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Six dice are thrown simultaneously. The probability that all of them show the same face, is |
$\frac{1}{6^5}$ $\frac{6!}{6^6}$ $\frac{1}{6!}$ $\frac{5!}{6^6}$ |
$\frac{6!}{6^6}$ |
The total number of ways in which all dice show different faces is same as the number of arrangements of 6 numbers 1, 2, 3, ,4 , 5, 6 by taking all at a time. So, favourable number of elementary events = 6! Hence, required probability = $\frac{6!}{6^6}$ |
