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(B) Let $p$ be the probability that $A$ wins a single game,so $p = 0.6 = \frac{3}{5}$.Let $n = 3$ be the number of games played.We want to find the pr

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$\frac{81}{125}$

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(B) Let $p$ be the probability that $A$ wins a single game,so $p = 0.6 = \frac{3}{5}$.Let $n = 3$ be the number of games played.We want to find the probability that $A$ wins at least two games,which is $P(X \ge 2) = P(X = 2) + P(X = 3)$,where $X$ follows a binomial distribution $B(n, p)$.The probability mass function is $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$.For $k = 2$: $P(X = 2) = \binom{3}{2} (0.6)^2 (0.4)^1 = 3 	imes 0.36 	imes 0.4 = 0.432$.For $k = 3$: $P(X = 3) = \binom{3}{3} (0.6)^3 (0.4)^0 = 1 	imes 0.216 	imes 1 = 0.216$.Summing these probabilities: $P(X \ge 2) = 0.432 + 0.216 = 0.648$.Converting to a fraction: $0.648 = \frac{648}{1000} = \frac{81}{125}$.

$A$ and $B$ are playing a chess game with each other. The probability that $A$ wins the game is $0.6$,the probability that he loses is $0.3$,and the probability that it is a draw is $0.1$. If they play three games,what is the probability that $A$ wins at least two games?

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