Statistics Question?

For Events J and K P(J)=0.4 P(J|K)=0.64

P(K|J)=0.2

a) Find P(J U K)

b) Find P(J'nK')

1 Answer

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  • 1 month ago

    PART A:

    Conditional probability says:

    P(A | B) = P(A ∩ B) / P(B)

    So in your case:

    P(K | J) = P(K ∩ J) / P(J)

    Plug in your known values and we have:

    0.2 = P(K ∩ J) / 0.4

    0.08 = P(K ∩ J)

    We can turn that around and we have:

    P(J ∩ K) = 0.08

    But that's not what we wanted. Just use another rule of probability:

    P(A ⋃ B) = P(A) + P(B) - P(A ∩ B)

    In your case:P(J ⋃ K) = P(J) + P(K) - P(J ∩ K)

    = 0.4 + P(K) - 0.08

    = 0.32 + P(K)

    Bayes' Law says:

    P(A | B) = P(B | A) * P(A) / P(B)

    So now we should have what we need:

    P(J | K) = P(K | J) * P(J) / P(K)

    0.64 = 0.2 * 0.4 / P(K)

    0.64 = 0.08 / P(K)

    P(K) = 0.08 / 0.64

    P(K) = 8/64 = 1/8 = 0.125

    Finally put that together from your result before:

    P(J ⋃ K) = 0.32 + P(K) 

    P(J ⋃ K) = 0.32 + 0.125

    P(J ⋃ K) = 0.445

    PART B:

    De Morgan's Laws say:

    J ⋃ K = (J' ∩ K')'

    So J' ∩ K' = (J ⋃ K)'

    So just take the complement of your given probability:

    P(J' ∩ K') = 1 - P(J ⋃ K)

    P(J' ∩ K') = 1 - 0.445

    P(J' ∩ K') = 0.555

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