Here is a set of 14 GMAT probability questions, all in the Problem Solving style on the test, collected from a series of blog articles. Answers and links to explanations to these these GMAT probability problems are at the end of set.

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*The scenario below is relevant to questions #1-#3.*

There are two sets of letters, and you are going to pick exactly one letter from each set.

Set #1 = {A, B, C, D, E}

Set #2 = {K, L, M, N, O, P}

1) What is the probability of picking a C and an M?

- (A) 1/30

(B) 1/15

(C) 1/6

(D) 1/5

(E) 1/3

2) What is the probability of picking a C or an M?

- (A) 1/30

(B) 1/15

(C) 1/6

(D) 1/5

(E) 1/3

3) What is the probability of picking two vowels?

- (A) 1/30

(B) 1/15

(C) 1/6

(D) 1/5

(E) 1/3

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4) In a certain corporation, there are 300 male employees and 100 female employees. It is known that 20% of the male employees have advanced degrees and 40% of the females have advanced degrees. If one of the 400 employees is chosen at random, what is the probability this employee has an advanced degree and is female?

- (A) 1/20

(B) 1/10

(C) 1/5

(D) 2/5

(E) 3/4

5) In a certain corporation, there are 300 male employees and 100 female employees. It is known that 20% of the male employees have advanced degrees and 40% of the females have advanced degrees. If one of the 400 employees is chosen at random, what is the probability this employee has an advanced degree or is female?

- (A) 1/20

(B) 1/10

(C) 1/5

(D) 2/5

(E) 3/4

————————————————————————————————————————————

Set #1 = {A, B, C, D, E}

Set #2 = {K, L, M, N, O, P}

6) There are these two sets of letters, and you are going to pick exactly one letter from each set. What is the probability of picking at least one vowel?

- (A) 1/6

(B) 1/3

(C) 1/2

(D) 2/3

(E) 5/6

7) Suppose you flip a fair coin six times. What is the probability that, in six flips, you get at least one head?

- (A) 5/8

(B) 13/16

(C) 15/16

(D) 31/32

(E) 63/64

8) In a certain game, you pick a card from a standard deck of 52 cards. If the card is a heart, you win. If the card is not a heart, the person replaces the card to the deck, reshuffles, and draws again. The person keeps repeating that process until he picks a heart, and the point is to measure: *how many draws did it take before the person picked a heart and won*? What is the probability that one will have at least three draws before one picks a heart?

- (A) 1/2

(B) 9/16

(C) 11/16

(D) 13/16

(E) 15/16

9) Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelred, sit randomly in five chairs in a row. What is the probability that Childeric and Ethelred sit next to each other?

- (A) 1/30

(B) 1/15

(C) 1/5

(D) 2/5

(E) 7/20

10) A division of a company consists of seven men and five women. If two of these twelve employees are randomly selected as representatives of the division, what is the probability that both representatives will be female?

- (A) 1/6

(B) 2/5

(C) 2/9

(D) 5/12

(E) 5/33

11) John has on his shelf four books of poetry, four novels, and two reference works. Suppose from these ten books, we were to pick two books at random. What is the probability that we pick one novel and one reference work?

- (A) 1/2

(B) 2/5

(C) 3/10

(D) 7/20

(E) 8/45

12) In the diagram above, the sides of rectangle ABCD have a ratio AB:BC = 1:2, and the circle is tangent to three sides of the rectangle. If a point is chosen at random inside the rectangle, what is the probability that it is *not* inside the circle?

13) Region R is a square in the x-y plane with vertices J = (–1, –2), K = (–1, 4), L = (5, 4), and M = (5, –2). What is the probability that a randomly selected point in region R lies below the line 3x – 5y = 10?

- (A) 5/12

(B) 5/18

(C) 5/24

(D) 5/36

(E) 5/72

14) In the diagram above, WZ = XZ, and circular arc XY has a center at W. If a point is selected from anywhere within this figure, what is the probability that it is selected from the shaded region?

## Summary

If you have comments or questions, please let us know in the comment section at the bottom! 🙂

## Answers

1) **A**

2) **E**

3) **B**

4) **B**

5) **D**

6) **C**

7) **E**

8) **B**

9) **D**

10) **E**

11) **E**

12) **D**

13) **C**

14) **A**

## Explanations

For the explanations for #1-5, see this post.

For the explanations for #6-8, see this post.

For the explanations for #9-11, see this post.

For the explanations for #12-14, see this post.

Check out this post if you want a wider range of Problem Solving sample questions.

