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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) `=1/16dot` Then the probability that at least one of the events occurs, is :

A

`(3)/(16)`

B

`(7)/(32)`

C

`(7)/(16)`

D

`(7)/(64)`

Text Solution

Verified by Experts

The correct Answer is:
C
P(Exactly one of A or B occurs)
`=P(A) + P(B) - 2P(A nn B) = (1)/(4)`
P(Exactly one of B or C occurs)
`=P(B) + P(C) - 2P(B nn C) = (1)/(4)`
P(Exactly one of C or A occurs)
`=P(C) + P(A) - 2P(C nn A) = (1)/(4)`
Adding all, we get
`P(A) + P(B) + P(A) - P(Ann B)-P(B nn C) - P(A nn C)= (3)/(8)`
`therefore SigmaP(A) - SigmaP(A nn B) = (3)/(8)`
Now, `P(A uu B uu C)`
`=SigmaP(A) - SigmaP(A nn B) + P(A nn B nn C)`
`=(3)/(8) + (1)/(16) = (7)/(16)`
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