Let E and F be two independent events. The probability that exactly one of them occurs is `11//25` and the probability of none of them occurring is `2//25.` If P(T) denotes the probability of occurrence of the event T, then
A. `P(E)=4/5,P(F)=3/5`
B. `P(E)=1/5,P(F)=2/5`
C. `P(E)=2/5,P(F)=1/5`
D. `P(E)=3/5,P(F)=4/5`


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Correct Answer - A::D
Let `P(E)=eand P(F)=f`
`P(EuuF)-P(EnnF)=11/25`
`impliese+f-2ef=11/25" "(1)`
`P(barEnnbarF)=2/25`
`implies(1-e)(1-f)=2/25" "(2)`
From (1) and (2),
`ef=12/25and e+f=7/5`
Solving, we get
`e=4/5,f=3/5or e=3/5,f=4/5`