GRE probability questions: 10 worked examples, easy to hardest

How much probability is on the GRE?

A little, and reliably. Probability sits inside Data Analysis, one of the four Quantitative Reasoning content areas alongside Arithmetic, Algebra and Geometry, and it comes paired with counting methods: combinations and permutations. With 27 Quant questions across the two sections, you should expect a small number of probability items rather than a block of them.

ETS does not publish a count per topic, so any exact number you read is an estimate. Ours: in our own question bank, which is written to the published content mix, roughly one Quant item in eleven is probability or counting. On a 27-question test that is two or three questions. Treat it as a planning figure, not an official one.

What matters more than the count is where these questions sit. Probability is rarely the easiest question in a section: in our bank, about three quarters of probability items are rated medium difficulty or above. And because the Quant measure is adaptive by section, a strong first section routes you to a harder second one, which is where the without-replacement and committee questions tend to live. If you are aiming above 160 in Quant, this is not a topic to skip. The full structure is on our GRE exam pattern page.

The five probability rules the GRE actually uses

Almost every GRE probability question is one of these, or two of them chained. None needs more than arithmetic once you have picked the right one.

One counting tool sits underneath the harder questions. When the order of selection does not matter, the number of ways to choose $k$ items from $n$ is the combination $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}$. Committee and “two marbles at once” questions are a count of favorable groups divided by a count of all groups, and both counts are combinations. The formulas you need for the rest of Quant are on our GRE math formulas cheat sheet.

GRE probability questions with worked solutions

Ten questions, ordered from easy to hardest, in the three formats probability shows up in: multiple choice, numeric entry and quantitative comparison. Work each one before opening the solution. These are original questions written to GRE style and difficulty, not ETS items, and the difficulty labels are our own editorial ratings.

1. Question 1Multiple choice · Easy · Probability

   A fair six-sided die, with faces numbered $1$ through $6$, is rolled once. What is the probability that the number rolled is greater than $4$?

   • (A)$\frac{1}{6}$
   • (B)$\frac{1}{3}$
   • (C)$\frac{1}{2}$
   • (D)$\frac{2}{3}$
   • (E)$\frac{5}{6}$
   Show the answer and worked solution

   Answer: (B) $\frac{1}{3}$

   The outcomes greater than $4$ are $5$ and $6$, which is $2$ of the $6$ equally likely outcomes. The probability is $\frac{2}{6}=\frac{1}{3}$. Choice (a) corresponds to a single outcome. Choice (c) corresponds to $3$ outcomes. Choice (d) is the complement error, and choice (e) corresponds to $5$ outcomes.

   Do

   Count the favorable outcomes precisely and divide by the total number of equally likely outcomes.

   Avoid

   Do not include the value 4 itself when counting outcomes greater than 4.

2. Question 2Numeric entry · Easy to medium · Probability

   A jar contains $3$ red marbles, $4$ blue marbles, and $5$ green marbles. If one marble is selected at random from the jar, what is the probability that it is not green?

   Numeric entry: you type the answer as a fraction. There are no choices to work back from.

   Show the answer and worked solution

   Answer: $\frac{7}{12}$

   The jar contains $3+4+5=12$ marbles in total. The marbles that are not green are the red and blue marbles: $3+4=7$. The probability of selecting a marble that is not green is $\frac{7}{12}$. Answer: $\frac{7}{12}$.

   Do

   Count the favorable outcomes (red and blue marbles) directly and divide by the total number of marbles.

   Avoid

   Don't divide by the number of colors (3) instead of the total count of marbles (12).

3. Question 3Numeric entry · Easy to medium · Probability

   A fair six-sided die is rolled twice. What is the probability that the sum of the two rolls equals 7?

   Numeric entry: you type the answer as a fraction. There are no choices to work back from.

   Show the answer and worked solution

   Answer: $\frac{1}{6}$

   There are $6\times 6=36$ equally likely outcomes for the two rolls. The outcomes summing to 7 are $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$, giving 6 favorable outcomes. The probability is $\frac{6}{36}=\frac{1}{6}$. Answer: $\frac{1}{6}$ (approximately 0.1667)

   Do

   List the ordered pairs that produce the target sum systematically, then divide the count by the total number of equally likely outcomes, 36.

   Avoid

   Don't assume every sum from 2 to 12 is equally likely; sums near 7 have more combinations than extreme sums like 2 or 12.

4. Question 4Numeric entry · Medium · Probability

   Two fair coins are flipped at the same time. What is the probability that at least one of them lands on heads? Enter your answer as a fraction.

   Numeric entry: you type the answer as a fraction. There are no choices to work back from.

   Show the answer and worked solution

   Answer: $\frac{3}{4}$

   The four equally likely outcomes are HH, HT, TH, TT.

