GRE probability questions

Make the possible outcomes clear. Try six probability questions, then work through the reasoning behind each answer.

Try the 6 questions

Free to practice. Answers and explanations included.

Start with the event you are counting.

Before calculating, describe the outcome the question asks for. “Either,” “both,” “none,” and “at least one” each define a different event.

Count each outcome once.

When outcomes are equally likely, probability is the number of favorable outcomes divided by the total. For “A or B,” an outcome in both groups still counts only once: P(A or B)P(A \text{ or } B) =P(A)+P(B)P(A and B)= P(A) + P(B) - P(A \text{ and } B).

Check what changes after a selection.

Without replacement, both the remaining total and the number of favorable items can change. For the playlist in Question 3, the second selection is from 14 remaining tracks, not the original 15.

Separate independence from overlap.

Independent events satisfy P(A and B)P(A \text{ and } B) =P(A)P(B)= P(A)P(B). They can still happen together. In Question 5, use independence to find the overlap before calculating the probability of either event.

6 GRE probability questions

This set includes five single-answer questions and one quantitative comparison. Select one answer for each question, then open its explanation.

Question format: ETS Quantitative Reasoning overview.

Question 1

A fair six-sided die is rolled once. What is the probability that the number rolled is even or greater than 4?

Select one answer.

Question 1 answers
Show answer and explanationfor question 1

Answer: A

The possible outcomes are 1, 2, 3, 4, 5, and 6. The even outcomes are 2, 4, and 6, and the outcomes greater than 4 are 5 and 6. Since 6 belongs to both sets, it is counted only once. Thus, the favorable outcomes are 2, 4, 5, and 6, so the probability is 46\frac{4}{6}, or 23\frac{2}{3}. The correct answer is Choice (A).

Question 2

A shelf contains 32 books, of which 14 are fiction and 12 are paperbacks. If 5 of the books are both fiction and paperbacks, what is the probability that a book selected at random from the shelf is fiction or a paperback?

Select one answer.

Question 2 answers
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Answer: B

Of the 32 books, 14 are fiction and 12 are paperbacks. The 5 books that are both fiction and paperbacks are included in both of these counts. Therefore, the number of books that are fiction or paperbacks is 14+12514 + 12 - 5 =21= 21. Since each book is equally likely to be selected, the probability is 2132\frac{21}{32}. Thus, the correct answer is Choice (B).

Question 3

A playlist consists of 12 instrumental tracks and 3 vocal tracks. The tracks are arranged in a random order, with each ordering equally likely. What is the probability that the first two tracks are both instrumental?

Select one answer.

Question 3 answers
Show answer and explanationfor question 3

Answer: E

The probability that the first track is instrumental is 1215\frac{12}{15}. If the first track is instrumental, 11 of the remaining 14 tracks are instrumental. Therefore, the probability that the first two tracks are both instrumental is (1215)(1114)(\frac{12}{15})(\frac{11}{14}) =2235= \frac{22}{35}. The correct answer is Choice E.

Question 4

A container holds 16 tiles, of which 6 are triangles and 10 are circles. Three tiles are selected at random, one at a time, without replacement. What is the probability that none of the three selected tiles is a triangle?

Select one answer.

Question 4 answers
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Answer: B

For none of the selected tiles to be triangles, all three selected tiles must be circles. The probability that the first tile is a circle is 1016\frac{10}{16}. After a circle is selected, 9 of the 15 remaining tiles are circles, and after a second circle is selected, 8 of the 14 remaining tiles are circles. Therefore, the required probability is (1016)(915)(814)\left(\frac{10}{16}\right)\left(\frac{9}{15}\right)\left(\frac{8}{14}\right) =314= \frac{3}{14}. The correct answer is Choice B.

Question 5

Events JJ and KK are independent. If the probability that JJ does not occur is 35\frac{3}{5} and P(K)P(K) =13= \frac{1}{3}, what is P(J or K)P(J \text{ or } K)?

Select one answer.

Question 5 answers
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Answer: C

Since P(J does not occur)P(J \text{ does not occur}) =35= \frac{3}{5}, P(J)P(J) =135= 1 - \frac{3}{5} =25= \frac{2}{5}. Because JJ and KK are independent, P(J and K)P(J \text{ and } K) =(25)(13)= (\frac{2}{5})(\frac{1}{3}) =215= \frac{2}{15}. By the addition rule, P(J or K)P(J \text{ or } K) =P(J)+P(K)P(J and K)= P(J) + P(K) - P(J \text{ and } K) =25+13215= \frac{2}{5} + \frac{1}{3} - \frac{2}{15} =615+515215= \frac{6}{15} + \frac{5}{15} - \frac{2}{15} =915= \frac{9}{15} =35= \frac{3}{5}.

Question 6

Quantitative comparison

An archive contains 160 grant applications. If one application is selected at random, the probability that it satisfies at least one of requirements RR and SS is 1316\frac{13}{16}, and the probability that it satisfies both requirements is 310\frac{3}{10}. Quantity A: The number of applications that satisfy exactly one of requirements RR and SS Quantity B: Three times the number of applications that satisfy neither requirement RR nor requirement SS

Select one answer.

Question 6 answers
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Answer: B

Of the 160 applications, the number that satisfy at least one requirement is 160×(1316)160 \times \left(\frac{13}{16}\right) =130= 130, and the number that satisfy both requirements is 160×(310)160 \times \left(\frac{3}{10}\right) =48= 48. Therefore, 13048130 - 48 =82= 82 applications satisfy exactly one requirement. Also, 160130160 - 130 =30= 30 applications satisfy neither requirement, so Quantity B is 3×303 \times 30 =90= 90. Since 9090 is greater than 8282, Quantity B is greater.

Something unclear? Report a question or explanation with this page link and the question number.

Find the step that changed your answer.

  • For Questions 1 and 2, list or count the overlapping outcomes before adding the groups.
  • For Questions 3 and 4, write a separate fraction for each selection. Check its numerator and denominator against what remains.
  • For Questions 5 and 6, distinguish “at least one” from “exactly one.” Draw two overlapping circles if the wording feels abstract.

Give your next session a direction.

In StudyPattern, continue with Probability, Counting and Combinations, or Venn Diagrams & Set Theory. Read the lesson, practice the concept, and revisit the questions you missed.

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