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two persons A and B appear in an interview for two vacancies. if the probability of their selection is ¼ and 1/6 respectively. then find the probability that none of them is selected?

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4 years ago

```							the question asked is independent eventprobability of selection of a is ¼ and probability of b is 1/6 probability of their union is probability of a+probability of b-probability of a intersection bit is from set theoryprobability of a intersection b is product of probability of a and probability of bp(a union b)=(1/4)+(1/6)-(1/4)*(1/6)=9/24probability of their not being selected is (1 – p(a union b))=1 -(9/24)=15/24
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4 years ago
```							@manoj dixitLet P=the event that A is selected andQ = the event that B is selected.Given that P(P) = 1/4 and P(Q) = 1/6We know that P' is the event that A does not get selected and Q' is the event that B does not get selectedProbability that none of them are selected =P(P'∩Q')    (∵ Reference : Algebra of Events) =P(P').P(Q') (∵ Here A and B are Independent Events and refer theorem on independent events) =[ 1 – P(P)][ 1 – P(Q)]=1−(1/4) x 1−(1/6) =(3/4)×(5/6)=5/8HENCE THE ANSWER IS 5/8I think this answer is correct,nice question,Please approve the solution ,so that I can know my answer is correct.
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4 years ago
```							@SAMBIT,THE ANSWER POSTED BY ME WAS 15/24 WHICH IS REDUCED TO 5/8THEN WHY IS IT DISAPPROVED?THE ANSWER IS DEFINITELY CORRECT WITH PROPER AND METHODICAL EXPLANATION
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4 years ago
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