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A die is rolled 8 times. given that there were 3 sixes in the 8 rolls, what is the probability that there were 2 sixes in the first five rolls

User Cana
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1 Answer

3 votes
Answer: 15/28
Suppose the following:
event A: the event that there were 3 sixes in the 8 roles,
event B: the event that there were 2 sixes in the first 5 roles.
Then the conditional probability to find is P(B|A)=P(A∩B)/P(A).
P(A)=C(3,8)*(1/6)^3*(5/6)^5,
P(A∩B)=C(2,5)*(1/6)^2*(5/6)^3*C(1,3)*(1/6)*(1/6)^2
∴P(B|A)=15/28
User Jeremy Pope
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