   "At least one head" includes HH, HT, and TH, which is $3$ of the $4$ outcomes, so the probability is $\frac{3}{4}$.

   Alternatively, use the complement: the only outcome with no heads is TT, with probability $\frac{1}{4}$, so $P(\text{at least one head})=1-\frac{1}{4}=\frac{3}{4}$.

   Do

   List all four outcomes, or use the complement $1-P(\text{no heads})$ since "at least one" pairs naturally with "none."

   Avoid

   Don't answer $\frac{1}{2}$ by treating the two coins as one; there are four outcomes, not two.

5. Question 5Numeric entry · Easy to medium · Probability

   A fair coin is flipped 3 times. What is the probability that exactly 2 of the 3 flips result in heads?

   Numeric entry: you type the answer as a fraction. There are no choices to work back from.

   Show the answer and worked solution

   Answer: $\frac{3}{8}$

   There are $2^3=8$ equally likely outcomes for three flips. The number of outcomes with exactly 2 heads is $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$ (HHT, HTH, THH). The probability is therefore $\frac{3}{8}$. Answer: $\frac{3}{8}$.

   Do

   Count favorable outcomes using combinations, then divide by the total number of possible outcomes.

   Avoid

   Don't forget to divide by the total number of outcomes (8); listing only the favorable arrangements is not sufficient.

6. Question 6Numeric entry · Medium · Probability

   A jar contains 4 white balls and 6 black balls. Two balls are drawn at random, one after the other, without replacement. What is the probability that both balls drawn are black?

   Numeric entry: you type the answer as a fraction. There are no choices to work back from.

   Show the answer and worked solution

   Answer: $\frac{1}{3}$

   The probability that the first ball is black is $\frac{6}{10}=\frac{3}{5}$. Given that the first ball drawn was black, 5 black balls remain out of 9 total, so the probability the second ball is also black is $\frac{5}{9}$. Multiplying these dependent probabilities: $\frac{3}{5}\times \frac{5}{9}=\frac{15}{45}=\frac{1}{3}$.

   Do

   Multiply the sequential conditional probabilities, adjusting both the black-ball count and the total count after the first draw since there is no replacement.

   Avoid

   Don't treat the draws as independent by using $\frac{6}{10}\times \frac{6}{10}$; the second draw's denominator and black-ball count must both decrease by one.

7. Question 7Numeric entry · Medium · Probability

   A box contains 6 red chips and 4 blue chips. Two chips are drawn at random, one after the other, without replacement. What is the probability that the two chips drawn are of different colors?

   Numeric entry: you type the answer as a fraction. There are no choices to work back from.

   Show the answer and worked solution

   Answer: $\frac{8}{15}$

   The two chips are different colors in two mutually exclusive ways: red then blue, or blue then red. $P(\text{red then blue})=\frac{6}{10}\times \frac{4}{9}=\frac{24}{90}$. $P(\text{blue then red})=\frac{4}{10}\times \frac{6}{9}=\frac{24}{90}$. Total probability: $\frac{24}{90}+\frac{24}{90}=\frac{48}{90}=\frac{8}{15}\approx 0.533$.

   Do

   Add the probabilities of both orderings (red-then-blue and blue-then-red), adjusting the denominator for the second draw since there is no replacement.

   Avoid

   Don't use the same denominator (10) for both draws, which ignores that only 9 chips remain for the second draw, and don't count only one ordering of the two colors.

8. Question 8Quantitative comparison · Hard · Probability

   A bag contains 4 red marbles, 3 blue marbles, and 5 green marbles. Two marbles are drawn at random from the bag without replacement.

   Quantity A

   The probability that both marbles drawn are red

   Quantity B

   $\frac{1}{9}$
   Show the answer and worked solution

   Answer: B. Quantity B is greater.

   The probability that the first marble is red is $4/12$. Given that the first marble drawn was red, the probability that the second is also red is $3/11$. Thus the probability that both are red is $\frac{4}{12}\times \frac{3}{11}=\frac{12}{132}=\frac{1}{11}$. Comparing $\frac{1}{11}$ to $\frac{1}{9}$, since $11>9$, it follows that $\frac{1}{11}<\frac{1}{9}$. Therefore Quantity B is greater. Both quantities are fixed numbers determined entirely by the given composition of the bag, so there is no case for D. Answer: B.

   Do

   Multiply the sequential conditional probabilities for drawing without replacement, adjusting the denominator after the first draw.

   Avoid

   Don't treat the draws as independent with replacement (which would give $(4/12{)}^2=1/9$), since that ignores that the marble count decreases after the first draw.

9. Question 9Numeric entry · Hard · Data analysis

   Three fair six-sided dice are rolled. What is the probability that all three dice show the same number? Enter your answer as a fraction.

