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

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

## 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**.

This probability is just less than 1/2.

Suppose **n = 3**.

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 —

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**.

### Most Popular Resources

Hi mike ,

Could you elaborate in detail explanation for question 3 statement 2 i dint understand the equally likely case

Hi Gayathri,

So, if we know that N (employees selected) = M (total employees), then we know that all employees will be selected to sit. Next, the question asks, “What is the probability that the employee Andrew is seated somewhere to the right of employee Georgia?” In this question, we’re focusing on two employees.

The testmakers want you to be overwhelmed by this question and think about all the potential orders that the employees can sit. But, you don’t need to do this! Note that no matter what order you make, 50% of the time Andrew will sit SOMEWHERE to the right of Georgia, and 50% of the time Andrew will sit SOMEWHERE to the left of Georgia.

For example, let’s say N = M = 8:

For every scenario, you can make the opposite true:

_ A _ _ _ _ G _

_ G _ _ _ _ A _

So, we know that half the time Andrew will be on the left of Georgia, while the other half he’ll be on the right of Georgia.

Hi Mike,

In statement #1 of question #7, what is the logical sense of P(B)?

Shouldn’t we talk about P (A/B) because the two events A and B are “not independent” according to the introduction (that means that they both depend on each other).

Please could you clarify the meaning?

Thanks

Dear Rodrigo,

I’m happy to respond. 🙂 It’s true that Events A & B are not independent, so they have some kind of influence or effect on each other. If we wanted to calculate the

exact valueof P(A and B), we most certainly would need a conditional probability — for example, we could calculate that exact value from the values of P(A) and P(B|A), for example.BUT— and this is a big “BUT” — this is GMAT DS, and the question isNOT ASKINGfor an exact value of P(A and B). We are only being asked to determine of P(A and B) is in a range, and through another line of reasoning, one I explain in the official explanation to the problem, we can easily determine something about its range.Remember, not all probability is about the formulas. A formula-based approach to probability is doomed to failure. You have to understand these ideas from a few different angles, and visual/spatial reasoning is one of those angles.

Does all this make sense?

Mike 🙂

Thank you Mike,

I understand the diagram approach and the importance of not using formulas as the only approach. The doubt that I have is different: what P(B) means in a world where probabilities in events A and B are dependent.

Is P(B) the probability of occurring event B when A is not occurring? Or Is P(B) the probability of ocurring event B when A is also ocurring? In a world in which there is some dependence, both probabilities are different numbers and I am wondering what P(B) alone means or represents.

Based on that, when we draw P(A) and P(B) in the diagram I think that we cannot use the only value of P(B) in both parts of the diagram (in the part where both events A and B overlap or occure simultaneously, and in the part where event A or event B occur separately), because we are not considering how dependence affects the probabilities and the diagram.

I hope this clarifies my question?

Thanks again!

Dear Rodrigo,

Ah, yes! I’m sorry: I understand now. 🙂 What you are asking is an excellent and subtle question. Let’s address this with a real world example. Consider the American adult population. Let A =

. We survey everyone in the country, ask them the simple question “Do you play chess regularly?” (Perhaps we have some quantitative way to clarify what we mean by “regularly.”) We get everyone’s answer, and then P(A) is the probability that, when we pick a random American, that randomly chosen person plays chess regularly.plays chess regularlyLet B =

. Again, survey the whole population, ask every single person, “Have you graduated from a four-year college?” Then P(B) would be the probability that, when we pick a randomly American, we pick someone who has a college degree.has a college educationNotice that P(A) and P(B) are NOT independent: there is a definite effect — folks who play chess regularly, as a general tendency, are more likely to go to college. The fact that the probabilities are dependent doesn’t mean the ideas are conceptually dependent. It doesn’t means that the probabilities are inseparable. We can have a long discussion about the value of getting a college education or the likelihood of getting a college education and never once mention chess, or vice versa. If we are picking random Americans, we could look for or measure one, and completely ignore the other. Being dependent or independent has nothing to do with that.

Here’s what being dependent means —- if I go to place (such as Magoosh) where everyone has a college education, where having a college degree is already a given, then if I ask the chess question, I am likely to get a high percentage of yeses than I would get from the general population; similarly, if I go to chess tournament, a place where everyone in the room is an avid chess player, and I ask all the adults about college degrees, then I am likely to get a much higher percentage of yeses than would get in the general population. Technically, those are conditional probabilities, P(A|B) and P(B|A) respectively. The conditional probability P(B|A) means I am going to focus only on folks who play chess regularly, and among this limited pool, I am going to ask about whether they have college degrees. By contrast, P(B) means we totally ignore chess, we don’t even think about it, and we just go out into the general population and ask about college degrees.

You see, the funny thing about dependent probabilities is that relationships will just arise unintentionally. Suppose I totally ignore chess, and just ask a very large number of people about college degrees, and I get a large “yes” group and a large “no” group — P(B) would be ratio of the (size of the “yes” group) to (everyone I asked). Now, someone else, who knows that A and B are dependent could come along and probably demonstrate that the the proportion of regular chess players is higher in my “yes” group than it is in my “no” group. A makes B more likely, and vice versa. Even though we can measure P(B), the chance among everyone that they have a college degree, if we dig a little in to the data, we will see the association with chess.

So, absolutely, P(B) has a number and value and meaning that is totally separable from A. We can consider B without A. We can considerout A with B. It’s just that, if we do happen to know one, that knowledge gives us information about the other.

Does all this make sense?

Mike 🙂

Mike, thank you very much for your awesome explanation! Now it is very clear!

Dear Rodrigo,

You are quite welcome, my friend. 🙂 I wish you the very best of luck!

Mike 🙂

Dear Mike

In qs 7, in statement 1 can we say that P(A and B) = P(A) * P(B) = 0.56 which is greater than 1/3… hence, sufficient

statement 2, P(A or B) = P(A) + P(B) – P(A and B)… so we dont have info about P(A) and P(B)… Hence, Sufficient.

Answer: A

I mean can these formulas be applied directly.?? instead of the diagram technique which i didnt understand much.

Dear Nandini,

I’m happy to respond. 🙂

As you will see, I made a slight clarification to the problem. If events A and B are

, then we can use the formula P(A and B) = P(A)*P(B). But, in the general case, many pairs of events are not independent, and that formula is no longer applicable. With the information given in that problem, there’s no way to solve it with formulas. As with many problems on the real GMAT, that problem was designed to frustrate folks who rely exclusively on formulas to get them through GMAT math. That’s why we had to use the diagram,and why the diagram is really the only way to solve it. It’s important to understand that thoroughly.independentDoes all this make sense?

Mike 🙂