GRE Probability Word Problems Practice Questions with Answers

A single roll of a fair six-sided die offers a 1 in 6 chance of landing on a specific number, a fundamental reality that governs the logic of GRE Probability Word Problems. These questions assess your ability to calculate the likelihood of specific outcomes by translating complex scenarios into mathematical ratios. Success on the Quantitative Reasoning section requires more than just memorizing formulas; you must understand how to combine events, recognize mutually exclusive scenarios, and handle independent trials within a timed environment. By integrating these concepts into your GRE Prep, you can transform intimidating word problems into manageable equations.

Concept Explanation

GRE Probability Word Problems measure the likelihood of a specific event occurring, expressed as the ratio of the number of successful outcomes to the total number of possible outcomes. The basic formula for any probability $P(E)$ is:

$$P(E) = \frac{ \text{Number of Desired Outcomes}}{ \text{Total Number of Possible Outcomes}}$$

Probabilities are always bounded between 0 and 1, where 0 represents an impossibility and 1 represents a certainty. For the GRE, you must be comfortable with three primary types of probability scenarios:

• Independent Events: The outcome of one event does not affect the other. To find the probability of both occurring, multiply their individual probabilities: $P(A \text{ and } B) = P(A) \times P(B)$.
• Mutually Exclusive Events: Events that cannot happen at the same time. To find the probability of either occurring, add their individual probabilities: $P(A \text{ or } B) = P(A) + P(B)$.
• Complementary Events: The probability of an event not happening is $1 - P(E)$. This is often the fastest way to solve "at least one" problems.

When dealing with sequences or selections where order doesn't matter, you may need to use combinations, whereas permutations are used when order is significant. For more complex quantitative practice, you might find similarities in logic when reviewing Hard USMLE Renal Physiology Practice Questions, which also require precise ratio-based reasoning.

Solved Examples

Review these step-by-step solutions to understand how to apply probability theory to word problems.

1. Example 1: A bag contains 4 red marbles, 3 blue marbles, and 5 green marbles. If two marbles are drawn at random without replacement, what is the probability that both are green?
   1. Calculate the total number of marbles: $4 + 3 + 5 = 12$.
   2. Find the probability of the first green marble: $\frac{5}{12}$.
   3. Find the probability of the second green marble (given one green is gone): $\frac{4}{11}$.
   4. Multiply the probabilities: $\frac{5}{12} \times \frac{4}{11} = \frac{20}{132}$.
   5. Simplify the fraction: $\frac{5}{33}$.
2. Example 2: A fair coin is flipped 3 times. What is the probability of getting exactly two heads?
   1. Identify total outcomes: $2^3 = 8$ (HHH, HHT, HTH, THH, TTH, THT, HTT, TTT).
   2. List desired outcomes: {HHT, HTH, THH}. There are 3 outcomes.
   3. Apply the formula: $P = \frac{3}{8}$.
3. Example 3: If the probability of rain tomorrow is 0.4 and the probability of a high wind is 0.7, assuming the events are independent, what is the probability that it will neither rain nor be windy?
   1. Find the probability of no rain: $1 - 0.4 = 0.6$.
   2. Find the probability of no wind: $1 - 0.7 = 0.3$.
   3. Multiply for independent events: $0.6 \times 0.3 = 0.18$.

Practice Questions

Test your skills with these GRE probability word problems. Use the AI Question Generator to create more sets if you need additional practice.

1. A jar contains 6 black beans and 4 white beans. If two beans are chosen at random without replacement, what is the probability that one is black and the other is white?

2. 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 will consist of 2 women and 1 man?

3. Two fair six-sided dice are rolled. What is the probability that the sum of the numbers shown is exactly 9?

4. The probability that Event A occurs is 0.5, and the probability that Event B occurs is 0.3. If A and B are independent events, what is the probability that at least one of the events occurs?

5. In a group of 10 students, 6 are female. If 3 students are selected at random for a presentation, what is the probability that all 3 are female?

6. A drawer contains 10 socks: 6 are blue and 4 are red. If you pick two socks at random, what is the probability they are the same color?

7. A spinner is divided into 8 equal sections numbered 1 through 8. What is the probability that the spinner lands on a prime number or an even number?

8. Three friends, Alex, Ben, and Carl, are in a race. If the probability of Alex winning is $\frac{1}{4}$ and the probability of Ben winning is $\frac{1}{3}$, what is the probability that Carl wins? (Assume no ties and only these three are racing).

9. A box contains 5 light bulbs, 2 of which are defective. If 2 bulbs are tested at random, what is the probability that neither is defective?

