GRE Quantitative Quiz: Probability

What this quiz covers

This quiz focuses on Probability, giving you a quick way to practice the rules, question types, and explanations that matter most for GRE Quantitative.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\frac{1}{3}$
2. $\frac{2}{3}$
3. $\frac{1}{2}$ (correct answer)
4. $\frac{1}{6}$
5. $\frac{1}{4}$

Explanation: This question tests conditional probability using the definition P(A|B) = P(A and B)/P(B). The sample space is the 6 equally likely outcomes from a die roll: 1 through 6. Event B (greater than 4) has outcomes 5 and 6, so P(B) = 2/6 = 1/3; A and B (even and >4) is just 6, so P(A and B) = 1/6, yielding P(A|B) = (1/6)/(1/3) = 1/2. Counting within B, 1 of 2 outcomes is even. This is justified as conditional probability restricts the space to B. A tempting incorrect option is 1/3, perhaps from P(A) overall instead of conditional. This fails because it ignores the conditioning on B, which changes the relevant sample space.

Question 2

1. $\frac{2}{3}$ (correct answer)
2. $\frac{1}{2}$
3. $\frac{1}{3}$
4. $\frac{5}{6}$
5. $\frac{1}{6}$

Explanation: This question tests the probability of a union of two events when rolling a die. The sample space consists of the six equally likely outcomes: {1, 2, 3, 4, 5, 6}. Event A (multiples of 3) = {3, 6} and event B (even numbers) = {2, 4, 6}. To find P(A ∪ B), we need the outcomes in either A or B: {2, 3, 4, 6}, which contains 4 elements. Therefore, P(A ∪ B) = 4/6 = 2/3. A common mistake would be to add P(A) + P(B) = 2/6 + 3/6 = 5/6, which double-counts the outcome 6 that appears in both events.

Question 3

1. $\frac{3}{28}$
2. $\frac{9}{64}$ (correct answer)
3. $\frac{3}{8}$
4. $\frac{9}{56}$
5. $\frac{6}{16}$

Explanation: This question tests probability in sampling with replacement, where draws are independent. The sample space is all possible outcomes of two draws from 8 balls (3 red, 5 yellow), with 88 = 64 equally likely results. The probability of red on first draw is 3/8, and again on second is 3/8, so (3/8)(3/8) = 9/64. This matches the proportion of red-red outcomes. The result is justified because replacement ensures independence, allowing multiplication of probabilities. A tempting incorrect option is 3/28, perhaps from using without replacement formula incorrectly. This fails as it assumes dependence, which does not apply here with replacement.

Question 4

1. $\frac{25}{144}$ (correct answer)
2. $\frac{5}{12}$
3. $\frac{1}{6}$
4. $\frac{10}{132}$
5. $\frac{5}{11}$

Explanation: This question tests probability with independent events (with replacement). The sample space contains 12 balls: 5 red and 7 blue. Since we replace the ball after each draw, the probability remains constant: P(red) = 5/12 for each draw. The probability of drawing two red balls is (5/12) × (5/12) = 25/144. A common mistake would be to treat this as without replacement, calculating (5/12) × (4/11) = 20/132 = 5/33, which incorrectly assumes the first ball isn't replaced.

Question 5

1. $\frac{1}{8}$
2. $\frac{3}{8}$
3. $\frac{7}{8}$ (correct answer)
4. $\frac{1}{2}$
5. $\frac{5}{8}$

Explanation: This question tests the complement rule in probability. The sample space consists of 2³ = 8 equally likely sequences: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. To find P(at least one head), it's easier to use the complement: P(at least one H) = 1 - P(no heads) = 1 - P(TTT). Since P(TTT) = 1/8, we get P(at least one H) = 1 - 1/8 = 7/8. A common error would be to think P(at least one H) = 3 × (1/2) = 3/2, which incorrectly adds probabilities and exceeds 1.

Question 6

1. $\frac{4}{25}$
2. $\frac{2}{15}$ (correct answer)
3. $\frac{8}{45}$
4. $\frac{1}{6}$
5. $\frac{16}{100}$

Explanation: This question tests probability with dependent events (without replacement). The sample space initially contains 10 balls: 4 black and 6 white. For the first draw, P(black) = 4/10. After removing one black ball, 9 balls remain with 3 black, so P(second black | first black) = 3/9. The probability of both events is (4/10) × (3/9) = 12/90 = 2/15. A tempting error would be to use (4/10) × (4/10) = 16/100, which incorrectly assumes replacement and independence between draws.

Question 7

1. $\frac{2}{5}$
2. $\frac{4}{10}$
3. $\frac{1}{3}$ (correct answer)
4. $\frac{3}{10}$
5. $\frac{1}{4}$

Explanation: This question tests conditional probability in sequential selection without replacement. The sample space initially has 10 employees: 4 managers and 6 non-managers. Given that the first selected is a manager, we now have 9 employees left: 3 managers and 6 non-managers. The probability that the second is a manager, given the first is a manager, is 3/9 = 1/3. A common mistake would be to use the original probability 4/10 = 2/5, which fails to account for the changed composition after the first selection.

