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प्रश्न
There are three social media groups on a mobile: Group I, Group II and Group III. The probabilities that Group I, Group II and Group III sending the messages on sports are `2/5, 1/2`, and `2/3` respectively. The probability of opening the messages by Group I, Group II and Group III are `1/2, 1/4` and `1/4` respectively. Randomly one of the messages is opened and found a message on sports. What is the probability that the message was from Group III
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उत्तर
Let E1, E2, E3 be the events that Group I, Group II, and Group III open the messages
Then, P(E1) = `1/2`, P(E2) = `1/4`, P(E3) = `1/4`
E1, E2, E3 are mutually exclusive and exhaustive
Let M ≡ the event that message on sports is sent
`"P"("M"//"E"_1)` = Probability that sports message is sent given that it is from Group I
= `2/5`
Similarly, `"P"("M"//"E"_2) = 1/2, "P"("M"//"E"_3) = 2/3`
By Baye's Theorem, the required probability
= `"P"("E"_3//"M")`
`=("P"("E"_3)*"P"("M"//"E"_3))/("P"("E"_1)*"P"("M"//"E"_1) + "P"("E"_2)*"P"("M"//"E"_2) + "P"("E"_3)*"P"("M"//"E"_3))`
= `((1/4)*(2/3))/((1/2)*(2/5) + (1/4)*(1/2) + (1/4)*(2/3))`
= `((1/6))/((1/5) + (1/8) + (1/6))`
= `((1/6))/((59/120))`
= `20/59`
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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"("A")*"P"("L"//"A") + "P"("C")*"P"("L"//"C") + "P"("T")*"P"("L"//"T")`
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`= square`
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= `("P"("C") * "P"("L"//"C"))/("P"("L"))`
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