Hard GRE Probability Practice Test Practice Questions

Concept Explanation

Probability measures the likelihood that a specific event will occur, expressed as a ratio between 0 and 1. In a Hard GRE Probability Practice Test, questions often move beyond simple ratios and require the application of combinatorics, conditional logic, and the complement rule. To calculate the probability of an event $E$, you use the formula $P(E) = \frac{ \text{Number of Favorable Outcomes}}{ \text{Total Number of Possible Outcomes}}$. However, high-level GRE problems frequently involve multiple events that are either independent (the outcome of one does not affect the other) or dependent. For independent events, you multiply their individual probabilities: $P(A \text{ and } B) = P(A) \times P(B)$. For mutually exclusive events, you add them: $P(A \text{ or } B) = P(A) + P(B)$. A common trap on the GRE involves the "at least one" scenario, which is most efficiently solved using the complement: $P( \text{at least one}) = 1 - P( \text{none})$. Understanding these nuances is essential for success in GRE Prep, as it allows you to break down complex word problems into solvable algebraic parts.

Solved Examples

Review these detailed solutions to understand the logic required for challenging probability scenarios.

1. Example 1: The Complement Rule
   A bag contains 4 red marbles and 6 blue marbles. If 3 marbles are drawn at random without replacement, what is the probability that at least one marble is red?
   
   1. Find the total outcomes: $\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
   2. Find the probability of the complement (no red marbles, meaning all are blue): $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
   3. Calculate the probability of no red: $\frac{20}{120} = \frac{1}{6}$.
   4. Subtract from 1: $1 - \frac{1}{6} = \frac{5}{6}$.
2. Example 2: Conditional Probability
   In a group of 100 students, 60 study Math, 40 study Physics, and 20 study both. If a student is chosen at random and they study Math, what is the probability they also study Physics?
   
   1. Identify the condition: The student must be among the 60 Math students.
   2. Identify the favorable outcome: The 20 students who study both.
   3. Apply the formula: $P( \text{Physics} | \text{Math}) = \frac{P( \text{Math and Physics})}{P( \text{Math})}$.
   4. Calculate: $\frac{20}{60} = \frac{1}{3}$.
3. Example 3: Independent Events with Replacement
   A fair six-sided die is rolled four times. What is the probability of rolling a 6 exactly twice?
   
   1. Probability of rolling a 6: $p = \frac{1}{6}$. Probability of not rolling a 6: $q = \frac{5}{6}$.
   2. Use the binomial probability formula: $\binom{n}{k} p^k q^{n-k}$, where $n=4$ and $k=2$.
   3. Calculate: $\binom{4}{2} (\frac{1}{6})^2 (\frac{5}{6})^2 = 6 \times \frac{1}{36} \times \frac{25}{36} = \frac{25}{216}$.

Practice Questions

1. 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?
2. Two cards are drawn from a standard 52-card deck without replacement. What is the probability that both cards are Aces?
3. An archer has an 80% chance of hitting a target. If they fire 3 arrows, what is the probability they hit the target at least twice?

4. A box contains 5 red balls, 3 green balls, and 2 yellow balls. If two balls are picked at random without replacement, what is the probability that they are of different colors?
5. If $P(A) = 0.4$, $P(B) = 0.5$, and events A and B are independent, what is the probability that neither A nor B occurs?
6. A fair coin is tossed 5 times. What is the probability of getting more heads than tails?
7. In a certain city, 30% of people own a dog, 20% own a cat, and 10% own both. If a person is selected at random, what is the probability they own neither a dog nor a cat?
8. A bag contains 10 light bulbs, 3 of which are defective. If 2 bulbs are selected at random without replacement, what is the probability that at least one is defective?

Answers & Explanations

1. Answer: 5/14
   Total ways to choose 3 people from 9: $\binom{9}{3} = 84$. Ways to choose 2 women from 4: $\binom{4}{2} = 6$. Ways to choose 1 man from 5: $\binom{5}{1} = 5$. Favorable outcomes: $6 \times 5 = 30$. Probability: $\frac{30}{84} = \frac{5}{14}$.
2. Answer: 1/221
   Probability of first Ace: $\frac{4}{52} = \frac{1}{13}$. Probability of second Ace: $\frac{3}{51} = \frac{1}{17}$. Multiply: $\frac{1}{13} \times \frac{1}{17} = \frac{1}{221}$.
3. Answer: 0.896
   Calculate exactly 2 hits: $\binom{3}{2}(0.8)^2(0.2)^1 = 3(0.64)(0.2) = 0.384$. Calculate exactly 3 hits: $(0.8)^3 = 0.512$. Add: $0.384 + 0.512 = 0.896$.
4. Answer: 31/45
   Total outcomes: $\binom{10}{2} = 45$. Probability of same color: Red-Red $\binom{5}{2}=10$, Green-Green $\binom{3}{2}=3$, Yellow-Yellow $\binom{2}{2}=1$. Total same: $10+3+1 = 14$. Probability different: $1 - \frac{14}{45} = \frac{31}{45}$.
5. Answer: 0.3
   Probability of not A: $1 - 0.4 = 0.6$. Probability of not B: $1 - 0.5 = 0.5$. Since independent, multiply: $0.6 \times 0.5 = 0.3$.
6. Answer: 1/2
   In 5 tosses, total outcomes are $2^5 = 32$. Because the coin is fair, the probability of getting more heads than tails is equal to the probability of getting more tails than heads. Since there can't be a tie with 5 tosses, the probability is precisely $\frac{1}{2}$.
7. Answer: 0.6
   Use the inclusion-exclusion principle: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.2 - 0.1 = 0.4$. Probability of neither: $1 - 0.4 = 0.6$.
8. Answer: 8/15
   Total outcomes: $\binom{10}{2} = 45$. Complement (no defective): $\binom{7}{2} = 21$. Probability no defective: $\frac{21}{45} = \frac{7}{15}$. Probability at least one: $1 - \frac{7}{15} = \frac{8}{15}$.

1. If a jar contains 3 red, 4 green, and 5 blue marbles, and two are drawn without replacement, what is the probability that both are green?

• A1/11
• B4/33
• C1/9
• D1/22

Frequently Asked Questions

What is the difference between independent and mutually exclusive events?

Independent events are those where the outcome of one does not affect the other, while mutually exclusive events cannot occur at the same time. On the GRE, you multiply probabilities for independent events and add them for mutually exclusive ones.

How do I know when to use combinations versus permutations?

Use combinations when the order of selection does not matter, such as picking a committee. Use permutations when order is significant, such as arranging books on a shelf or assigning specific roles like President and Vice President.

What is the complement rule in probability?

The complement rule states that the probability of an event occurring plus the probability of it not occurring must equal 1. It is frequently used to solve "at least one" problems by calculating the probability of "none" and subtracting from 1.

Can probability on the GRE be greater than 1?

No, probability is always a value between 0 and 1, inclusive. If your calculation results in a number larger than 1, you have likely added probabilities that were not mutually exclusive or made a calculation error.

What are the odds versus probability?

Probability is the ratio of favorable outcomes to total outcomes, whereas odds is the ratio of favorable outcomes to unfavorable outcomes. While the GRE primarily tests probability, understanding that odds of 1:3 equals a probability of 1/4 can be helpful.