Question

P and Q play chess frequently. P has won 80\% of the matches, drawn 15\%, and lost 5\%. If they play 3 more matches, what is the probability that P wins exactly 2 of these 3 matches?

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

Hint:

Whenever the problem asks for "exactly \(k\) successes in \(n\) trials," use the binomial probability formula. Remember to include all "not success" cases (loss + draw here) together.

Answer 1

The Correct Option is A

Step 1: Probabilities.
- Probability of winning (\(p\)) = 0.8. - Probability of not winning (\(q\)) = 1 - 0.8 = 0.2 (this includes both draws and losses). Step 2: Use Binomial distribution.
We want exactly 2 wins out of 3 trials. Formula: \[ P(k) = \binom{n}{k} p^k q^{n-k}. \] Step 3: Apply values.
\[ P(2) = \binom{3}{2} (0.8)^2 (0.2)^1. \] \[ = 3 \times 0.64 \times 0.2. \] \[ = 3 \times 0.128 = 0.384. \] Step 4: Convert to fraction.
\[ 0.384 = \frac{48}{125}. \] Final Answer: \[ \boxed{\tfrac{48}{125}} \]