If `A` and `B` are two independent events, prove that `P(AuuB).P(A'nnB')<=P(C)`, where `C` is an event defined that exactly one of `A` and `B` occurs.
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Given that A and B are two independent events. C is the event in which exactly one of A and B occurs. Let P (A) = x, P(B)=y. Then, `P(C)=P(AnnbarB)+P(barAnnB)` `=P(A)P(barB)+P(barA)P(B)" "[because"if A and B are independent so are A"and barB and barA adn B]` `=x(1-y)+y(1-x)" "(1)` Now consider, `P(AuuB)P(barAnnbarB)` `=[P(A)+P(B)-P(A)P(B)][P(barA)P(barB)]` `=(x+y-xy)(1-x)(1-y)` `=(x+y(1-x)(1-y)-xy(1-x)(1-y)` `lex(1-x)(1-y)+y(1-x)(1-y)" "[becausex, y in(0,1)]` `lex(1-y)+y(1-x)-x^(2)(1-y)-y^(2)(1-x)` `leP(C)" "["Using Eq."(1)]` Thus, `P(C)geP(AuuB)P(barAnnbarB)`
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