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प्रश्न
If E1 and E2 are equally likely, mutually exclusive and exhaustive events and `"P"("A"/"E"_1 )` = 0.2, `"P"("A"/"E"_2)` = 0.3. Find `"P"("E"_1/"A")`
उत्तर
E1, E2 are equally likely, mutually exclusive and exhaustive events
∴ P(E1) = P(E2) = `1/2`
It is given that `"P"("A"/"E"_1 )` = 0.2, `"P"("A"/"E"_2)` = 0.3
By Baye's Theorem,
`"P"("E"_1/"A") = ("P"("E"_1)*"P"("A"/"E"_1))/("P"("E"_1)*"P"("A"/"E"_1) + "P"("E"_2)*"P"("A"/"E"_2)`
= `(1/2 xx 0.2)/(1/2 xx 0.2 + 1/2 xx 0.3)`
= `(0.2)/(0.5)`
= `2/5`
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Solution: Let A, C and T be the events that Mr. X goes to office by Auto, Car and Train respectively. Let L be event that he is late.
Given that P(A) = `square`, P(C) = `square`
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= `("P"("C") * "P"("L"//"C"))/("P"("L"))`
`= (square * square)/square`
`= square`
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