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Let A ⊂ B with P(A) > 0 and P(B) < 1. Show that P(B|A) =...

Let A ⊂ B with P(A) > 0 and P(B) < 1. Show that P(B|A) = 1 and P(A|B) = P(A)/P(B).

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Answer #1

We are given here that A B that is A is a subset of B here which means that all the outcomes in A are also there in B ( there could be elements in B not in A though )

Also as we are given here that: P(B) < 1, therefore B does not contain all the elements which means there are some elements in neither B and hence not in A.

  • Given that an outcome of A has happened, that outcome is also in B because A is a subset of B, therefore B also has to happen here. And therefore given A has happened B also happens. P(B | A) = 1. Hence probed.
  • Now given that B has happened, the required probability here is computed from the Bayes theorem as:
    P(A | B) = P(A and B) / P(B)
    P(A and B) = P(A) because all elements in A are in both A and B as A is the subset of B.
    Therefore, P(A | B) = P(A)/ P(B). Hence proved.
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