InterviewSolution
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InterviewSolution
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If `E_(1)` and `E_(2)` are two events such that `P(E_(1))=1//4`, `P(E_(2)//E_(1))=1//2` and `P(E_(1)//E_(2))=1//4`, thenA. then `E_(1)` and `E_(2)` are independentB. `E_(1)` and `E_(2)` are exhaustiveC. `E_(2)` is twice as likely to occur as `E_(1)`D. Probabilites of the events `E_(1) nn E_(2)`, `E_(1)` and `E_(2)` are in `G.P.` |
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Answer» Correct Answer - A::C::D `(a,c,d)` `P(E_(2)//E_(1))=(P(E_(1)nnE_(2)))/(P(E_(1)))` `(1)/(2)=(P(E_(1)nnE_(2)))/(1//4)` `impliesP(E_(1)nnE_(2))=(1)/(8)=P(E_(2))*P(E_(1)//E_(2))` `=P(E_(2))*(1)/(4)` `impliesP(E_(2))=(1)/(2)` Since `P(E_(1) nnE_(2))=(1)/(8)=P(E_(1))*P(E_(1))*P(E_(2))`, events are independent Also `P(E_(1)uuE_(2))=(1)/(2)+(1)/(4)-(1)/(8)=(5)/(8)` `impliesE_(1)` and `E_(2)` are not exhaustive. |
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