Hard GRE Probability Set 2 Practice Questions

Calculate the odds of three independent events occurring simultaneously, and you will quickly realize why probability is a frequent hurdle for graduate school applicants. These Hard GRE Probability Set 2 Practice Questions challenge your ability to navigate complex scenarios involving combinations, conditional outcomes, and mutually exclusive events. While the GRE Prep journey involves many disciplines, mastering the nuances of probability is essential for securing a top-tier quantitative score. High-level problems often mask simple rules behind wordy descriptions, requiring a disciplined approach to identify whether you should multiply probabilities for independent events or subtract the complement from 1 to find the desired result.

Concept Explanation

GRE probability measures the likelihood of a specific event occurring, expressed as a ratio between 0 and 1. To solve advanced problems, you must understand the distinction between independent events, where one outcome does not affect the next, and dependent events, where the sample space changes. For independent events $A$ and $B$, the probability of both occurring is $P(A \text{ and } B) = P(A) \times P(B)$. Conversely, for mutually exclusive events, the probability of either occurring is $P(A \text{ or } B) = P(A) + P(B)$. A common "hard" tactic on the GRE is the "at least one" rule; instead of calculating several successful outcomes, it is often faster to calculate the probability of the event never happening and subtracting that from 1: $P( \text{at least one}) = 1 - P( \text{none})$. For more complex scenarios, you may need to integrate GRE practice questions with explanations to refine your counting techniques, such as permutations and combinations, which define the total number of possible outcomes in the denominator. You can further explore these concepts using Probability Theory resources to deepen your mathematical intuition.

Solved Examples

1. Example 1: A bag contains 4 red marbles and 6 blue marbles. If two marbles are picked at random without replacement, what is the probability that both marbles are red?
   1. Identify the probability of picking the first red marble: $P(R1) = \frac{4}{10} = \frac{2}{5}$.
   2. Identify the probability of picking the second red marble, given the first was red: $P(R2|R1) = \frac{3}{9} = \frac{1}{3}$.
   3. Multiply the probabilities: $\frac{2}{5} \times \frac{1}{3} = \frac{2}{15}$.
2. Example 2: In a certain group, 60% of people like coffee, 40% like tea, and 20% like both. If a person is chosen at random, what is the probability they like neither?
   1. Use the formula for the union of two sets: $P(C \cup T) = P(C) + P(T) - P(C \cap T)$.
   2. Calculate: $0.60 + 0.40 - 0.20 = 0.80$. This is the probability they like at least one.
   3. Subtract from 1 to find those who like neither: $1 - 0.80 = 0.20$.
3. Example 3: A fair six-sided die is rolled three times. What is the probability that the sum of the outcomes is 4?
   1. Determine the total outcomes: $6 \times 6 \times 6 = 216$.
   2. List the combinations that sum to 4: (1, 1, 2), (1, 2, 1), and (2, 1, 1). There are 3 such outcomes.
   3. Calculate the probability: $\frac{3}{216} = \frac{1}{72}$.

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 will consist of exactly 2 women and 1 man?

2. Two integers are selected at random from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} without replacement. What is the probability that the product of the two integers is even?

3. A shared target is fired upon by two archers. Archer A has a $\frac{2}{3}$ chance of hitting the target, and Archer B has a $\frac{3}{5}$ chance. If both fire once, what is the probability the target is hit at least once?

4. Box A contains 3 red and 2 white chips. Box B contains 2 red and 4 white chips. A chip is drawn from Box A and placed into Box B. Then, a chip is drawn from Box B. What is the probability the second chip drawn is red?

5. If three distinct numbers are chosen from the set {1, 2, 3, 4, 5, 6}, what is the probability that their sum is divisible by 3?

6. A fair coin is flipped 5 times. What is the probability of getting more heads than tails?

7. A bag contains 5 green balls and 7 yellow balls. If three balls are drawn simultaneously, what is the probability that at least one ball is green?

8. A password consists of 4 digits (0-9). If the digits cannot be repeated, what is the probability that the password consists only of odd digits?

