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(B) The total number of outcomes when throwing three dice is $6^3 = 216$.Let $E$ be the event that at least one die shows up $1$.It is easier to calcu

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$\frac{91}{216}$

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(B) The total number of outcomes when throwing three dice is $6^3 = 216$.Let $E$ be the event that at least one die shows up $1$.It is easier to calculate the probability of the complement event $E'$,which is the event that no die shows up $1$.For a single die,the probability of not getting a $1$ is $\frac{5}{6}$.Since the three dice are independent,the probability that none of the three dice show $1$ is $\left( \frac{5}{6} \right)^3 = \frac{125}{216}$.Therefore,the probability of the event $E$ is $P(E) = 1 - P(E') = 1 - \frac{125}{216} = \frac{216 - 125}{216} = \frac{91}{216}$.

In a throw of three dice,the probability that at least one die shows up $1$,is

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