The probability of event `A` is `3//4`. The probability of event `B`, given that event `A` occurs is `1//4`. The probability of event `A`, given that event `B` occurs is `2//3`. The probability that neither event occurs is

Options

A. `(1)/(6)`
B. `(27)/(112)`
C. `(5)/(32)`
D. `(1)/(8)`

Answer

C

Explanation

`(c )` `P(A)=(3)/(4)`, `P(B//A)=(1)/(4)`
`P(A//B)=(2)/(3)`
`P(B//A)=(P(BnnA))/(P(A))=(1)/(4)` (given)
`:.P(BnnA)=(3)/(4)*(1)/(4)=(3)/(16)`
Now `P(A//B)=(P(AnnB))/(P(B))=(2)/(3)`(given)
`:.P(B)=(3)/(2)*(3)/(16)=(9)/(32)`
`:. P(AuuB)=(3)/(4)+(9)/(32)-(3)/(16)=(24+9-6)/(32)=(27)/(32)`
`:. P(A^(C )nnB^(C ))=1-P(AuuB)=1-(27)/(32)=(5)/(32)`