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

Options

A. a) `7/(64)`
B. b) `3/(16)`
C. c) `7/(32)`
D. d) `7/(16)`

Explanation

Given, `P(A \or \ B)=P(A)+P(B)-2P(AnnB)=1/4`` \ \ \...(1)`


`P(A \ or \ C)=P(A)+P(C)-2P(AnnC)=1/4`` \ \ \...(2)`


`P(B \ or \ C)=P(B)+P(C)-2P(BnnC)=1/4`` \ \ \...(3)`


Adding all equation `(1), (2), (3)`


`3/4=P(A)+P(B)-2P(AnnB)+P(A)+P(C)-2P(AnnC)+P(B)+P(C)-2P(BnnC)`


`3/4=2P(A)+2P(B)+2P(C)-2P(AnnB)-2P(AnnC)-2P(BnnC)`


`P(A)+P(B)+P(C)-P(AnnB)-P(AnnC)-P(BnnC)=3/8`


`P`(Atleast one event)`=3/8+1/16=7/16`


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