2 Find the probability that in five tosses of a fair die, a three will appear at least two times
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But there is a 1/216 chance of getting three 2s in a row on a fair die. Therefore 55+36=91 possibilities, out of 6x6x6=216 possibilities. Therefore the answer is 91/216. Hope this helps!
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Answer:
Step-by-step explanation:
p=P(getting a 3 on one throw)
=16 and
q=P(not getting a 3 on one throw)
=1−16=56.
The two probabilities p and q are the possibilities of a Bernoulli trial. So, to answer the problem, the binomial probability distribution with n = 5, k = 2, p = 1/6, and q = 5/6 is applied. Formula is
(nk)pkqn−k
⟹
P(getting 3 twice in 5 Bernoulli trials)
=P(getting a 3 twice in 5 throws)
=(5/2) (1/6)^2 (5/6)^3
=10 × 1/36 × 125/216
=5×1/18×125/216
=5/18×125/108
=625/3888=0.160751029.
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