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Question
A box contains 2 blue and 3 pink balls and another box contains 4 blue and 5 pink balls. One ball is drawn at random from one of the two boxes and it is found to be pink. Find the probability that it was drawn from first box
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Solution
Let event A1: The ball is drawn from 1st box and event A2: The ball is drawn from the 2nd box.
∴ P(A1) = `1/2`, P(A2) = `1/2`
Let event B: The ball drawn is pink.
There are 5 balls in the 1st box, of which 3 are pink.
∴ `"P"("B"/"A"_1) = 3/5`
There are 9 balls in the 2nd box, of which 5 are pink.
∴ `"P"("B"/"A"_2) = 5/9`
By Bayes’ theorem, the probability that a pink ball is drawn from the first box, is given by
`"P"("A"_1/"B") = ("P"("A"_1)*"P"("B"/"A"_1))/("P"("A"_1)*"P"("B"/"A"_1) + "P"("A"_2)*"P"("B"/"A"_2))`
=`(1/2 xx 3/5)/(1/2 xx 3/5 + 1/2 xx 5/9)`
= `(3/10)/(3/10 + 5/18)`
= `(3/10)/((27 + 25)/90`
= `(3/10)/(52/90)`
= `27/52`
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= `("P"("C") * "P"("L"//"C"))/("P"("L"))`
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