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(A) Let $X$ be the number of dice showing a $3$ or a $5$ in a single throw of $4$ dice. The probability of getting a $3$ or a $5$ on a single die is $

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$11$

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(A) Let $X$ be the number of dice showing a $3$ or a $5$ in a single throw of $4$ dice. The probability of getting a $3$ or a $5$ on a single die is $p = \frac{2}{6} = \frac{1}{3}$.Thus,the probability of not getting a $3$ or a $5$ is $q = 1 - \frac{1}{3} = \frac{2}{3}$.Since the dice are independent,$X$ follows a binomial distribution $B(n=4, p=1/3)$.The probability that at least two dice show a $3$ or a $5$ is $P(X \geq 2) = 1 - P(X=0) - P(X=1)$.$P(X=0) = { }^4 C_0 (\frac{1}{3})^0 (\frac{2}{3})^4 = 1 	imes 1 	imes \frac{16}{81} = \frac{16}{81}$.$P(X=1) = { }^4 C_1 (\frac{1}{3})^1 (\frac{2}{3})^3 = 4 	imes \frac{1}{3} 	imes \frac{8}{27} = \frac{32}{81}$.$P(X \geq 2) = 1 - (\frac{16}{81} + \frac{32}{81}) = 1 - \frac{48}{81} = \frac{33}{81} = \frac{11}{27}$.The experiment is performed $27$ times. The expected number of times is $E = n 	imes P = 27 	imes \frac{11}{27} = 11$.

Four fair dice are thrown independently $27$ times. Then the expected number of times,at least two dice show up a $3$ or a $5$ is

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