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?
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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.
The probabilities for the outcomes of a match involving P are defined as follows, based on the provided data: \( P({Win}) = 0.80 \) \( P({Draw}) = 0.15 \) \( P({Loss}) = 0.05 \) We 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. The binomial probability formula is stated as: \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]In this context: \(n = 3\) represents the total number of matches played. \(k = 2\) denotes the desired number of wins. \(p = 0.80\) is the probability of P winning a single match. \(1 - p = 0.20\) is the probability of P not winning a single match. The calculation for the probability of exactly 2 wins is: \[P(X = 2) = \binom{3}{2} (0.80)^2 (0.20)^1 = 3 \times 0.64 \times 0.20 = \frac{48}{125}.\]Consequently, the definitive answer is \( \frac{48}{125} \).