Hard GRE Independent Events Questions Practice Questions

Two events are independent if the occurrence of one does not change the probability that the other will occur. Solving Hard GRE Independent Events Questions requires more than just knowing a formula; it involves identifying these relationships within complex word problems and applying the multiplication rule correctly under pressure. While basic probability is covered in introductory math, the GRE Prep curriculum tests your ability to handle multi-step scenarios where independence is either stated or must be inferred from the context.

Concept Explanation

Independent events are occurrences where the outcome of the first event has no statistical influence on the probability of the second event. Mathematically, two events $A$ and $B$ are independent if and only if the probability of both occurring is the product of their individual probabilities: $P(A \text{ and } B) = P(A) \times P(B)$. This principle extends to any number of events; for instance, if three events are mutually independent, the probability of all three happening is $P(A) \times P(B) \times P(C)$.

On the GRE, these problems often appear in the form of "at least one" scenarios or sequences of dice rolls and coin flips. A critical tool for solving Hard GRE Independent Events Questions is the Complement Rule. If you need to find the probability that an event happens "at least once," it is often easier to calculate the probability that it never happens and subtract that value from 1. For more foundational practice, you can explore GRE Practice Questions with Answers to build your confidence before tackling these advanced iterations.

Solved Examples

1. Example 1: The Triple Target. An archer has a 0.8 probability of hitting a target on any single shot. If the archer takes three independent shots, what is the probability that they hit the target exactly twice?
   1. Identify the probability of success $P(S) = 0.8$ and failure $P(F) = 0.2$.
   2. List the possible successful sequences: (Hit, Hit, Miss), (Hit, Miss, Hit), or (Miss, Hit, Hit).
   3. Calculate the probability of one sequence: $0.8 \times 0.8 \times 0.2 = 0.128$.
   4. Since there are 3 such sequences, multiply by 3: $3 \times 0.128 = 0.384$.
2. Example 2: Combined Probabilities. Box A contains 3 red marbles and 2 blue marbles. Box B contains 4 red marbles and 6 blue marbles. If one marble is drawn from each box, what is the probability that both marbles are red?
   1. Calculate the probability of drawing red from Box A: $P(A) = \frac{3}{5}$.
   2. Calculate the probability of drawing red from Box B: $P(B) = \frac{4}{10} = \frac{2}{5}$.
   3. Multiply the independent probabilities: $\frac{3}{5} \times \frac{2}{5} = \frac{6}{25}$.
   4. Convert to decimal if necessary: $0.24$.
3. Example 3: The "At Least One" Rule. A fair coin is flipped 5 times. What is the probability of getting at least one head?
   1. Identify the complement: The only way to not get "at least one head" is to get zero heads (all tails).
   2. Calculate the probability of all tails: $(\frac{1}{2})^5 = \frac{1}{32}$.
   3. Subtract from 1: $1 - \frac{1}{32} = \frac{31}{32}$.

Practice Questions

1. A machine has three independent components, A, B, and C. The probabilities of failure for each component are 0.1, 0.2, and 0.05, respectively. What is the probability that the machine operates without any component failing?

2. In a certain game, the probability of winning a single round is $\frac{1}{4}$. If 4 rounds are played independently, what is the probability of winning exactly one round?

3. Two students, X and Y, are trying to solve a logic puzzle. The probability that X solves it is 0.6, and the probability that Y solves it is 0.8. If they work independently, what is the probability that the puzzle is solved by at least one of them?

4. A bag contains 5 white balls and 3 black balls. A ball is drawn, its color noted, and then it is replaced before a second ball is drawn. What is the probability that the two balls drawn are of different colors?

5. If the probability of rain on any given day in a specific city is 0.3, and weather patterns are assumed to be independent across days, what is the probability that it rains on exactly two days out of a three-day period?

6. An electronic system consists of two subsystems, X and Y, connected in series. The system fails if either X or Y fails. If the probability of X failing is 0.15 and the probability of Y failing is 0.10, and they fail independently, what is the probability the system functions?

7. A fair six-sided die is rolled 4 times. What is the probability that a "6" appears for the first time on the fourth roll?

8. A marksman has a probability $p$ of hitting a target. If the probability of hitting the target at least once in three independent shots is $\frac{19}{27}$, what is the value of $p$?

9. Three independent events A, B, and C have probabilities $P(A) = 0.5$, $P(B) = 0.4$, and $P(C) = 0.2$. What is the probability that exactly two of these events occur?

10. A basket contains 10 apples, 3 of which are rotten. If 2 apples are selected one after the other with replacement, what is the probability that the first is rotten and the second is not?

