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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) `=1/6dot` Then the probability that at least one of the events occurs, is :

A

`(7)/(32)`

B

`(7)/(16)`

C

`(7)/(64)`

D

`(3)/(16)`

Text Solution

Verified by Experts

The correct Answer is:
B
We have, P (exactly one of A or B occurs)
`=P(uuB)-P(AnnB)`
`=P(A)+P(B)-2P(AnnB)`
According to the question,
`P(A)+P(B)-2P(AnnB)=(1)/(4)" "`….(i)
`P(B)+P(C )-2P(BnnC)=(1)/(4)" "`.......(ii)
and `P(C )+P(A)-2P(CnnA)=(1)/(4)" "`........(iii)
On adding Eqs. (i), (ii) and (iii), we get
`2[P(A)+P(B)+P(C )-P(AnnB)-P(BnnC)-P(CnnA)]=(3)/(4)`
`rArr P(A)+P(B)+P(C )-P(AnnB)-P(BnnC)-P(CnnA)=(3)/(8)`
`:. P` (atleast one event occurs)
`=P(AuuBuuC)`
`=P(A)+P(B)+P(C )-P(AnnB)-P(BnnC)-P(CnnA)+P(AnnBnnC)`
`=(3)/(8)+(1)/(16)=(7)/(16) " "[:' P(AnnBnnC)=(1)/(16)]`
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