First question: Someone has asked about #8 previously but has not gotten a response. I am having issues with that question as well. My solution was

P(three draws before a heart)=P(first three are not hearts)*P(Fourth is a heart)=[1-(1/4)^3]*(1/4)

Second question: Also, for #11, I came up with a few scenarios I can’t figure out. Please help!

Scenario 1: you have to pick 1 poetry, 1 novel, 1 reference.

Scenario 2: you have to pick 2 poetry and 1 novel

Scenario 3: you have to pick 2 poetry, 3 novel, 2 reference

Thanks for the help!

Great questions, ZT! Let’s take a look at Q8 first. (And sorry for the lack of response when the last student asked back in 2015!)

First, let’s look at the formula you came up with for the probability of winning in 3 or more draws:

P(three draws before a heart)=P(first three are not hearts)*P(Fourth is a heart)

This formula is not quite correct. Remember, Q8 is asking you for the odds of needing

at leastthree draws before you get a heart card. But your formula, by including the odds that the fourth is a heart and dealing with the odds of three draws without getting a heart, is dealing with exactly three draws before getting a heart on the fourth draw. In the actual problem, you could get a heart on the third draw, the fourth draw, or the fifth, or anything after from the third draw onward.So the formula you really want to go with is: P(at least three draws to win) = 1 – P(win in two or fewer draws).

And P(win in two or fewer draws) is the same as saying:

P(win in one draw) + P(win in two draws)

So:

P(at least three draws to win)

= 1 – P(win in two or fewer draws)

= 1 – P(win in one draw) + P(win in two draws)

So let’s look at what this means in terms of actual numbers. In a standard deck, hearts take up 1/4 of the deck. So one’s odds of winning in one draw are 1/4. Thus:

P(at least three draws to win) = 1 – P(win in one draw) + P(win in two draws) = 1 – (1/4) + P(win in two draws)

So what’s the numeric value of winning in two draws? Well, the odds of NOT drawing a heart on a given draw are 3/4, since 3/4 of the cards are not hearts. And the odds of drawing a heart in a given draw are 1/4, as mentioned. So you want to look at the 3/4 odds of not drawing a heart on the first draw, combined with the 1/4 odds of drawing a heart on the second. Since these events are independent, you can say you are looking at the odds of not drawing a heart in the first draw AND drawing a heart in the second. And remember, when you use AND in probability, you multiply the two odds together. So, P(win in two draws) = (3/4)*(1/4). Now we have all of our numbers:

1 – (1/4) + P(win in two draws)

= 1 – (1/4) + [(3/4)*(1/4)]

= 1 – (1/4) + (3/16)

= 1 – (4/16) + (3/16)

= 1 – (7/16)

= 9/16, or answer choice B.

Next, let’s look at Q11. For Q11, I’m going to challenge you a little bit before giving you answers on the new problems you’ve created based on Q11. (Basically, each scenario constitutes a new problem.) I’m going to ask you to review the nC2 formula based on the answer page for Q11. (Q11 is labelled as Q3 on the answer page; confusing, I know.) And you’ll also want to base your work on the broader nCr formula the nC2 is based on. If you’re still not sure how to do those other scenarios, come back, show your work in a follow-up comment, and I can give you some additional pointers. 🙂

Hi,

For question no 4, I got the answer through the approach you mentioned. I was just curious to try another method and got it wrong somewhere, may be concept it self is wrong from beginning or way I implemented it is wrong:

Probability of selecting a woman: 100/400=1/4

Probability of selecting advance degree: 100/400=1/4.

AND implies answer would be 1/4*1/4=1/16.

Is it wrong because both events are not exclusive.

You are correct here–this answer is wrong because the two events (being a woman and having an advanced degree) are NOT independent events. The simplified “and” rule only applies to independent events, so we can’t use it in this case!

Hello, question on #2,

I wanted to be creative by applying the following logic (assuming that it should come down to the same probability).

What is the probability of picking a C and then a non-M (1/5 x 5/6) = 1/6

What is the probability of picking a non C and then an M (4/5 x 1/6) = 4/30 = 2/15

If I add the above two probability, I would expect to get the same result (3/10) as picking C or M, but I am getting 9/30 instead of 10/30.

Please, can you see where my logic fails?

