Understanding the Chess Match Probability Problem

This question asks for the probability of a specific outcome in a series of independent events. We are given the probabilities of P winning, drawing, and losing against Q in a single chess match. We need to find the probability that P wins exactly 2 out of the next 3 matches played.

Defining Probabilities and Outcomes

Let's define the probabilities for a single match:

• Probability of P winning (let's call this 'p'): $p = 80\% = 0.80$
• Probability of a draw: $15\% = 0.15$
• Probability of P losing: $5\% = 0.05$

For this problem, we are interested in whether P wins or does not win. The event "P does not win" includes draws and losses.

Probability of P not winning (let's call this 'q'): $q = P(\text{Draw}) + P(\text{P loses})$ $q = 0.15 + 0.05 = 0.20$ So, $q = 0.20$.

We can check that $p + q = 0.80 + 0.20 = 1.00$, which accounts for all possible outcomes for P (win or not win).

Applying the Binomial Probability Formula

This scenario fits the binomial distribution criteria:

• There are a fixed number of trials ($n=3$ matches).
• Each trial is independent (the outcome of one match doesn't affect others).
• Each trial has two possible outcomes: P wins (success) or P does not win (failure).
• The probability of success ($p=0.80$) is the same for each trial.

The binomial probability formula is:

Where:

• $n$ = total number of trials
• $k$ = number of successful outcomes
• $p$ = probability of success on a single trial
• $q$ = probability of failure on a single trial
• $\binom{n}{k}$ = the binomial coefficient, calculated as $\frac{n!}{k!(n-k)!}$

Calculating the Probability

We want to find the probability of P winning exactly 2 matches ($k=2$) out of 3 played ($n=3$).

1. Calculate the binomial coefficient $\binom{n}{k}$: $\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3!}{2!1!} = \frac{3 \times 2 \times 1}{(2 \times 1)(1)} = 3$ This means there are 3 possible ways P can win exactly 2 out of 3 matches (e.g., WWL, WLW, LWW).
2. Calculate $p^k$: $p^2 = (0.80)^2 = 0.64$
3. Calculate $q^{n-k}$: $q^{3-2} = q^1 = (0.20)^1 = 0.20$
4. Combine the parts using the formula: $P(X=2) = \binom{3}{2} p^2 q^1$ $P(X=2) = 3 \times (0.80)^2 \times (0.20)^1$ $P(X=2) = 3 \times 0.64 \times 0.20$ $P(X=2) = 3 \times 0.128$ $P(X=2) = 0.384$
5. Convert the result to a fraction: $0.384 = \frac{384}{1000}$ Simplify the fraction: $\frac{384}{1000} \div \frac{8}{8} = \frac{48}{125}$

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