Suppose A and B are two equally strong table tennis players . The probability that A beats B in exactly 3 games out of 4 is

Options

A. `1//2`
B. `1//4`
C. `1//8`
D. `3//4`

Answer

To solve the problem of finding the probability that player A beats player B in exactly 3 games out of 4, we can use the binomial probability formula. Here’s a step-by-step solution:

### Step 1: Define the Problem
We know that A and B are equally strong players, which means the probability of A winning a single game (p) is 0.5, and the probability of B winning a single game (q) is also 0.5. We need to find the probability that A wins exactly 3 out of 4 games.

### Step 2: Identify the Binomial Distribution
Let X be the random variable representing the number of games A wins. Since A can win or lose each game, X follows a binomial distribution with parameters n = 4 (the total number of games) and p = 0.5 (the probability of A winning a single game).

### Step 3: Use the Binomial Probability Formula
The formula for the probability of getting exactly k successes (A wins) in n trials (games) is given by:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
In our case, we want to find \( P(X = 3) \) where \( n = 4 \) and \( k = 3 \).

### Step 4: Calculate the Binomial Coefficient
First, we calculate the binomial coefficient \( \binom{4}{3} \):
\[ \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4 \times 3!}{3! \times 1!} = 4 \]

### Step 5: Calculate the Probability
Now we can substitute into the binomial formula:
\[ P(X = 3) = \binom{4}{3} (0.5)^3 (0.5)^{4-3} \]
\[ P(X = 3) = 4 \times (0.5)^3 \times (0.5)^1 \]
\[ P(X = 3) = 4 \times (0.5)^4 \]
\[ P(X = 3) = 4 \times \frac{1}{16} = \frac{4}{16} = \frac{1}{4} \]

### Final Result
The probability that A beats B in exactly 3 games out of 4 is \( \frac{1}{4} \).

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