Answers & Explanations

1. 0.684: The probability of each component not failing is $0.9, 0.8,$ and $0.95$. Multiply them: $0.9 \times 0.8 \times 0.95 = 0.684$.
2. $\frac{27}{64}$: The probability of winning one round is $\frac{1}{4}$ and losing is $\frac{3}{4}$. The number of ways to win exactly one is $\binom{4}{1} = 4$. Calculation: $4 \times (\frac{1}{4})^1 \times (\frac{3}{4})^3 = 4 \times \frac{27}{256} = \frac{108}{256} = \frac{27}{64}$.
3. 0.92: Use the complement (neither solves it). $P( \text{not X}) = 0.4$, $P( \text{not Y}) = 0.2$. $P( \text{neither}) = 0.4 \times 0.2 = 0.08$. $P( \text{at least one}) = 1 - 0.08 = 0.92$.
4. $\frac{15}{32}$: Two scenarios: (White, Black) or (Black, White). $P(W, B) = \frac{5}{8} \times \frac{3}{8} = \frac{15}{64}$. $P(B, W) = \frac{3}{8} \times \frac{5}{8} = \frac{15}{64}$. Sum: $\frac{30}{64} = \frac{15}{32}$.
5. 0.189: $P( \text{Rain}) = 0.3$, $P( \text{No Rain}) = 0.7$. Ways to have 2 rainy days: $\binom{3}{2} = 3$. Calculation: $3 \times (0.3)^2 \times (0.7)^1 = 3 \times 0.09 \times 0.7 = 0.189$.
6. 0.765: The system functions only if both X and Y function. $P( \text{X functions}) = 0.85$, $P( \text{Y functions}) = 0.90$. Multiply: $0.85 \times 0.90 = 0.765$.
7. $\frac{125}{1296}$: This requires (Not 6, Not 6, Not 6, 6). Probability: $(\frac{5}{6})^3 \times \frac{1}{6} = \frac{125}{216} \times \frac{1}{6} = \frac{125}{1296}$.
8. $\frac{1}{3}$: $P( \text{at least one}) = 1 - P( \text{none}) = 1 - (1-p)^3 = \frac{19}{27}$. Then $(1-p)^3 = 1 - \frac{19}{27} = \frac{8}{27}$. Taking the cube root, $1-p = \frac{2}{3}$, so $p = \frac{1}{3}$.
9. 0.22: Scenarios: (A,B,notC) + (A,notB,C) + (notA,B,C). $(0.5 \times 0.4 \times 0.8) + (0.5 \times 0.6 \times 0.2) + (0.5 \times 0.4 \times 0.2) = 0.16 + 0.06 + 0.04 = 0.26$. Correction: $0.16 + 0.06 + 0.04 = 0.26$.
10. 0.21: $P( \text{Rotten}) = \frac{3}{10}$, $P( \text{Not Rotten}) = \frac{7}{10}$. Calculation: $\frac{3}{10} \times \frac{7}{10} = \frac{21}{100} = 0.21$.

1. If Event A and Event B are independent, which of the following must be true?

• AP(A or B) = P(A) + P(B)
• BP(A and B) = P(A) * P(B)
• CP(A) = P(B)
• DP(A and B) = 0

Frequently Asked Questions

How can I tell if two events are independent on the GRE?

Events are independent if the problem explicitly states it or if the outcome of the first event (like rolling a die or flipping a coin) physically cannot affect the second. In "with replacement" scenarios, events are always independent, whereas "without replacement" usually implies dependence.

What is the difference between mutually exclusive and independent events?

Mutually exclusive events cannot happen at the same time (e.g., a coin being both Heads and Tails), whereas independent events can happen together but don't influence each other. In fact, if two events with non-zero probabilities are mutually exclusive, they cannot be independent because knowing one happened tells you the other definitely did not.

Why is the "at least one" rule so common in hard GRE questions?

The "at least one" rule is a favorite for test-makers because it forces students to use the complement rule to save time. Calculating the probability of exactly 1, exactly 2, and exactly 3 successes separately is much more prone to error than simply subtracting the probability of zero successes from 1.

Can I use the multiplication rule for dependent events?

No, the simple multiplication rule $P(A) \times P(B)$ only applies to independent events. For dependent events, you must use conditional probability: $P(A \text{ and } B) = P(A) \times P(B|A)$, where $P(B|A)$ is the probability of B occurring given that A has already occurred.

Does the order of events matter in independent probability?

Order matters when you are calculating the probability of a specific sequence, but if the question asks for a general outcome (like "one red and one blue"), you must account for all possible orders. You can use tools like the AI Flashcard Generator to memorize these distinctions and improve your speed on test day.