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(B) When three $6$-faced dice are thrown,the total number of possible outcomes is $6 	imes 6 	imes 6 = 6^3 = 216$.To find the number of favorable ou

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$\frac{5}{9}$

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(B) When three $6$-faced dice are thrown,the total number of possible outcomes is $6 	imes 6 	imes 6 = 6^3 = 216$.To find the number of favorable outcomes where no two dice show the same number:The first die can show any of the $6$ numbers ($6$ choices).The second die must show a number different from the first die ($5$ choices).The third die must show a number different from both the first and second dice ($4$ choices).Therefore,the number of favorable outcomes is $6 	imes 5 	imes 4 = 120$.The required probability is $\frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{120}{216} = \frac{5}{9}$.

Three $6$-faced dice are thrown together. The probability that no two dice show the same number on them is

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