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प्रश्न
Suppose you have two coins which appear identical in your pocket. You know that one is fair and one is 2-headed. If you take one out, toss it and get a head, what is the probability that it was a fair coin?
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उत्तर
Let E1 = Event that the coin is fair
E2 = Event that the coin is 2-headed
And H = Event that the tossed coin gets head.
P(E1) `1/2`
P(E2) `1/2`
`"P"("H"/"E"_1) = 1/2`
`"P"("H"/"E"_2)` = 1
∴ Using Bayes’ Theorem, we get
`"P"("E"_1/"H") = ("P"("E"_1)*"P"("H"/"E"_1))/("P"("E"_1)*"P"("H"/"E"_1) + "P"("E"_2)*"P"("H"/"E"_2))`
= `(1/2*1/2)/(1/2*1/2 + 1/2*1)`
= `(1/4)/(1/4 + 1/2)`
= `(1/4)/(3/4)`
= `1/3`
Hence the required probability is `1/3`.
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Mr. X goes to office by Auto, Car, and train. The probabilities him travelling by these modes are `2/7, 3/7, 2/7` respectively. The chances of him being late to the office are `1/2, 1/4, 1/4` respectively by Auto, Car, and train. On one particular day, he was late to the office. Find the probability that he travelled by car.
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.
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P(L) = P(A ∩ L) + P(C ∩ L) + P(T ∩ L)
`="P"("A")*"P"("L"//"A") + "P"("C")*"P"("L"//"C") + "P"("T")*"P"("L"//"T")`
`= square * square + square * square + square * square`
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`= square`
`"P"("C"//"L") = ("P"("L" ∩ "C"))/("P"("L"))`
= `("P"("C") * "P"("L"//"C"))/("P"("L"))`
`= (square * square)/square`
`= square`
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