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(B) Let event $A$ be the event that the numbers appearing on both the dice are different.Then,the complementary event $\overline{A}$ is the event that

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$\frac{5}{6}$

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(B) Let event $A$ be the event that the numbers appearing on both the dice are different.Then,the complementary event $\overline{A}$ is the event that the numbers appearing on both the dice are the same.The total number of possible outcomes when two dice are rolled is $6 	imes 6 = 36$.The outcomes favorable to event $\overline{A}$ are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$.Thus,the number of favorable outcomes for $\overline{A}$ is $m = 6$.The probability of event $\overline{A}$ is $P(\overline{A}) = \frac{m}{n} = \frac{6}{36} = \frac{1}{6}$.Since $P(A) + P(\overline{A}) = 1$,the probability that the numbers are different is $P(A) = 1 - P(\overline{A}) = 1 - \frac{1}{6} = \frac{5}{6}$.

Two balanced dice are rolled simultaneously. Then,the probability that the numbers appearing on both the dice are different is ........

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