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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?
2 years ago

the question asked is independent event
probability 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 b
it is from set theory
probability of a intersection b is product of probability of a and probability of b
p(a union b)=(1/4)+(1/6)-(1/4)*(1/6)=9/24
probability of their not being selected is (1 – p(a union b))=1 -(9/24)=15/24
2 years ago

@manoj dixit
Let P=the event that A is selected and

Q = the event that B is selected.

Given that P(P) = 1/4 and P(Q) = 1/6

We know that P' is the event that A does not get selected and Q' is the event that B does not get selected

Probability 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)=5/8

HENCE THE ANSWER IS 5/8

I think this answer is correct,nice question,

Please approve the solution ,so that I can know my answer is correct.

2 years ago

@SAMBIT,THE ANSWER POSTED BY ME WAS 15/24 WHICH IS REDUCED TO 5/8
THEN WHY IS IT DISAPPROVED?THE ANSWER IS DEFINITELY CORRECT WITH PROPER AND METHODICAL EXPLANATION
2 years ago
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