If three dice are thrown together,then the pr | vedclass

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(D) The total number of outcomes when three dice are thrown is $6^3 = 216$. Let $X$ be the number of dice showing $5$. The probability of getting a $5

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$\frac{91}{216}$

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(D) The total number of outcomes when three dice are thrown is $6^3 = 216$. Let $X$ be the number of dice showing $5$. The probability of getting a $5$ on a single die is $p = \frac{1}{6}$,and the probability of not getting a $5$ is $q = 1 - \frac{1}{6} = \frac{5}{6}$. We want the probability of getting $5$ on at least one die,which is $P(X \ge 1) = 1 - P(X = 0)$. The probability of getting no $5$ on any of the three dice is $P(X = 0) = \left(\frac{5}{6}\right)^3 = \frac{125}{216}$. Therefore,the required probability is $1 - \frac{125}{216} = \frac{216 - 125}{216} = \frac{91}{216}$.

If three dice are thrown together,then the probability of getting $5$ on at least one of them is

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