are two events that are not independent.  Is the probability P(A and B) > 1/3?

Statement #1: P(A) = 0.8 and P(B) = 0.7

Statement #2: P(A or B) = 0.9

8) A group of N students will be randomly seated in a row of N chairs.  What is the probability that Beth, one of the students, will be at the extreme right-hand end of the row?

Statement #1: N is an odd prime number

Statement #2: the probability that Steve, another one of the students, is at the extreme left-hand end of the row is 1/13

gdspqop_img1

 

Practice problem solutions

1) D

2) B

3) B

4) C

5) E

6) A

7) A

8) B

 

Practice problem explanations

1) We know there are nine trials.  To figure out the probability that Outcome A happens at least once, we would need a way to figure out the probability of A, P(A).

Statement #1: this statement gives us the complement.  The complement of (Outcome A happens at least once) is (Outcome A does not happen at all).   Complementary probabilities have a sum of 1 — P(not K) = 1 – P(K).  Therefore, using the complement rule, we could figure out that

P(A happens at least once) = 1 – P(A doesn’t happen at all) = 1 – 0.026 = 0.974

This information allows us to answer the prompt question.  This statement, alone and by itself, is sufficient.

Statement #2: this statement gives us P(A), which would allow us to calculate the probability that A happens at least once in nine trials.   This statement, alone and by itself, is sufficient.

Both statement sufficient.  Answer = D

 

2) Fundamentally, Probability = (desired options)/(total options).  The “desired options” are the number of boys and the “total options” is the number of students.

Statement #1: we know here the “total options”, but we don’t know how many of these 35 students are boys.  This statement, alone and by itself, is insufficient.

Statement #2: Ignore the information in the first statement.  From any ratio among parts, we can calculate the ratio to the whole.   If boys are “3 parts” and girls are “4 parts” , so the whole must be 3 + 4 = 7 parts.  Thus, the ratio of boys to whole is 3/7 — that’s the probability.  Using the properties of ratios, we can answer the prompt question.   This statement, alone and by itself, is sufficient.

First not sufficient, second sufficient.  Answer = B.

3) Statement #1: from this, we know there are 15 employees selected.  We don’t know the value of M, which is clearly greater than or equal to 15.  If M is very large, it may be highly unlikely that either Andrew or Georgia is seated.  If M is closer to 15, then that would change the probability. Without knowing the value of M, we can’t complete the calculation.   This statement, alone and by itself, is insufficient.

Statement #2: Ignore the information in the first statement.  If M = N, then all employees, everyone in the pool of selection, takes a seat.  It is simply a matter of seating a group of M employees randomly in M seats.

Think about this.  For every seating arrangement with Andrew to the right of Georgia, there will be exactly one seating arrangement, the mirror image, that has Andrew to the left of Georgia.   Since there’s a one-to-one correspondence between these two, there must be just as many arrangements with Andrew on one side of Georgia as on the other side.   Both are equally likely, so the probability must be 1/2.

This information was enough to calculate a numerical answer to the prompt question.   This statement, alone and by itself, is sufficient.

First not sufficient, second sufficient.  Answer = B.

4) Statement #1: given this, we know K = 15 employees are seated, but we have no idea what the size of the larger pool is.  If J is much larger than K, then it becomes unlikely that either Lisa or Phillip is among those seated.  Both would have to be among the seated in order for them to have a chance of sitting next to each other.   Even if J is closer to K, then it becomes more likely that both are among the seated, but we would need to know the number for J to complete the calculation.   This statement, alone and by itself, is insufficient.

Statement #2: Ignore the information in the first statement.  If M = N, then all employees, everyone in the pool of selection, takes a seat.  The problem is —- if M = N is a small number, say 4 or 5, then it would be more likely that Lisa and Phillip would wind up next to each other, but if M = N is large, say 200, then its quietly likely that two specific employees wind up nowhere near each other.  The size of M = N would make a big different in the calculation, and without knowing that, we can’t calculate.   This statement, alone and by itself, is insufficient.

Combined statements: Now, M = N = 15.  We are seating a group of 15 employees, and we want to know whether two of the people seated, Lisa and Andrew, are next to each other.   This is a calculation we can perform, using counting techniques.   With this combined information, we can give a definitive numerical answer to the prompt question.  Together, the statements are sufficient.

