For any events `A` and `B`. Given `P(AuuB)=0.6`, `P(A)=P(B)`, `P(B//A)=0.8`. Then the value of `P[AnnbarB)uu(barAnnB)]` is

Options

A. `1//3`
B. `1//2`
C. `1//4`
D. `1//5`

Answer

D

Explanation

`(d)` Let `P(A)=P(B)=p` (say)
`P(B//A)=(P(BnnA))/(P(A))`
`:.(0.8)p=P(BnnA)`………..`(i)`
`P(AuuB)=p+p-P(AnnB)`
`0.6=2p-p(0.8)=(1.2)p`
`:.p=(0.6)/(1.2)=(1)/(2)`
`:.P(A)=P(B)=(1)/(2)`
Now, `P[AnnbarB) uu (barAnnB)]`
`=P(A)+P(B)-2P(AnnB)`
`=1-2(0.8)(1)/(2)=0.2=(1)/(5)`