Originally Posted by
tbl9301 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?
I got an answer but I wanted to see if I am setting it up correctly. Its a conditional probability P(2 sixes in 5 rolls | 3 sixes in 8 rolls)
And I think the P(2 sixes in 5 rolls) = (5 choose 2)*(1/6)^2*(5/6)^3 = .16075
And P(3 sixes in 8 rolls) = (8 choose 3)*(1/6)^3*(5/6)^5 = .10419
But then I'm not sure ... is it [(.16075)*(.10419)]/(.10419) so the answer is just .16075?