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Math Help - Binomial Distribution Twice

  1. #1
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    Binomial Distribution Twice

    Problem: 100 Cars are going to be randomly assigned to inspection team A or inspection team B (but not both). They are assigned to team A with probability p=.6 and team B with probability q=.4
    Team A randomly passes cars with probability p_a=.35 and Team B randomly passes cars with probability p_b=.6
    How many cars do you expect to pass inspection?


    I'm not sure how easy this is. Here's what I did:

    I would expect 100(.6)=60 cars to go to Team A and 40 cars to go to team B.

    Then of the 60 Team A inspects, 60(.35)=21 should pass
    Of the 40 Team B inspects, 40(.6)=24 should pass

    So I would expect 21+24=45 to pass?

    Is this alright?
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  2. #2
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    Quote Originally Posted by artvandalay11 View Post
    Problem: 100 Cars are going to be randomly assigned to inspection team A or inspection team B (but not both). They are assigned to team A with probability p=.6 and team B with probability q=.4
    Team A randomly passes cars with probability p_a=.35 and Team B randomly passes cars with probability p_b=.6
    How many cars do you expect to pass inspection?


    I'm not sure how easy this is. Here's what I did:

    I would expect 100(.6)=60 cars to go to Team A and 40 cars to go to team B.

    Then of the 60 Team A inspects, 60(.35)=21 should pass
    Of the 40 Team B inspects, 40(.6)=24 should pass

    So I would expect 21+24=45 to pass?

    Is this alright?
    Yes.
    Another way to look at it, a bit more formally,

    C = Car passes
    A = Car assigned to A
    B = Car assigned to A

    P(C) = P(C and A) + P(C and B) ---- Event A and B are exhaustive and mutually exclusive
    = P(C|A).P(A) + P(C|B).P(B)

    expected num is just 100*P(C)
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