A and B are two independent events. C is event in which exactly one of

1.	A and B are two independent events. C is event in which exactly one of A or B occurs. Prove that P (C) ≥ P (A ∪ B) P (bar A ∩ B)
Answer» 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.

A and B are two independent events. C is event in which exactly one of A or B occurs. Prove that P (C) ≥ P (A ∪ B) P (bar A ∩ B)

Answer»

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.

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