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(B) The total number of outcomes when three $6$-faced dice are thrown is $6 	imes 6 	imes 6 = 216$.To find the number of outcomes where exactly two

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$\frac{5}{12}$

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(B) The total number of outcomes when three $6$-faced dice are thrown is $6 	imes 6 	imes 6 = 216$.To find the number of outcomes where exactly two dice show the same number,we follow these steps:$1$. Choose which two dice show the same number: There are $^3C_2 = 3$ ways to choose the two dice.$2$. Choose the number that appears on these two dice: There are $6$ choices $(1, 2, 3, 4, 5, 6)$.$3$. Choose the number for the third die: It must be different from the number chosen for the first two dice,so there are $5$ choices.Total favourable outcomes $= 3 	imes 6 	imes 5 = 90$.Required probability $= \frac{90}{216} = \frac{5}{12}$.

Three $6$-faced dice are thrown together. The probability that exactly two dice show the same number on them is:

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