Probability on the GRE at a Glance
- Where it appears: Quantitative Reasoning, under data analysis
- Probability topics include elementary probability (compound and independent events), conditional probability, and counting methods such as combinations and permutations
- Question types it turns up in: Quantitative Comparison, Multiple-choice and Numeric Entry
- The core formula: probability = favourable outcomes ÷ total equally likely outcomes
How Is Probability Tested on the GRE?
Probability sits in the data analysis part of GRE Quantitative Reasoning. ETS's probability topics include elementary probability, such as compound and independent events, conditional probability, and counting methods such as combinations, permutations and Venn diagrams. The maths stays at high-school level; what the questions test is whether you count the outcomes correctly.
You will meet probability in all four Quant question types: Quantitative Comparison, Multiple-choice with one answer, Multiple-choice with one or more answers, and Numeric Entry, where you type a fraction or a number into a box.
Source: ETS, GRE General Test Quantitative Reasoning content, read 22 October 2026.
GRE Probability Formulas You Need
| Rule | Formula | When to use it |
|---|---|---|
| Basic probability | P(A) = favourable ÷ total | All outcomes are equally likely |
| Complement | P(not A) = 1 − P(A) | “At least one” questions |
| Independent events | P(A and B) = P(A) × P(B) | One event does not change the other |
| Either event | P(A or B) = P(A) + P(B) − P(A and B) | Events that can happen together |
| Conditional | P(A given B) = P(A and B) ÷ P(B) | You are told something has already happened |
| Combinations | nCr = n! ÷ (r! (n − r)!) | Choosing a group where order does not matter |
| Permutations | nPr = n! ÷ (n − r)! | Arrangements where order matters |
The questions below are written by Galvanize in the style of the GRE. They are not official ETS questions. Work each one out first, then open the solution.
Single-Event Probability Questions
Question 1 (Multiple-choice)
A bag holds 5 red, 7 blue and 8 green marbles. One marble is drawn at random. What is the probability that it is blue?
(A) 1/4 (B) 7/20 (C) 7/13 (D) 2/5 (E) 13/20
Show answer and solution
Answer: (B) 7/20.
There are 5 + 7 + 8 = 20 marbles, and 7 of them are blue, so the probability is 7/20. Choice (C), 7/13, is the trap: it compares blue with the marbles that are not blue instead of with all 20.
Question 2 (Numeric Entry)
A two-digit positive integer is chosen at random. What is the probability that it is divisible by 7? Give your answer as a fraction.
Show answer and solution
Answer: 13/90.
The two-digit integers run from 10 to 99, which is 90 numbers. The multiples of 7 in that range run from 14 (7 × 2) to 98 (7 × 14), which is 14 − 2 + 1 = 13 numbers. The probability is 13/90.
Question 3 (Quantitative Comparison)
One of the positive divisors of 36 is chosen at random.
Quantity A: the probability that the divisor chosen is odd
Quantity B: 1/3
(A) Quantity A is greater. (B) Quantity B is greater. (C) The two quantities are equal. (D) The relationship cannot be determined from the information given.
Show answer and solution
Answer: (C) The two quantities are equal.
The positive divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36: nine in all. The odd ones are 1, 3 and 9. The probability is 3/9 = 1/3, which equals Quantity B.
Combined and Independent Events Questions
Question 4 (Multiple-choice)
A fair coin is tossed 4 times. What is the probability of getting at least one head?
(A) 1/16 (B) 1/4 (C) 3/4 (D) 7/8 (E) 15/16
Show answer and solution
Answer: (E) 15/16.
Use the complement. The only way to get no heads is four tails, with probability (1/2)4 = 1/16. So the probability of at least one head is 1 − 1/16 = 15/16. “At least one” is almost always fastest through the complement.
Question 5 (Numeric Entry)
Two fair six-sided dice are rolled. What is the probability that the sum of the two numbers is 8?
Show answer and solution
Answer: 5/36.
There are 6 × 6 = 36 equally likely outcomes. A sum of 8 comes from (2, 6), (3, 5), (4, 4), (5, 3) and (6, 2): five outcomes. The probability is 5/36. Count (3, 5) and (5, 3) separately; they are different outcomes.
Question 6 (Multiple-choice)
Events A and B are independent. The probability of A is 0.4 and the probability of B is 0.5. What is the probability that A or B, or both, occur?
(A) 0.2 (B) 0.45 (C) 0.7 (D) 0.8 (E) 0.9
Show answer and solution
Answer: (C) 0.7.
Because A and B are independent, P(A and B) = 0.4 × 0.5 = 0.2. Then P(A or B) = 0.4 + 0.5 − 0.2 = 0.7. Choice (E), 0.9, adds the two probabilities without removing the overlap.
Question 7 (Numeric Entry)
For two events A and B, P(A) = 0.30, P(B) = 0.45 and P(A and B) = 0.15. What is the probability that neither A nor B occurs?
Show answer and solution
Answer: 0.4.
