`A` and `B` are `2` events such that `P(A)=(3)/(4)` and `P(B)=(5)/(8)`. If `x` and `y` are the possible minimum and maximum values of `P(AnnB)`, then the value of `a+b` is

Options

A. `0.5`
B. `0.8`
C. `0.9`
D. `1`

Answer

D

Explanation

`(d)` `AnnBsubB`
`P(AnnB) le P(B)`……….`(i)`
`P(AnnB) ,e (5)/(8)`
Now `P(AuuB) =P(A)+P(B)-P(AnnB)`
`P(AuuB) le 1`
`impliesP(A)+P(B)-P(AnnB) le 1`
`impliesP(AnnB) ge (3)/(4)+(5)/(8)-1`
i.e., `P(AnnB) ge (3)/(8)`……`(ii)`
From `(i)` and `(ii)`
`(3)/(8) le P(AnnB) le (5)/(8)`
`:.a+b=(3)/(8)+(5)/(8)=1`