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.

Explanation

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)`