Hard GRE Probability Set 1 Practice Questions

Calculate the likelihood of a specific event occurring by dividing the number of favorable outcomes by the total number of possible outcomes in a sample space. This Hard GRE Probability Set 1 Practice Questions guide targets the most complex permutations, combinations, and conditional probability scenarios found on the Quantitative Reasoning section. Success on the GRE requires more than just knowing basic formulas; it demands the ability to visualize overlapping sets and independent events under time pressure.

Concept Explanation

Probability is the mathematical measure of the likelihood that an event will occur, expressed as a number between 0 and 1. To solve advanced problems, you must master the relationship between independent and dependent events, as well as the fundamental counting principle. For high-level GRE Prep, students often encounter "at least" problems, where calculating the complement (the probability of the event not happening) is much faster than calculating each successful outcome individually.

Key concepts include:

• Independent Events: The outcome of one does not affect the other. Calculate using $P(A \text{ and } B) = P(A) \times P(B)$.
• Dependent Events: The outcome of the first affects the second. Calculate using $P(A \text{ and } B) = P(A) \times P(B|A)$.
• Mutually Exclusive Events: Events that cannot happen at the same time. The probability is $P(A \text{ or } B) = P(A) + P(B)$.
• Combinations and Permutations: Used to determine the total sample space when order either matters or does not matter.

For those looking to refine their skills further, using a AI Question Generator can provide endless variations of these difficult logic puzzles. Understanding the fundamentals of probability and statistics at a deep level is essential for reaching the 165+ score range.

Solved Examples

1. Example 1: The Complement Rule
   A bag contains 4 red marbles, 3 blue marbles, and 2 green marbles. If three marbles are drawn at random without replacement, what is the probability that at least one marble is green?
   Solution:
   1. Calculate the total number of ways to choose 3 marbles from 9: $\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
   2. Calculate the ways to choose 3 marbles that are NOT green (choosing from the 7 red/blue marbles): $\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.
   3. Find the probability of NO green marbles: $\frac{35}{84} = \frac{5}{12}$.
   4. Subtract from 1 to find the probability of "at least one": $1 - \frac{5}{12} = \frac{7}{12}$.
2. Example 2: Conditional Probability
   In a certain group, 60% of people own a car, 30% own a bike, and 20% own both. If a person is chosen at random and they own a car, what is the probability they also own a bike?
   Solution:
   1. Use the formula for conditional probability: $P(B|A) = \frac{P(A \cap B)}{P(A)}$.
   2. Identify the values: $P( \text{Car}) = 0.60$ and $P( \text{Car and Bike}) = 0.20$.
   3. Apply the formula: $\frac{0.20}{0.60} = \frac{1}{3}$.
3. Example 3: Geometric Probability
   A square target has a side length of 10. Inside the square is a circle with a radius of 3. If a point is chosen at random inside the square, what is the probability it is NOT inside the circle?
   Solution:
   1. Calculate the area of the square: $10^2 = 100$.
   2. Calculate the area of the circle: $\pi r^2 = \pi(3^2) = 9\pi$.
   3. Find the area outside the circle: $100 - 9\pi$.
   4. The probability is the ratio of the areas: $\frac{100 - 9\pi}{100} = 1 - 0.09\pi$.

Practice Questions

1. A committee of 4 is to be selected from a group of 6 men and 4 women. What is the probability that the committee contains exactly 2 men and 2 women?

2. Two fair six-sided dice are rolled. What is the probability that the sum of the numbers shown is a prime number?

3. A box contains 10 light bulbs, 3 of which are defective. If a sample of 2 bulbs is chosen at random without replacement, what is the probability that both bulbs are defective?

4. Five people (A, B, C, D, E) are to be seated in a row. What is the probability that A and B are seated next to each other?

5. A fair coin is flipped 6 times. What is the probability of getting exactly 3 heads and 3 tails?

6. Set X consists of integers {1, 2, 3, 4, 5}. Set Y consists of integers {6, 7, 8, 9, 10}. One number is picked from X and one from Y. What is the probability that the product of the two numbers is even?