Question 8

1. $\frac{2}{3}$
2. $\frac{3}{10}$
3. $\frac{11}{15}$ (correct answer)
4. $\frac{4}{9}$
5. $\frac{7}{15}$

Explanation: This question tests probability of complementary events in categorized sampling. The sample space is all combinations of 2 chips from 10 (4A, 3B, 3C), totaling C(10,2) = 45 equally likely pairs. Same labels: C(4,2) + C(3,2) + C(3,2) = 6 + 3 + 3 = 12, so different is 45 - 12 = 33, probability 33/45 = 11/15. This is justified as the complement simplifies counting diverse pairs. A tempting incorrect option is 4/9, perhaps from averaging probabilities incorrectly. This fails because it might assume equal group sizes, ignoring the actual distributions.

Question 9

1. $0.08$
2. $0.92$ (correct answer)
3. $0.88$
4. $0.12$
5. $0.80$

Explanation: This question tests basic probability in a binary outcome scenario. The sample space consists of all items produced, where each is either acceptable or defective, with defects occurring at 8%. The probability of selecting a defective item is 0.08, so the probability of acceptable is 1 - 0.08 = 0.92. This complementary approach is efficient here. The result is justified as the events are exhaustive and mutually exclusive, based on historical data assuming stability. A tempting incorrect option is 0.08, confusing acceptable with defective. This fails because it directly misreads the question, emphasizing the need to distinguish between the event and its complement.

Question 10

1. $\frac{1}{12}$
2. $\frac{1}{4}$
3. $\frac{1}{3}$
4. $\frac{1}{6}$ (correct answer)
5. $\frac{1}{2}$

Explanation: This question tests probability of joint events from independent experiments. The sample space consists of 12 equally likely pairs: (H or T) with (1 through 6). Favorable outcomes are tails with 5 or 6, giving 2 outcomes. Thus, the probability is 2/12 = 1/6. This is justified as the coin and die are independent, so P(tails and >4) = P(tails) * P(>4) = (1/2)*(2/6) = 1/6. A tempting incorrect option is 1/12, perhaps from miscounting only one die outcome >4. This fails because >4 includes both 5 and 6, underlining the need to accurately define the event.

Question 11

1. $\frac{1}{2}$ (correct answer)
2. $\frac{2}{5}$
3. $\frac{2}{3}$
4. $\frac{1}{3}$
5. $\frac{1}{4}$

Explanation: This question tests conditional probability in a drawing without replacement scenario. The sample space after drawing a red chip first consists of the 4 remaining chips: 2 red and 2 blue. Given that the first chip drawn is red, we need to find the probability that the second chip is blue. With 2 blue chips remaining out of 4 total chips, the probability is 2/4 = 1/2. This is a straightforward conditional probability where the condition (first chip is red) has already occurred. A common error would be to calculate the joint probability P(first red AND second blue) = (3/5) × (2/4) = 6/20 instead of the conditional probability.

Question 12

1. $\frac{4}{10}$
2. $\frac{2}{15}$ (correct answer)
3. $\frac{1}{4}$
4. $\frac{8}{25}$
5. $\frac{1}{5}$

Explanation: This question tests the probability of drawing two white balls without replacement from a jar. The sample space for the first draw consists of 10 balls (4 white and 6 black). The probability of drawing a white ball first is 4/10. Given that the first ball drawn is white, there are now 3 white balls remaining out of 9 total balls for the second draw. The probability of drawing a second white ball is 3/9. Since both events must occur, we multiply: (4/10) × (3/9) = 12/90 = 2/15. A common error would be to use 4/10 for both draws, incorrectly assuming replacement or independence when the problem explicitly states "without replacement."

Question 13

1. $\frac{1}{13}$
2. $\frac{1}{4}$
3. $\frac{4}{52}$
4. $\frac{4}{13}$ (correct answer)
5. $\frac{4}{13}+\frac{1}{13}$

Explanation: This question tests probability with the inclusion-exclusion principle for non-mutually exclusive events. The sample space is a standard deck of 52 equally likely cards. There are 13 hearts and 4 kings, but the king of hearts is counted in both, so P(heart or king) = (13 + 4 - 1)/52 = 16/52 = 4/13. This calculation avoids double-counting the overlapping card. The result is justified because inclusion-exclusion ensures all favorable outcomes are counted exactly once. A tempting incorrect option is 4/13 + 1/13 = 5/13, which overcounts by ignoring the overlap. This fails because it assumes the events are disjoint when they are not, leading to an inflated probability.

Question 14

1. $\frac{16}{100}$
2. $\frac{2}{15}$ (correct answer)
3. $\frac{1}{6}$
4. $\frac{4}{25}$
5. $\frac{1}{15}$

Explanation: This question tests probability in sampling without replacement, focusing on sequential draws. The sample space is all possible sequences of two balls from 10 (4 white, 6 black), but combinations can also be used for efficiency. The probability of first drawing a white is 4/10, and then another white is 3/9, so multiplying gives (4/10)*(3/9) = 12/90 = 2/15. Using combinations, it's C(4,2)/C(10,2) = 6/45 = 2/15, confirming the result. This is justified because without replacement, the draws are dependent, and the calculation accounts for the reduced total after the first draw. A tempting incorrect option is 4/25, which might come from incorrectly assuming independence and using (4/10)^2. Such an assumption fails because it ignores the change in the jar's composition after the first draw.

