Question

P and Q play chess frequently against each other. Of these matches, P has won 80% of the matches, drawn 15% of the matches, and lost 5% of the matches.
If they play 3 more matches, what is the probability of P winning exactly 2 of these 3 matches?

• A. \( \frac{48}{125} \)
• B. \( \frac{16}{125} \)
• C. \( \frac{16}{25} \)
• D. \( \frac{25}{48} \)

Hint:

When solving binomial probability problems, remember the formula \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \) and carefully calculate the combinations, probabilities, and powers.

Answer 1

The Correct Option is A

Let’s define the possible outcomes of a match. From the given data:
The probability of P winning a match is \( P({Win}) = 0.80 \).
The probability of P drawing a match is \( P({Draw}) = 0.15 \).
The probability of P losing a match is \( P({Loss}) = 0.05 \).
We are asked to find the probability of P winning exactly 2 out of the 3 matches. Since the outcome of each match is independent, this is a binomial probability problem, where we need to calculate the probability of 2 wins out of 3 trials. The binomial probability formula is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where:
\(n = 3\) (number of matches),
\(k = 2\) (number of wins),
\(p = 0.80\) (probability of winning),
\(1 - p = 0.20\) (probability of not winning).
The probability of exactly 2 wins is: \[ P(X = 2) = \binom{3}{2} (0.80)^2 (0.20)^1 = 3 \times 0.64 \times 0.20 = \frac{48}{125}. \] Thus, the correct answer is \( \frac{48}{125} \).