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(C) Let $n = 5$ be the number of dice thrown. For a specific face (e.g.,$1$), the probability of it appearing on a single die is $p = \frac{1}{6}$ and

Vedclass · Accepted Answer

$\frac{276}{6^4}$

Vedclass · Answer

(C) Let $n = 5$ be the number of dice thrown. For a specific face (e.g.,$1$), the probability of it appearing on a single die is $p = \frac{1}{6}$ and the probability of it not appearing is $q = \frac{5}{6}$. Since there are $6$ possible faces, the probability that at least $3$ dice show the same face is $6 	imes P(X \geq 3)$, where $X$ follows a binomial distribution $B(5, \frac{1}{6})$. $P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5)$. $P(X = 3) = \binom{5}{3} (\frac{1}{6})^3 (\frac{5}{6})^2 = 10 	imes \frac{25}{6^5} = \frac{250}{6^5}$. $P(X = 4) = \binom{5}{4} (\frac{1}{6})^4 (\frac{5}{6})^1 = 5 	imes \frac{5}{6^5} = \frac{25}{6^5}$. $P(X = 5) = \binom{5}{5} (\frac{1}{6})^5 = 1 	imes \frac{1}{6^5} = \frac{1}{6^5}$. Total probability $= 6 	imes (\frac{250 + 25 + 1}{6^5}) = 6 	imes \frac{276}{6^5} = \frac{276}{6^4}$.

If five dice are thrown simultaneously, then the probability that at least three of them show the same numbered face is

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