Let A and B be events with 0 Solution0 PA intersection B P

Solution

0 < P(A intersection B) < P(A) < P(B) < P(A U B) < 1

This shows that the P(A) is the probability value that is next highest to P(A n B).
So, we\'d be happiest if A occurred ---> ANSWER

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

And this is given to be <= P(A)

P(A and B) / P(B) <= P(A)

P(A)*P(B) >= P(A and B)

Now, lets find P(A | Bc)

P(A and Bc) / P(Bc)

P(A and Bc) / (1 - P(B))

(1 - P(A and B)) / (1 - P(B))

Divide all over by P(B) :

(1/P(B) - P(A and B)/P(B)) / (1/P(B) - 1)

Let 1/P(B) be a constant value of c

(c - P(A and B)/P(B)) / (c - 1)

(c - (some value less than or equal to P(A)) / (c - 1)

This has to be greater than or equal to P(A)

Hence proved!