7. A bag contains 5 red, 4 blue, and 3 yellow balls. Two balls are picked at random without replacement. What is the probability that they are of different colors?

8. If the probability of event A occurring is 0.4 and the probability of event B occurring is 0.5, and the probability of at least one occurring is 0.7, are events A and B independent?

Answers & Explanations

1. Answer: 3/7
   Total ways to choose 4 from 10: $\binom{10}{4} = 210$. Ways to choose 2 men from 6: $\binom{6}{2} = 15$. Ways to choose 2 women from 4: $\binom{4}{2} = 6$. Favorable outcomes: $15 \times 6 = 90$. Probability: $90/210 = 3/7$. Compare this to GRE Practice Questions with Explanations for similar combinatorics challenges.
2. Answer: 5/12
   Total outcomes when rolling two dice: 36. Prime sums possible: 2 (1 way), 3 (2 ways), 5 (4 ways), 7 (6 ways), 11 (2 ways). Total favorable: $1+2+4+6+2 = 15$. Probability: $15/36 = 5/12$.
3. Answer: 1/15
   Total ways to choose 2 bulbs from 10: $\binom{10}{2} = 45$. Ways to choose 2 defective bulbs from 3: $\binom{3}{2} = 3$. Probability: $3/45 = 1/15$.
4. Answer: 2/5
   Total permutations of 5 people: $5! = 120$. Treat A and B as a single unit. There are now 4 units (AB, C, D, E). Permutations of 4 units: $4! = 24$. Since A and B can switch places (AB or BA), multiply by 2: $24 \times 2 = 48$. Probability: $48/120 = 2/5$.
5. Answer: 5/16
   Total outcomes for 6 flips: $2^6 = 64$. Favorable outcomes (choosing 3 spots for heads out of 6): $\binom{6}{3} = 20$. Probability: $20/64 = 5/16$.
6. Answer: 4/5
   A product is odd only if both numbers are odd. Odds in X: {1, 3, 5} (3 total). Odds in Y: {7, 9} (2 total). Ways to get an odd product: $3 \times 2 = 6$. Total possible products: $5 \times 5 = 25$. Probability of odd: $6/25$. Probability of even: $1 - 6/25 = 19/25$. Wait, checking Set Y: {6, 7, 8, 9, 10} has two odds (7, 9). Correct. $19/25 = 0.76$. For more logic-based sets, see our Unlimited GRE Practice Questions.
7. Answer: 47/66
   Easier to find probability of same color and subtract from 1. Total balls: 12. Total pairs: $\binom{12}{2} = 66$. Same color pairs: Red $\binom{5}{2}=10$, Blue $\binom{4}{2}=6$, Yellow $\binom{3}{2}=3$. Total same color: $10+6+3=19$. Different color: $1 - 19/66 = 47/66$.
8. Answer: Yes
   Use the addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. $0.7 = 0.4 + 0.5 - P(A \cap B) \implies P(A \cap B) = 0.2$. If independent, $P(A) \times P(B) = P(A \cap B)$. $0.4 \times 0.5 = 0.2$. Since they match, they are independent.

1. If a fair coin is tossed 3 times, what is the probability that the result is three heads?

• A1/2
• B1/4
• C1/8
• D1/6

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 event has no effect on the likelihood of the second, such as multiple coin flips. Dependent events occur when the first outcome changes the sample space for the next, commonly seen in "without replacement" marble 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 friends for a trip. Use permutations when the order is significant, such as assigning specific roles like President and Vice President or arranging books on a shelf.

What is the "at least" rule in 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. This avoids the tedious process of summing the probabilities of one occurrence, two occurrences, and so on.

Can probability on the GRE be greater than 1?

No, probability must always be between 0 and 1, inclusive. If your calculation results in a number greater than 1, you likely added probabilities that were not mutually exclusive without subtracting their intersection.

How does the GRE test conditional probability?

The GRE tests conditional probability by providing a restricted subset of the total group, such as "Given that a student passed Math, what is the probability they passed Science?" You must use the number of people who passed both as the numerator and only the people who passed Math as the denominator.