First find the probability that at least one occurs: P(A or B) = 0.30 + 0.45 − 0.15 = 0.60. “Neither” is the complement: 1 − 0.60 = 0.40.
Question 8 (Quantitative Comparison)
Quantity A: the probability of getting at least 2 heads in 3 tosses of a fair coin
Quantity B: the probability of getting 2 heads in 2 tosses of a fair coin
(A) Quantity A is greater. (B) Quantity B is greater. (C) The two quantities are equal. (D) The relationship cannot be determined from the information given.
Show answer and solution
Answer: (A) Quantity A is greater.
Quantity A: of the 8 outcomes of 3 tosses, HHT, HTH, THH and HHH have at least 2 heads, so the probability is 4/8 = 1/2. Quantity B: both tosses must be heads, (1/2) × (1/2) = 1/4. Since 1/2 > 1/4, Quantity A is greater.
Where Do You Stand in GRE Quant?
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Take the Free GRE Mock TestConditional Probability Questions
Question 9 (Multiple-choice)
A box holds 4 red and 6 blue pens. Two pens are taken out at random, one after the other, without replacement. What is the probability that both are red?
(A) 2/15 (B) 4/25 (C) 1/5 (D) 2/5 (E) 3/5
Show answer and solution
Answer: (A) 2/15.
The first pen is red with probability 4/10. Given that it was red, 3 red pens are left out of 9, so the second is red with probability 3/9. Multiply: (4/10) × (3/9) = 12/90 = 2/15. Choice (B), 4/25, is what you would get with replacement.
Question 10 (Numeric Entry)
In a class of 30 students, 18 study French, 12 study Spanish and 5 study both. A student who studies French is chosen at random. What is the probability that this student also studies Spanish?
Show answer and solution
Answer: 5/18.
You are told the student studies French, so the 18 French students are the whole sample. Of them, 5 also study Spanish. The probability is 5/18. The class size, 30, is not needed; it is there to tempt you into 5/30.
Question 11 (Multiple-choice)
A company has 120 employees, of whom 70 work full time. Of the full-time employees, 40 have a master’s degree; of the part-time employees, 20 have one. An employee with a master’s degree is chosen at random. What is the probability that this employee works full time?
(A) 1/3 (B) 4/7 (C) 7/12 (D) 2/3 (E) 3/4
Show answer and solution
Answer: (D) 2/3.
The condition is “has a master’s degree”, so the sample is everyone with one: 40 + 20 = 60 employees. Of those, 40 work full time. The probability is 40/60 = 2/3. Choice (B), 4/7, answers the reverse question: the chance that a full-time employee has a master’s degree.
Probability with Combinations and Permutations
Question 12 (Multiple-choice)
A committee of 3 is chosen at random from 5 men and 4 women. What is the probability that the committee has exactly 2 women?
(A) 1/6 (B) 5/21 (C) 5/14 (D) 4/9 (E) 1/2
Show answer and solution
Answer: (C) 5/14.
Total committees: 9C3 = 84. Committees with exactly 2 women: choose the women in 4C2 = 6 ways and the man in 5C1 = 5 ways, so 6 × 5 = 30. The probability is 30/84 = 5/14.
Question 13 (Numeric Entry)
Five different books are placed on a shelf in a random order. What is the probability that two particular books end up next to each other?
Show answer and solution
Answer: 2/5.
All orders: 5! = 120. To keep the two books together, treat them as one block. That gives 4 items to arrange in 4! = 24 ways, and the two books can sit inside the block in 2 ways, so 24 × 2 = 48 orders. The probability is 48/120 = 2/5.
Question 14 (Multiple-choice)
A 4-digit PIN is formed at random from the digits 0 to 9, and digits may repeat. What is the probability that all four digits are different?
(A) 21/100 (B) 1/2 (C) 63/125 (D) 3/5 (E) 189/250
Show answer and solution
Answer: (C) 63/125.
All PINs: 104 = 10,000. PINs with four different digits: 10 × 9 × 8 × 7 = 5,040. The probability is 5,040/10,000 = 63/125, which is 0.504.
Question 15 (Numeric Entry)
Two different integers are chosen at random from the integers 1 to 10. What is the probability that their product is even?
Show answer and solution
Answer: 7/9.
A product is odd only when both numbers are odd. There are 10C2 = 45 pairs in all, and 5C2 = 10 pairs of odd numbers (from 1, 3, 5, 7 and 9). So the probability of an odd product is 10/45 = 2/9, and of an even product 1 − 2/9 = 7/9.
Tips for GRE Probability Questions
1. Name the sample before you count
Most wrong answers divide by the wrong total. In a conditional question, the total is only the group you were told about, as in Questions 10 and 11.
2. Reach for the complement on “at least”
“At least one” usually means 1 minus the probability of none. It turns many cases into one.
3. Check replacement and order
Without replacement, the second probability changes. Use combinations when order does not matter, permutations when it does.
4. Leave answers as fractions
When a Numeric Entry question gives you a fraction box, ETS accepts equivalent forms of the correct answer.
For the wider maths syllabus, see our GRE math guide.