   Numeric entry: you type the answer as a fraction. There are no choices to work back from.

   Show the answer and worked solution

   Answer: $\frac{1}{36}$

   Approach: fix what the first die shows, then require the second and third to match it. The first die can show anything; the probability it 'sets the target' is $1$. Each of the remaining two dice independently matches the first with probability $\frac{1}{6}$.

   $P(\text{all three equal})=1\cdot \frac{1}{6}\cdot \frac{1}{6}=\frac{1}{36}$.

   Count check: there are $6^3=216$ equally likely outcomes; the favorable ones are $(1,1,1),(2,2,2),\dots ,(6,6,6)$, which is $6$. So $\frac{6}{216}=\frac{1}{36}$.

   Why the wrong approaches fail: $\frac{1}{6^3}=\frac{1}{216}$ would be the probability of one specific triple like $(4,4,4)$, not any matching triple. $\frac{1}{6}\cdot \frac{1}{5}\cdot \frac{1}{4}$ treats the dice like cards drawn without replacement, but dice are independent. $\frac{1}{6}$ alone forgets that two of the three dice need to match.

   Do

   Let the first die set the target, then require each of the next two to match it independently.

   Avoid

   Don't use $\frac{1}{216}$; that's the chance of a specific triple, not any matching triple.

10. Question 10Multiple choice · Hardest · Probability

    A committee of $4$ people is to be selected at random from a group of $6$ men and $5$ women. What is the probability that the committee includes at least $2$ women?

    • (A)$\frac{43}{66}$
    • (B)$\frac{23}{33}$
    • (C)$\frac{115}{330}$
    • (D)$\frac{1}{2}$
    • (E)$\frac{41}{66}$
    Show the answer and worked solution

    Answer: (A) $\frac{43}{66}$

    The total number of ways to choose $4$ people from $11$ is $\left(\genfrac{}{}{0ex}{}{11}{4}\right)=330$. It is easier to find the complement: committees with $0$ or $1$ women. With $0$ women, all $4$ are men: $\left(\genfrac{}{}{0ex}{}{6}{4}\right)=15$ ways. With exactly $1$ woman: $\left(\genfrac{}{}{0ex}{}{5}{1}\right)\left(\genfrac{}{}{0ex}{}{6}{3}\right)=5\times 20=100$ ways. The complement total is $15+100=115$, so the number of committees with at least $2$ women is $330-115=215$. The probability is $\frac{215}{330}=\frac{43}{66}$. Choice (c) is the unreduced complement probability, not its complement; (b), (d), and (e) come from miscounting one of the two complementary cases. Answer: a

    Do

    Compute the complement (fewer than 2 women) since it involves only two cases, then subtract from the total to get the 'at least' probability.

    Avoid

    Don't try to sum the cases of exactly 2, 3, and 4 women directly without checking the complement approach; it is far more error-prone for an 'at least' condition.

What makes GRE probability questions hard

Not the arithmetic. The numbers are small on purpose. The difficulty is in reading which situation you are in, and four misreadings account for most wrong answers.

• Missing “without replacement”. If the first marble is not put back, the second draw has one fewer marble in both the numerator and the denominator. Question 6 turns on exactly this: $\frac{6}{10}\times \frac{5}{9}$, not $\frac{6}{10}\times \frac{6}{10}$.
• Attacking “at least” head on.“At least two women” on a committee of four means two, three or four, which is three separate counts. The complement, zero or one, is two counts. Question 10 is far shorter that way.
• Confusing one specific outcome with any matching outcome. The probability that three dice all show a 4 is $\frac{1}{216}$. The probability that they all show the same number, whatever it is, is six times that. Question 9 is built on the difference.
• Forgetting that an outcome can happen in more than one order.“One red and one blue” is red then blue, or blue then red. Count one order and you get exactly half the right answer, which is usually sitting among the choices. See question 7.

In quantitative comparison, add one more discipline: compute both quantities exactly before you compare them. Probability comparisons are designed so that the intuitive guess and the real answer disagree. More on that format in our quantitative comparison practice, and the wider set of traps is in the 12 GRE Quant traps.

How to practice probability for the GRE

Learn the five rules first, then practice recognizing which one a question wants. That second skill is the one the test measures, and it only comes from volume.

• Drill by format. Our free problem solving and numeric entry sets mix probability in with the other Quant topics, the way the real test does.
• Move up in difficulty deliberately. Once these ten feel routine, the committee and counting questions in our hardest GRE math questions are the next step.
• Practice the translation too. Many probability questions are word problems first. The method for turning a paragraph into an equation is on our GRE word problems page.
• Then test it under time. A free full-length mock shows whether you can still pick the right rule with a clock running, which is a different skill from doing it on a quiet afternoon.

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