GRE Probability Set 2 Practice Questions with Answers

Probability measures the likelihood of a specific event occurring, expressed as a ratio between 0 and 1. For many test-takers, the GRE Probability Set 2 Practice Questions represent a significant step up in complexity from basic coin-toss scenarios, requiring a deeper grasp of independent events, mutually exclusive outcomes, and combinatorics. Success on the Quantitative Reasoning section often hinges on whether you can accurately distinguish between "and" (multiplication) and "or" (addition) logic in word problems.

To prepare effectively, students should integrate these concepts into a broader GRE Prep strategy. By moving beyond simple fractions and into conditional probability and multi-stage experiments, you can secure those high-percentile scores needed for competitive graduate programs. This guide provides the rigorous practice needed to refine your intuition and calculation speed.

Concept Explanation

GRE probability involves calculating the chance of an event by dividing the number of successful outcomes by the total number of possible outcomes in the sample space. In more advanced sets, you will encounter the Multiplication Rule for independent events, which states that the probability of both Event A and Event B occurring is $P(A) \times P(B)$. Conversely, the Addition Rule applies to mutually exclusive events, where the probability of either Event A or Event B occurring is $P(A) + P(B)$.

Key concepts included in this practice set are:

• Independent Events: The outcome of one event does not affect the outcome of another.
• Dependent Events: The outcome of the first event changes the probability of the second (often seen in "without replacement" problems).
• Complementary Events: The probability that an event does not occur, calculated as $1 - P( \text{Event})$.
• Combinations and Permutations: Used to determine the total number of outcomes when order does or does not matter.

Understanding these rules is essential for solving complex data interpretation questions, similar to how standardized medical exams test statistical significance in clinical trials. For a more structured approach, you might use an AI MasterPlan to schedule your practice sessions across different math topics.

Solved Examples

Review these worked-out problems to understand the logic required for high-level GRE questions.

1. Example 1: Independent Events
   A fair six-sided die is rolled twice. What is the probability that the first roll is a 4 and the second roll is an even number?
   
   1. Find the probability of the first event: $P( \text{rolling a 4}) = \frac{1}{6}$.
   2. Find the probability of the second event: The even numbers are {2, 4, 6}, so $P( \text{even}) = \frac{3}{6} = \frac{1}{2}$.
   3. Multiply the probabilities: $\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$.
   4. Answer: $\frac{1}{12}$.
2. Example 2: Complementary Probability
   A bag contains 5 red marbles and 3 blue marbles. If two marbles are drawn at random without replacement, what is the probability that at least one marble is blue?
   
   1. It is easier to find the probability that no marbles are blue (both are red) and subtract from 1.
   2. $P( \text{1st is Red}) = \frac{5}{8}$.
   3. $P( \text{2nd is Red}) = \frac{4}{7}$ (since one red is gone).
   4. $P( \text{Both Red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$.
   5. $P( \text{At least one blue}) = 1 - \frac{5}{14} = \frac{9}{14}$.
   6. Answer: $\frac{9}{14}$.
3. Example 3: Combinations in Probability
   A committee of 3 people is to be chosen from a group of 5 men and 4 women. What is the probability that the committee consists of exactly 2 women and 1 man?
   
   1. Calculate the total ways to choose 3 people from 9: $\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
   2. Calculate the ways to choose 2 women from 4: $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$.
   3. Calculate the ways to choose 1 man from 5: $\binom{5}{1} = 5$.
   4. Multiply the successful outcomes: $6 \times 5 = 30$.
   5. Divide by total outcomes: $\frac{30}{84} = \frac{5}{14}$.
   6. Answer: $\frac{5}{14}$.

Practice Questions

Test your knowledge with these GRE Probability Set 2 Practice Questions. Ensure you read the constraints of each problem carefully.

1. If a jar contains 4 green, 6 yellow, and 2 red candies, what is the probability of picking a candy that is not green?
2. Two cards are drawn from a standard 52-card deck without replacement. What is the probability that both cards are Aces?
3. A box contains 10 light bulbs, 3 of which are defective. If 2 bulbs are selected at random, what is the probability that neither is defective?

