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(A) For a single die,the total number of outcomes is $6$. The probability of getting the number $1$ on the face is $P(1) = \frac{1}{6}$.Therefore,the

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$\frac{625}{1296}$

Vedclass · Answer

(A) For a single die,the total number of outcomes is $6$. The probability of getting the number $1$ on the face is $P(1) = \frac{1}{6}$.Therefore,the probability of not getting the number $1$ on the face of a single die is $P(	ext{not } 1) = 1 - \frac{1}{6} = \frac{5}{6}$.Since $4$ dice are thrown simultaneously,the events are independent.Thus,the probability that none of the $4$ dice shows the number $1$ is $\left(\frac{5}{6}ight) 	imes \left(\frac{5}{6}ight) 	imes \left(\frac{5}{6}ight) 	imes \left(\frac{5}{6}ight) = \left(\frac{5}{6}ight)^4 = \frac{625}{1296}$.

If four dice are thrown simultaneously,then the probability that none of the dice shows the number $1$ on its face is:

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