Six dice are thrown simultaneously. The probability that all of them show the same face, is
Answer & explanation
Correct answer: option 2
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 = $\frac{6!}{6^6}$