Answers & Explanations

1. Answer: $\frac{10}{21}$. Total ways to choose 3 people is $\binom{9}{3} = 84$. Ways to choose 2 women is $\binom{4}{2} = 6$. Ways to choose 1 man is $\binom{5}{1} = 5$. Successful outcomes = $6 \times 5 = 30$. Probability = $\frac{30}{84} = \frac{5}{14}$. (Correction: $\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$; $30/84 = 5/14$).
2. Answer: $\frac{7}{9}$. The product is even unless both numbers are odd. Total ways to pick 2 numbers is $\binom{10}{2} = 45$. There are 5 odd numbers {1, 3, 5, 7, 9}. Ways to pick 2 odd numbers is $\binom{5}{2} = 10$. Probability of odd product = $\frac{10}{45} = \frac{2}{9}$. Probability of even product = $1 - \frac{2}{9} = \frac{7}{9}$.
3. Answer: $\frac{13}{15}$. Use the complement: $P( \text{at least one hit}) = 1 - P( \text{both miss})$. $P( \text{A miss}) = 1 - \frac{2}{3} = \frac{1}{3}$. $P( \text{B miss}) = 1 - \frac{3}{5} = \frac{2}{5}$. $P( \text{both miss}) = \frac{1}{3} \times \frac{2}{5} = \frac{2}{15}$. Result = $1 - \frac{2}{15} = \frac{13}{15}$.
4. Answer: $\frac{13}{35}$. Case 1: Red from A ($\frac{3}{5}$), then red from B (now 3R, 4W): $\frac{3}{5} \times \frac{3}{7} = \frac{9}{35}$. Case 2: White from A ($\frac{2}{5}$), then red from B (now 2R, 5W): $\frac{2}{5} \times \frac{2}{7} = \frac{4}{35}$. Total = $\frac{9}{35} + \frac{4}{35} = \frac{13}{35}$.
5. Answer: $\frac{2}{5}$. Total combinations = $\binom{6}{3} = 20$. Sums divisible by 3: (1,2,3), (1,2,6), (1,3,5), (1,5,6), (2,3,4), (2,4,6), (3,4,5), (4,5,6). There are 8 combinations. Probability = $\frac{8}{20} = \frac{2}{5}$.
6. Answer: $\frac{1}{2}$. In 5 flips, you cannot have an equal number of heads and tails. Because the coin is fair, the probability of having more heads is equal to the probability of having more tails. Thus, $\frac{1}{2}$.
7. Answer: $\frac{37}{44}$. Total ways = $\binom{12}{3} = 220$. Ways to get no green (all yellow) = $\binom{7}{3} = 35$. $P( \text{no green}) = \frac{35}{220} = \frac{7}{44}$. $P( \text{at least one green}) = 1 - \frac{7}{44} = \frac{37}{44}$.
8. Answer: $\frac{1}{42}$. Total ways to pick 4 digits = $10 \times 9 \times 8 \times 7 = 5040$. Ways to pick 4 odd digits (1,3,5,7,9) = $5 \times 4 \times 3 \times 2 = 120$. Probability = $\frac{120}{5040} = \frac{1}{42}$.

Frequently Asked Questions

How do I know when to use combinations versus permutations in probability?

Use combinations when the order of selection does not matter, such as picking a committee or a handful of marbles. Use permutations when the specific sequence or arrangement is important, such as assigning seats or creating a multi-digit password.

What does it mean for two events to be mutually exclusive?

Mutually exclusive events are events that cannot happen at the same time, meaning their intersection is zero. For example, a single roll of a die cannot result in both a 2 and a 5 simultaneously.

Why do we subtract from 1 in "at least one" problems?

Subtracting from 1 uses the complement rule, which is often mathematically simpler than calculating every possible successful scenario. It is easier to find the single case where the event does not happen at all than to sum the probabilities of it happening once, twice, or more.

Does "without replacement" change the denominator?

Yes, in "without replacement" scenarios, the total number of available items decreases with each selection. This makes the events dependent, as the probability of the second draw depends on what was removed during the first draw.

How is probability tested differently on the GRE compared to SAT?

GRE probability questions often involve larger sets and multi-step logic, frequently requiring you to combine probability with advanced counting or algebraic variables. The GRE also utilizes more data interpretation contexts where you must extract probabilities from tables or charts.