Hard GRE Probability Exam Questions Practice Questions

Concept Explanation

Probability measures the likelihood that a specific event will occur, expressed as a ratio between 0 and 1. For those tackling Hard GRE Probability Exam Questions, the focus shifts from simple coin tosses to complex scenarios involving combinatorics, conditional probability, and independent versus dependent events. At its core, the probability of an event $A$, denoted as $P(A)$, is calculated by dividing the number of successful outcomes by the total number of equally likely possible outcomes in the sample space. To succeed on the GRE Prep journey, you must understand that the sum of the probabilities of all possible mutually exclusive events is exactly 1. Advanced problems often require the use of the Complement Rule, which states that $P(A) = 1 - P( \text{not } A)$. This is particularly useful when calculating "at least one" scenarios. Furthermore, the Multiplication Rule for independent events, $P(A \cap B) = P(A) \times P(B)$, and the Addition Rule for non-mutually exclusive events, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, are essential tools for high-level quantitative reasoning.

Solved Examples

1. Example 1: Combinations and Probability
   A bag contains 5 red marbles and 7 blue marbles. If 3 marbles are selected at random without replacement, what is the probability that exactly 2 are red?
   1. Calculate the total ways to choose 3 marbles from 12: $$\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$$
   2. Calculate the ways to choose 2 red marbles from 5: $$\binom{5}{2} = 10$$
   3. Calculate the ways to choose 1 blue marble from 7: $$\binom{7}{1} = 7$$
   4. Multiply the successful outcomes: $10 \times 7 = 70$
   5. Find the probability: $\frac{70}{220} = \frac{7}{22}$
2. Example 2: The Complement Rule
   The probability that it will rain on any given day is 0.2. What is the probability that it will rain at least once in a 3-day period?
   1. Identify the complement: The complement of "at least once" is "never."
   2. Calculate the probability of no rain on one day: $1 - 0.2 = 0.8$
   3. Calculate the probability of no rain for 3 consecutive days: $0.8 \times 0.8 \times 0.8 = 0.512$
   4. Subtract from 1: $1 - 0.512 = 0.488$
3. Example 3: Conditional Probability
   In a group of 100 students, 60 study Math, 40 study Physics, and 20 study both. If a student who studies Math is chosen at random, what is the probability they also study Physics?
   1. Identify the restricted sample space: We are only looking at the 60 students who study Math.
   2. Identify the successful outcomes within that space: The 20 students who study both.
   3. Apply the formula $P(Physics|Math) = \frac{P(Physics \cap Math)}{P(Math)}$: $\frac{20}{60} = \frac{1}{3}$

Practice Questions

1. A box contains 4 black, 4 white, and 4 red balls. If three balls are drawn at random without replacement, what is the probability that all three balls are of the same color?
2. A fair six-sided die is rolled four times. What is the probability that the number 6 appears exactly twice?
3. If two integers are chosen randomly from the set $\{1, 2, 3, \dots, 10\}$, what is the probability that their product is an even number?

4. A committee of 3 is to be selected from a group of 5 men and 5 women. What is the probability that the committee will have more women than men?
5. A fair coin is flipped 5 times. What is the probability of getting more heads than tails?
6. Two cards are drawn from a standard 52-card deck without replacement. What is the probability that both cards are Aces?
7. An archer has a 75% chance of hitting a target. If they shoot 3 times, what is the probability they hit the target at least twice?
8. A jar contains 10 coins: 7 are fair and 3 are biased with a 0.8 probability of landing on heads. If a coin is chosen at random and flipped, what is the probability it lands on heads?
9. In a sequence of 10 independent trials with a success probability of 0.4, what is the probability of exactly 1 success?
10. Three people are chosen from a group of 6. What is the probability that a specific person, John, is included in the selection?