Thank you,

Hi Kevin,

Great question! You’re actually almost there. So, the question ask: “What is the probability of picking a C or an M?”. You’ve captured is the first probability

onlygetting a “C”, and in the second probabilityonlygetting a “M”. You have to also calculate picking a C and M. The question asks for the probability of picking a C or an M, but does not say toonlyone of those.So, you have:

= “9/30” + “probability of picking both C and M”

= 9/30 + (1/5)(1/6)

= 9/30 + 1/30

= 10/30

= 1/3

I hope this helps! 😀

I really appreciate your help!!

I am wondering what the answer to the following question would be:

“Tom has 32 ties, 5 are blue, 16 are yellow and 11 are red. What is the probability of picking one which is either red or yellow?”

I thought it may be :

numerator: disjoint events: 11/32+16/32

denominator: 32!

is this correct?

many many tnanks!!

Hi Maria,

Based on how the question is phrased, you may actually be overanalyzing the problem. It sounds like you have 32 ties (5 blue; 16 yellow; 11 red), and you’re going to pick one tie out. The question is asking what is the probability that this one tie will be red or yellow. So, it is simply:

Probability of Red/Yellow Tie = (Red Ties + Yellow Ties) / (Total Ties) = (11 + 16) / 32 = 27 / 32. You have a “27/32” probability that you will pick a red or yellow tie.

Hey Mike,

Can you explain why the answer to #6 is C and not B? Here’s my logic:

P(at least 1 vowel) = 1 – P(no vowels)

P(at least 1 vowel) = 1 – (4/5)*(5/6)

P(at least 1 vowel) = 1 – (20/30) = (10/30) = (1/3)

Hi Brian,

Your reasoning and methods are right, but you said that P(no vowels) for the first set was 4/5, when in fact it is 3/5 (there are two vowels: A and E). When you change this an multiply your fractions out, you will get the correct answer of 1/2. You can also see a more in-depth explanation to this question in this blog post: https://magoosh.com/gmat/2012/gmat-math-the-probability-at-least-question/ 🙂

Hello Mike

Can you please explain answer question number 8 . I could not figure out the “at least” scenario. We have not be given the number of try picker has to perform. so in simple words we can not apply the complementary rule as defined in the approach .

Now the question is as follows :-

“the person keeps repeating that process until he picks a heart, and the point is to measure: how many draws did it take before the person picked a heart and won? What is the probability that one will have at least three draws before one picks a heart?”

So he has to pick at least three times and we are not given the number of tries ,so lets imagine that he picked total of 4 times and in the 4th time he won ! so now the probability is p(getting heart)=1-p(not getting a heart once)

1-(3/4)(3/4)(3/4)= 47/64 .

request you to please explain the answer .

Thanks and regards

Ujjwal

Hi Mike,

Can you please explain why Answer 2 is not 1/5+1/6=11/30? Instead you say it is 1/3?

(Ref: What is the probability of picking a C or an M?)

Thanks

K

Dear K,

I’m happy to respond. 🙂 It’s very important not to confuse ordinary OR with exclusive OR — in computer science, the latter is sometimes denoted XOR. Ordinary OR includes the overlap case — “C or M” includes (1) C only, (2) M only, and (3) both C and M. Exclusive OR, XOR, would include the first two but not the third. Whenever you see “OR” given in a GMAT question, with no further restrictions or stipulations, you have to assume that it is an ordinary OR and never an exclusive OR.

In addition, I would say: you need to learn about the

generalized OR rule. See this post:https://magoosh.com/gmat/2012/gmat-math-probability-rules/

Mike 🙂

Wow! Now I know my mistake.. Thanks for the explanation.

Regards,

Srish

How come in question 11 the answer is not 4/45?

My logic is as follows:

1) the probability of choosing one novel out of 10 books = 4/10 times

2) the probability of choosing one reference work out of 9 books = 2/9

4/10*2/9=8/90 or 4/45

Best

Dear Leszek,

I’m happy to respond. 🙂 This is one of the tricky things about probability: it’s sometimes hard to see what assumptions the calculations implies. By saying novel = 4/10, then reference = 2/9, then you are implicitly assuming that the novel is

picked first, and the reference work is picked second. To get the result of a novel + reference work, it doesn’t matter which one was picked first, and this means we would have to calculate both orders and add them, which gives us the answer of 8/45.Does this make sense?

Mike 🙂

Sure – makes a lot of sense!

thanks

Dear Leszek,

You are quite welcome. Best of luck to you, my friend.

Mike 🙂

I think the #9 option D is 2/5 and not 2/15.

Meera,

Yes! You are perfectly correct. It should be 2/5 — I just made that correction. Thank you very much for pointing that out.

Mike 🙂