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If E1 and E2 are equally likely, mutually exclusive and exhaustive events and P(AE1) = 0.2, P(AE2) = 0.3. Find P(E1A) - Mathematics and Statistics

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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")`

योग

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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अध्याय 9: Probability - Exercise 9.4 [पृष्ठ २०९]

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