Let `E and F` be two independent events. The probability that exactly one of them occurs is `11/25` and the probability if 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`

Options

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 )=(6)/(5),P(F)=(1)/(5)`

Answer

A

Explanation

We have,
`P(E )+P(F)-2P(E cap F)=(11)/(25) " and" P(overline(E ) cap overline(F))=(2)/(25)`
`implies P(E )+P(F)-2P(E )P(F)=(11)/(25) " and "P(overline(E ))P(overline(F))=(2)/(25)`
`implies x+y-2xy=(11)/(25) " and "1-x+xy=(2)/(25)`, where P(E ) =x and P(F)=y
`implies x+y+2-2x-2y=(11)/(25)+2xx(2)/(25)` [ On eliminating xy]
`implies x+y=(7)/(5) implies y=(7)/(5) -x`
Substituting `y=(7)/(5)-x " in " 1-x-y+xy=(2)/(25)`, we get
`1-(7)/(5)+x((7)/(5)-x)=(2)/(25)`
`implies 25x^(2)-35x+12=0 implies x=(3)/(5), (4)/(5)`
When `x=(3)/(5), y=(4)/(5) " and " y=(3)/(5) " for " x=(4)/(5)`
Hence, `P(E )=(3)/(5), P(F)=(4)/(5) " or " P(E )=(4)/(5), P(F)=(3)/(5)`