Can some one please help with this question? Find the probability that in FOUR tosses of a fair die a 3 appears
a)exactly no (zero)time
b)exactly three times
a) (5/6)^4 = 625/1296 = 0.48225
b) (1/6)^3 *(5/6)*4
= 20/1296 = 0.01543
(The 4 comes from the fact that there are 4 possibilities for which toss is not a 3)