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

The probabilities for the outcomes of a match involving P are defined as follows, based on the provided data:
\n \( P({Win}) = 0.80 \)
\n \( P({Draw}) = 0.15 \)
\n \( P({Loss}) = 0.05 \)
\n\nWe are tasked with determining the probability of P achieving exactly 2 wins in a series of 3 matches. Given that each match's outcome is independent, this scenario aligns with a binomial probability calculation, specifically the probability of obtaining 2 successes in 3 trials.
\n\nThe binomial probability formula is stated as:
\n\[\nP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\n\]\nIn this context:
\n \(n = 3\) represents the total number of matches played.
\n \(k = 2\) denotes the desired number of wins.
\n \(p = 0.80\) is the probability of P winning a single match.
\n \(1 - p = 0.20\) is the probability of P not winning a single match.
\n\nThe calculation for the probability of exactly 2 wins is:
\n\[\nP(X = 2) = \binom{3}{2} (0.80)^2 (0.20)^1 = 3 \times 0.64 \times 0.20 = \frac{48}{125}.\n\]\n\nConsequently, the definitive answer is \( \frac{48}{125} \).