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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 success (getting $3$ or $5$) for one die is

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(A) Let $X$ be the number of dice showing a $3$ or a $5$ in a single throw of $4$ dice. The probability of success (getting $3$ or $5$) for one die is $p = \frac{2}{6} = \frac{1}{3}$.The probability of failure is $q = 1 - p = \frac{2}{3}$.Since there are $n = 4$ dice,the number of successes $X$ follows a binomial distribution $B(4, \frac{1}{3})$.We want the probability that at least two dice show a $3$ or a $5$,which is $P(X \ge 2) = 1 - P(X = 0) - P(X = 1)$.$P(X = 0) = \binom{4}{0} (\frac{1}{3})^0 (\frac{2}{3})^4 = 1 	imes 1 	imes \frac{16}{81} = \frac{16}{81}$.$P(X = 1) = \binom{4}{1} (\frac{1}{3})^1 (\frac{2}{3})^3 = 4 	imes \frac{1}{3} 	imes \frac{8}{27} = \frac{32}{81}$.$P(X \ge 2) = 1 - (\frac{16}{81} + \frac{32}{81}) = 1 - \frac{48}{81} = \frac{81 - 48}{81} = \frac{33}{81}$.The experiment is performed $N = 27$ times. The expected number of times is $E = N 	imes P(X \ge 2) = 27 	imes \frac{33}{81} = \frac{33}{3} = 11$.

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

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