Calculating P's Probability of Exactly 2 Wins

This problem involves calculating the probability of a specific number of successful outcomes (P winning) in a fixed number of independent trials (3 matches). This is a classic example of a binomial probability scenario.

Identifying Binomial Parameters

• Number of trials (matches played), n = 3
• Number of successful outcomes (P wins), k = 2
• Probability of success (P winning a match), p = 80% = 0.8
• Probability of failure (P not winning a match), q = 1 - p = 1 - 0.8 = 0.2

Applying the Binomial Probability Formula

The formula for binomial probability is:

$ P(X=k) = \binom{n}{k} p^k q^{n-k} $

Where:

• $\binom{n}{k}$ represents the number of ways to choose k successes from n trials.

Step-by-Step Calculation

1. Calculate the number of combinations: $\binom{n}{k}$ $ \binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3!}{2!1!} = \frac{3 \times 2 \times 1}{(2 \times 1)(1)} = 3 $ There are 3 possible ways for P to win exactly 2 out of 3 matches.
2. Calculate the probability of P winning 2 matches: $p^k$ $ p^2 = (0.8)^2 = 0.64 $
3. Calculate the probability of P not winning 1 match: $q^{n-k}$ $ q^{3-2} = q^1 = (0.2)^1 = 0.2 $
4. Multiply the results together: $ P(X=2) = \binom{3}{2} \times p^2 \times q^1 = 3 \times 0.64 \times 0.2 $ $ P(X=2) = 3 \times 0.128 = 0.384 $
5. Convert the decimal probability to a fraction: $ 0.384 = \frac{384}{1000} $ Simplify the fraction by dividing the numerator and denominator by their greatest common divisor (which is 8): $ \frac{384 \div 8}{1000 \div 8} = \frac{48}{125} $

Therefore, the probability of P winning exactly 2 of the 3 matches is $\frac{48}{125}$.