Given that A and B are two independent events. C is the event in which exactly of A or B occurs.
Let P (A) = x, P (B) = y
Then P(C) = p(A ∩ B) + P(A ∩ B)
P(A)P(B) + P(A)P(B)
[∵ If A and B are independent so are 'A and B' and 'A and B']
P(C) = x(1 - y) + y(1 - x) ........(1)
Now consider, P(A ∪ ∩ B) p(bar A ∩ bar B)
= [P(A) + P(B) - P(A)P(B)][P(bar A)P(bar B)]
= (x + y - xy)(1 - x)(1 - y)
= (x + y) (1 – x) (1 – y) – xy (1 – x) (1 – y) ≤ (x + y) (1 – x) (1 – x) [∵ x, y(0, 1)]
= x (1 – x) (1 – y) + y (1 – x) (1 – y)
= x(1 – y) + y(1 – x) – x2(1 – y) – y2(1 – x) ≤ x(1 – y) + y(1 – x)3
= P (C) [Using eqn (1)]
Thus P(C) ≥ P(A ∪ B) P(A ∩ B) is proved.