Statements sufficient together but no individually.   Answer = C

NOTE: I didn’t show the calculation for that case, because GMAT DS is all about skipping the calculation if it’s not necessary to determine sufficiency.  On the calculation of P, I’ll briefly say: denominator = 15!, and numerator is 14 positions for the Lisa-Phillip pair, times two possible orders, times 13! for everyone else.  P = 2/15

5) A reminder for non-American students — $1.00 = 100¢.  Thus

penny = $0.01 = 1¢

nickel = $0.05 = 5¢

dime = $0.10 = 10¢

quarter = $0.25 = 25¢

Statement #1: Bert has seven pennies, amounting to 7¢.  The other coins total $1.30.  This could be all nickels — 26 nickels — so that of the 7 + 26 = 33 coins in the pocket, 26 are nickel, and the probability of picking a nickel would be 26/33.  Or, that $1.30 could be 4 quarters and 3 dimes, so that there were no nickels, and the probability of picking a nickel would be zero!  Different possible choices lead to different answer to the prompt question, so this statement, alone and by itself, is insufficient.

Statement #2: Bert has three quarters, amounting to 75¢.  The other coins total 62¢ — this could be two pennies and twelve nickels, that of the 3 + 2 + 12 = 17 coins in the pocket, 12 are nickels, and the probability is 12/17. Or, that 62¢ could be entirely in pennies, without any nickels or dimes at all: then the probability of picking a nickel would be zero.   Different possible choices lead to different answer to the prompt question, so this statement, alone and by itself, is insufficient.

Combined statements: Now, we know Bert has seven pennies and three quarters, and these ten coins together account for 82¢.  The remaining 55¢ must be composed of dimes and nickels.  There could be eleven nickels and no dimes, so that of the 10 + 11 = 21 coins in the pocket, 11 are nickels, and P = 11/21.  OR, there could be five dimes and one nickel in the pocket, so of the 10 + 5 + 1 = 16 coins in the pocket, only one is a nickel, and P = 1/16.  Different possible choices lead to different answer to the prompt question, so both statements combined are insufficient.

Nothing is sufficient.  Answer = E

6) So, let’s play with this scenario first a little.  Remember, we calculate the “at least” scenario using the complement rule.

Suppose n = 1.  Then the probability that T happens at least once (ie. at all!) is 1/4.  Of course, this is less than 1/2.

Suppose n = 2.

gdspqop_img2

This probability  is just less than 1/2.

Suppose n = 3.

gdspqop_img3

This probability  is greater than 1/2.

Now, think about it.   As n increases, there are more and more chances for at least one occurrence of T to happen, so as n increase, the probability of at least one T must also increase.  Thus, for all values of n greater than 3, the probability will be even higher, so it must be greater than 1/2.

Statement #1: as we just said, for all larger values of n, at long as we are at n = 3 or higher, the answer to the prompt question will be yes.  This statement, alone and by itself, is sufficient.

Statement #2: the problem here, if n < 6, then we could have n = 0 or n = 1, in which case the answer to the prompt question is “no”, or we could have n = 3 or higher, in which case the answer is “yes.”  Different possible choices give different answers.   This statement, alone and by itself, is insufficient.

First sufficient, second not sufficient.  Answer = A.

7) Statement #1: this is very tricky.  There is no cut-and-dry probability rule for this.  we have to think about overlap.  The total probability space, which encompasses anything that possibly could happen, has a size of 1, and P(A) and P(B) have to fit in this space.   These two have a size of 0.8 and 0.7 respectively, so they are going to overlap.  Think about it visually —

gdspqop_img4

Push the P(A) = 0.8 all the way to the left (whatever that means!), leaving the 0.2 outside of A on the right.  Now, push P(B) = 0.7 all the way to the right, leaving the 0.3 outside of B on the left.  Suppose the 0.2 outside of A is inside B, and the 0.03 outside of B is inside A.  This would be the minimum possible overlap, and even then the overlap, P(A and B), equals 0.5.  Thus, P(A and B) ≥ 0.5, so it must be greater than 1/3.  this statement allows us to give a definitive answer to the prompt question.   This statement, alone and by itself, is sufficient.

Statement #2: forget everything we learned in the analysis of statement (1).  Now, all we know is P(A or B) = 0.9, and we know absolutely nothing about P(A) or P(B).  We can calculate nothing else.   This statement, alone and by itself, is insufficient.

First sufficient, second not sufficient.  Answer = A.

8) Statement #1: the chance that Beth will be on the right end depends on the number of students.  Obviously, as the number of students increases, it becomes less and less likely that any given student is at the end of the row.   With this statement, we know only that N is an odd prime number — it could be 3 or 5 or 7, or it could be 109. It could be N = 524,287 (you do not need to know how to find prime number this big!!)  Obviously, if N is very big the chances that Beth will be on the right end of this half-million row would be just about zero.  Different values of N give different probabilities,  and different answers to the prompt question.   This statement, alone and by itself, is insufficient.

Statement #2: If the probability that Steve is on the left end is 1/13, this must mean that N = 13.  Thus, the probability that any given student is on any given end is 1/13.  This allows us to give a definitive answer to the prompt question.   This statement, alone and by itself, is sufficient.

First not sufficient, second sufficient.  Answer = B.

 

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