4. In a certain class, 60% of students pass Math, 50% pass English, and 30% pass both. What is the probability that a randomly selected student passes Math or English?
5. A fair coin is flipped 4 times. What is the probability of getting exactly 3 heads?
6. If three people are chosen at random, what is the probability that at least two of them were born in the same month? (Assume 12 months with equal probability).
7. A bag has 4 red and 4 blue balls. If you draw 3 balls without replacement, what is the probability they are all the same color?
8. Quantity A: The probability of rolling a sum of 7 with two fair 6-sided dice.
   Quantity B: The probability of rolling a sum of 11 with two fair 6-sided dice.
9. A spinner is divided into 8 equal sectors numbered 1 through 8. If spun twice, what is the probability that the sum of the numbers is greater than 14?
10. A code consists of 2 letters followed by 1 digit. If letters and digits can repeat, what is the probability that the code ends in the number 7?

Answers & Explanations

1. Answer: $\frac{2}{3}$. Total candies = 12. Green candies = 4. Not green candies = $12 - 4 = 8$. Probability = $\frac{8}{12} = \frac{2}{3}$.
2. Answer: $\frac{1}{221}$. There are 4 Aces in 52 cards. $P( \text{1st Ace}) = \frac{4}{52} = \frac{1}{13}$. $P( \text{2nd Ace}) = \frac{3}{51} = \frac{1}{17}$. Multiply: $\frac{1}{13} \times \frac{1}{17} = \frac{1}{221}$.
3. Answer: $\frac{7}{15}$. Total ways to pick 2 bulbs = $\binom{10}{2} = 45$. Non-defective bulbs = 7. Ways to pick 2 non-defective = $\binom{7}{2} = 21$. Probability = $\frac{21}{45} = \frac{7}{15}$.
4. Answer: 0.8 (or 80%). Use the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. So, $0.6 + 0.5 - 0.3 = 0.8$.
5. Answer: $\frac{1}{4}$. Total outcomes = $2^4 = 16$. Ways to get 3 heads = $\binom{4}{3} = 4$. Probability = $\frac{4}{16} = \frac{1}{4}$.
6. Answer: $\frac{17}{72}$. Calculate the complement (all different months). Total = $12^3$. Different = $12 \times 11 \times 10$. $P( \text{diff}) = \frac{1320}{1728} = \frac{55}{72}$. $P( \text{at least 2 same}) = 1 - \frac{55}{72} = \frac{17}{72}$.
7. Answer: $\frac{1}{7}$. Ways to get 3 red: $\binom{4}{3} = 4$. Ways to get 3 blue: $\binom{4}{3} = 4$. Total success = 8. Total combinations = $\binom{8}{3} = 56$. $\frac{8}{56} = \frac{1}{7}$.
8. Answer: Quantity A is greater. Sum of 7 outcomes: (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) — 6 ways. Sum of 11 outcomes: (5,6), (6,5) — 2 ways.
9. Answer: $\frac{3}{64}$. Possible sums > 14: (7,8), (8,7), (8,8). Total outcomes = $8 \times 8 = 64$. Probability = $\frac{3}{64}$.
10. Answer: $\frac{1}{10}$. The letters don't affect the digit. There are 10 possible digits (0-9). The probability of any specific digit is $\frac{1}{10}$.

1. If the probability of event A is 0.3 and the probability of event B is 0.4, and A and B are independent, what is the probability that neither occurs?

• A0.12
• B0.70
• C0.42
• D0.58

Frequently Asked Questions

What is the difference between independent and dependent events on the GRE?

Independent events are those where the outcome of the first does not change the probability of the second, such as rolling a die twice. Dependent events occur when the first outcome changes the sample space for the second, typically seen in "without replacement" marble or card problems.

How do I know when to use combinations versus permutations?

Use combinations when the order of selection does not matter, such as picking a group of three friends for a trip. Use permutations when the order or specific roles matter, such as assigning a gold, silver, and bronze medal to race finishers.

Can a probability value be greater than 1?

No, probability values must always fall within the range of 0 to 1, inclusive. If your calculation results in a number greater than 1, you have likely added probabilities that are not mutually exclusive or made an arithmetic error.

What is the "at least one" rule in GRE probability?

The "at least one" rule suggests that it is often easier to calculate the probability of the event never happening and subtracting that from 1. For example, $P( \text{at least one head}) = 1 - P( \text{no heads})$.

Are probability questions common on the GRE?

Yes, you can expect to see 2-4 probability and counting questions per Quantitative section. These often appear as multiple-choice, quantitative comparison, or numeric entry questions, sometimes linked to Probability Theory concepts.