Question 15

1. $\frac{5}{16}$
2. $\frac{1}{4}$
3. $\frac{1}{2}$
4. $\frac{3}{8}$ (correct answer)
5. $\frac{6}{16}$

Explanation: This question tests probability in binomial distributions for exact counts. The sample space consists of 16 equally likely sequences of H and T from 4 coin flips. The number of sequences with exactly 2 heads is C(4,2) = 6. Thus, the probability is 6/16 = 3/8. This is justified using the binomial coefficient, which counts the ways to choose positions for heads. A tempting incorrect option is 1/4, perhaps from averaging or misapplying uniformity. This fails because it does not account for the varying numbers of ways to achieve different head counts, unlike the binomial approach.

Question 16

1. $\frac{1}{8}$
2. $\frac{3}{8}$
3. $\frac{7}{8}$ (correct answer)
4. $\frac{1}{2}$
5. $\frac{5}{8}$

Explanation: This question tests probability of complementary events in repeated independent trials. The sample space consists of 8 equally likely sequences of H and T from 3 coin flips. The probability of all tails is 1/8, so the probability of at least one head is 1 - 1/8 = 7/8. Counting directly, there are 7 sequences with at least one H. This result is justified because the complement (all tails) is a single outcome, making the calculation straightforward and efficient for 'at least' problems. A tempting incorrect option is 3/8, perhaps from confusing it with exactly one head (which has 3 outcomes). This fails because 'at least one' includes exactly one, two, and three heads, not just one.

Question 17

1. $\frac{1}{2}$
2. $\frac{2}{3}$ (correct answer)
3. $\frac{1}{3}$
4. $\frac{5}{6}$
5. $\frac{1}{6}$

Explanation: This question tests probability of unions using inclusion-exclusion. The sample space is 6 equally likely die outcomes: 1 through 6. Event C (at most 2): 1,2 (2 outcomes); D (even): 2,4,6 (3 outcomes); intersection: 2 (1 outcome), so union has 2 + 3 - 1 = 4 outcomes. The probability is 4/6 = 2/3. This is justified as inclusion-exclusion prevents double-counting the overlap. A tempting incorrect option is 5/6, from adding without subtracting the intersection. This fails because it overcounts outcome 2, inflating the union size.

Question 18

1. $\frac{1}{13}$
2. $\frac{1}{4}$
3. $\frac{1}{3}$ (correct answer)
4. $\frac{3}{13}$
5. $\frac{4}{13}$

Explanation: This question tests conditional probability in a deck of cards. The sample space, conditioned on drawing a face card, consists of 12 equally likely face cards (4 jacks, 4 queens, 4 kings). The favorable outcomes are the 4 queens. Thus, the probability is 4/12 = 1/3. This is justified by restricting to the conditional space and counting directly. A tempting incorrect option is 1/13, perhaps from unconditional P(queen) overall. This fails because it ignores the given condition, which changes the relevant denominator from 52 to 12.

Question 19

1. $\frac{1}{4}$
2. $\frac{3}{8}$
3. $\frac{5}{8}$
4. $\frac{1}{2}$ (correct answer)
5. $\frac{3}{4}$

Explanation: This question tests probability using the inclusion-exclusion principle for union of events. The sample space consists of 8 equally likely sectors labeled 1 through 8. Multiples of 3 are 3 and 6 (2 outcomes), multiples of 4 are 4 and 8 (2 outcomes), with no overlap, so the union has 4 outcomes. The probability is 4/8 = 1/2. This result is justified as the events are mutually exclusive in this case, allowing simple addition, though inclusion-exclusion handles overlaps generally. A tempting incorrect option is 3/8, perhaps from counting only multiples of 3 and forgetting multiples of 4. Another error might involve subtracting overlaps incorrectly, but here there are none, emphasizing the need to check for intersections.

Question 20

1. $\frac{1}{9}$
2. $\frac{5}{36}$
3. $\frac{1}{12}$
4. $\frac{1}{6}$ (correct answer)
5. $\frac{1}{36}$

Explanation: This question tests probability by requiring the enumeration of favorable outcomes for a sum in dice rolls. The sample space consists of 36 equally likely ordered pairs (a, b) where a and b range from 1 to 6. The favorable outcomes for a sum of 7 are (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1), totaling 6 outcomes. Thus, the probability is 6/36 = 1/6. This result is justified as each pair is equally likely, and the count of favorable pairs directly gives the proportion. A tempting incorrect option is 5/36, perhaps from overlooking one of the pairs like (6,1) due to incomplete listing. Another error might involve assuming unordered pairs, which would incorrectly reduce the sample space and favorable outcomes.