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Math Help - conditional probability

  1. #1
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    conditional probability

    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?
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    Quote Originally Posted by tbl9301 View Post
    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?
    The answer is:

    Pr(3 sixes in 8 rolls AND 2 sixes in first 5 rolls)/Pr(3 sixes in 8 rolls)

    = Pr(2 sixes in first 5 rolls AND 1 six in last 3 rolls)/Pr(3 sixes in 8 rolls).
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