Answers & Explanations

1. Answer: 3/55. Total ways to choose 3 balls from 12 is $\binom{12}{3} = 220$. Ways to choose 3 balls of one color (e.g., black) is $\binom{4}{3} = 4$. Since there are three colors, total successful outcomes = $4 \times 3 = 12$. Probability = $12/220 = 3/55$.
2. Answer: 25/216. This is a binomial probability problem: $\binom{4}{2} \times (\frac{1}{6})^2 \times (\frac{5}{6})^2$. This equals $6 \times \frac{1}{36} \times \frac{25}{36} = \frac{150}{1296} = \frac{25}{216}$.
3. Answer: 13/18. Use the complement: The product is odd only if both numbers are odd. Odd numbers are {1, 3, 5, 7, 9} (5 total). Ways to pick 2 odds: $\binom{5}{2} = 10$. Total ways to pick 2 numbers: $\binom{10}{2} = 45$. $P( \text{odd}) = 10/45 = 2/9$. $P( \text{even}) = 1 - 2/9 = 7/9$. Wait, $7/9 = 14/18$. Let's re-verify: $1 - 10/45 = 35/45 = 7/9$.
4. Answer: 1/2. Total ways: $\binom{10}{3} = 120$. "More women than men" means 2 women/1 man or 3 women/0 men. Case 1: $\binom{5}{2} \times \binom{5}{1} = 10 \times 5 = 50$. Case 2: $\binom{5}{3} \times \binom{5}{0} = 10 \times 1 = 10$. Total successful = $50 + 10 = 60$. Probability = $60/120 = 1/2$.
5. Answer: 1/2. Since the coin is fair and 5 is odd, the probability of heads > tails is equal to the probability of tails > heads. By symmetry, the answer is 0.5.
6. Answer: 1/221. First card Ace: 4/52. Second card Ace: 3/51. Multiply: $\frac{4}{52} \times \frac{3}{51} = \frac{1}{13} \times \frac{1}{17} = \frac{1}{221}$.
7. Answer: 27/32. Hits: 2 or 3. Case 2 hits: $\binom{3}{2} \times (0.75)^2 \times (0.25)^1 = 3 \times \frac{9}{16} \times \frac{1}{4} = \frac{27}{64}$. Case 3 hits: $(0.75)^3 = \frac{27}{64}$. Total = $\frac{27}{64} + \frac{27}{64} = \frac{54}{64} = \frac{27}{32}$.
8. Answer: 0.59. Use law of total probability: $P(H) = P(H| \text{Fair})P( \text{Fair}) + P(H| \text{Biased})P( \text{Biased})$. $P(H) = (0.5 \times 0.7) + (0.8 \times 0.3) = 0.35 + 0.24 = 0.59$.
9. Answer: $10 \times (0.4)^1 \times (0.6)^9$. Using binomial formula $\binom{n}{k} p^k (1-p)^{n-k}$: $\binom{10}{1} (0.4)^1 (0.6)^9$.
10. Answer: 1/2. Total ways to choose 3 from 6: $\binom{6}{3} = 20$. Ways to choose 3 including John: John is fixed, choose 2 more from 5: $\binom{5}{2} = 10$. Probability = $10/20 = 1/2$.

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

• A0.12
• B0.70
• C0.42
• D0.58

Frequently Asked Questions

What is the difference between independent and mutually exclusive events?

Independent events are those where the occurrence of one does not affect the probability of the other, while mutually exclusive events cannot happen at the same time. On the GRE, distinguishing between these is vital for choosing whether to multiply or add probabilities correctly.

How do I handle "at least one" probability questions?

The most efficient way to solve "at least one" problems is to use the complement rule by calculating the probability of the event never happening. Subtracting this "none" probability from 1 gives you the probability of the event occurring at least once.

When should I use combinations instead of 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 ranking winners in a race.

Does the GRE provide a calculator for probability questions?

Yes, the GRE provides an on-screen calculator for the Quantitative section, which can help with large multiplication or long division. However, most probability questions are designed to be solved through fraction simplification or conceptual understanding of GRE practice questions with explanations.

What is conditional probability on the GRE?

Conditional probability refers to the likelihood of an event occurring given that another event has already occurred. It effectively shrinks the sample space to only those outcomes that satisfy the first condition, as seen in unlimited GRE practice questions.