10. What is the probability of drawing an Ace or a Heart from a standard deck of 52 playing cards?

Answers & Explanations

1. Answer: $\frac{8}{15}$
There are two ways to get one black and one white: (Black, then White) or (White, then Black).
P(B then W) = $\frac{6}{10} \times \frac{4}{9} = \frac{24}{90}$.
P(W then B) = $\frac{4}{10} \times \frac{6}{9} = \frac{24}{90}$.
Total Probability = $\frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}$.

2. Answer: $\frac{5}{14}$
Total ways to choose 3 from 9: $\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
Ways to choose 2 women from 4: $\binom{4}{2} = 6$.
Ways to choose 1 man from 5: $\binom{5}{1} = 5$.
Successful outcomes: $6 \times 5 = 30$.
Probability = $\frac{30}{84} = \frac{5}{14}$.

3. Answer: $\frac{1}{9}$
Total outcomes: $6 \times 6 = 36$.
Pairs that sum to 9: (3,6), (4,5), (5,4), (6,3). Total = 4.
Probability = $\frac{4}{36} = \frac{1}{9}$.

4. Answer: 0.65
Use the complement rule: P(At least one) = 1 - P(Neither).
P(Not A) = 0.5; P(Not B) = 0.7.
P(Neither) = $0.5 \times 0.7 = 0.35$.
P(At least one) = $1 - 0.35 = 0.65$.

5. Answer: $\frac{1}{6}$
Total ways to choose 3 from 10: $\binom{10}{3} = 120$.
Ways to choose 3 from 6 females: $\binom{6}{3} = 20$.
Probability = $\frac{20}{120} = \frac{1}{6}$.

6. Answer: $\frac{7}{15}$
Case 1 (Both Blue): $\frac{6}{10} \times \frac{5}{9} = \frac{30}{90}$.
Case 2 (Both Red): $\frac{4}{10} \times \frac{3}{9} = \frac{12}{90}$.
Total = $\frac{30+12}{90} = \frac{42}{90} = \frac{7}{15}$.

7. Answer: $\frac{3}{4}$
Prime numbers: {2, 3, 5, 7}. Even numbers: {2, 4, 6, 8}.
Combined set: {2, 3, 4, 5, 6, 7, 8}. Total = 6 unique numbers.
Probability = $\frac{6}{8} = \frac{3}{4}$.

8. Answer: $\frac{5}{12}$
The sum of probabilities of all possible outcomes must be 1.
P(Carl) = $1 - (\frac{1}{4} + \frac{1}{3}) = 1 - (\frac{3}{12} + \frac{4}{12}) = 1 - \frac{7}{12} = \frac{5}{12}$.

9. Answer: $\frac{3}{10}$
There are 3 non-defective bulbs.
P(1st non-defective) = $\frac{3}{5}$.
P(2nd non-defective) = $\frac{2}{4}$.
Total = $\frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$.

10. Answer: $\frac{4}{13}$
Aces = 4. Hearts = 13. Overlap (Ace of Hearts) = 1.
Total = $4 + 13 - 1 = 16$.
Probability = $\frac{16}{52} = \frac{4}{13}$.

1. If a bag contains 5 red and 5 blue balls, what is the probability of drawing two blue balls in a row without replacement?

• A1/4
• B2/9
• C1/2
• D5/18

Frequently Asked Questions

What is the difference between independent and dependent events in GRE probability?

Independent events are those where the outcome of the first does not change the likelihood of the second, such as flipping a coin twice. Dependent events occur when the first outcome alters the total pool of possibilities for the second, typically seen in "without replacement" scenarios.

How do I solve "at least one" probability questions efficiently?

To solve "at least one" problems, it is usually faster to calculate the probability of the event never happening and subtract that value from 1. This avoids the need to calculate multiple successful scenarios separately and add them together.

Are combinations and permutations tested within probability word problems?

Yes, the GRE frequently combines counting methods with probability by requiring you to use the combination formula to determine the total number of possible outcomes or the number of successful outcomes. If you are struggling with these, practicing Hard USMLE Cardiovascular Pathology Practice Questions can help build the stamina needed for multi-step quantitative reasoning.

What is the most common mistake students make on GRE probability questions?

The most frequent error is failing to account for "without replacement" conditions, which changes the denominator of the second fraction. Another common mistake is adding probabilities when they should be multiplied, or forgetting to subtract overlapping outcomes in "OR" problems.

Can a probability value on the GRE be greater than 1 or negative?

No, a probability must always range from 0 to 1 inclusive; any answer choice outside this range is mathematically impossible. If your calculation results in a number like 1.2 or -0.5, you have likely used the wrong operation or formula.