Conditional Probability?

How did the value for probability of A given B come about at bottom here? I have no idea how that equation was generated.

Thanks

Update:

This problem is solved in the chapter before distributions of discrete random variable. The section deals with Bayes' Rule.

Is there a manner to generate the same equation without a known distribution at hand?

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  • 2 years ago
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    A is the event that 1 of the 3 fuses was defective

    B is the event that the fuses came from line 1

    If the fuse came from line 1 (given as B), then we only have to consider those probabilities. We use the binomial probability:

    P(X=k) = C(n,k) * p^k * q^(n-k)

    n : number of total fuses tested (3)

    k : number of defective fuses (1)

    p : probability it is defective (0.05)

    q : probability it is not defective (0.95)

    P(X=1) = C(3,1) * 0.05 * 0.95²

    = 3 * 0.05 * 0.95²

    = 0.135375

    Basically the 3 comes from the number of ways to pick which of the 3 is the defective fuse.

    The probability of one being defective is 0.05

    The probability of two more being non-defective is 0.95²

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