Asked by Victoria
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)exactly no (zero)time
b)exactly three times
Answers
Answered by
drwls
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)
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)
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