For three events `A ,B` and `C ,P` (Exactly one of `A` or `B` occurs) `=P` (Exactly one of `B` or `C` occurs) `=P` (Exactly one of `C` or `A` occurs) `=1/4` and `P` (All the three events occur simultaneously) `=1/6dot` Then the probability that at least one of the events occurs, is : `7/(64)` (2) `3/(16)` (3) `7/(32)` (4) `7/(16)`

Options

A. `(3p+2p^(2))/(2)`
B. `(p+3 p^(2))/(2)`
C. `(3 p+ p^(2))/(2)`
D. `(3 p +2 p^(2))/(4)`

Answer

A

Explanation

We have,
`P(A)+P(B)-2P(A cap B)=P`
`P(B)+P(C )-2P(B cap C)=p`
`P(C )+P(A)-2P(C cap A)=p`
and, `P(A cap B cap C)=p^(2)`
Adding (i),(ii) and (iii), we get
`2[P(A)+P(B)+P(C )+P(A cap B)-P(B cap C)-P(A cap C)]=3p`
`implies P(A)+P(B)+P(C )-P(A cap B)-P(B cap C)`
`P(A cap C)=3p//2`
`therefore` Required probability
`=P(A cup B cup C)`
`=P(A)+P(B)+P(C )-P(A cap B)-P(B cap C)-P(A cap C)+P(A cap B cap C)`
`=(3p)/(2)+p^(2)=(3p+2p^(2))/(2)`