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Question
There are two boxes, namely box-I and box-II. Box-I contains 3 red and 6 black balls. Box-II contains 5 red and 5 black balls. One of the two boxes, is selected at random and a ball is drawn at random. The ball drawn is found to be red. Find the probability that this red ball comes out from box-II.
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Solution
Let E1 = Selecting Box-I
E2 = Selecting Box-II
A = Getting a red ball from the selected box
Here, P(E1) = `1/2`, P(E2) = `1/2`
`P(A/E_1) = 3/9 = 1/3`
`P(A/E_2) = 5/10 = 1/2`
Required probability = `P(E_2/A)`
= P(Red ball comes out from Box-II)
Using Bayes' theorem,
`P(E_2/A) = (P(E_2)P(A/E_2))/(P(E_1).P(A/E_1) + P(E_2).P(A/E_2))`
= `(1/2 xx 1/2)/(1/2 xx 1/3 + 1/2 xx 1/2)`
= `(1/4)/(1/6 + 1/4)`
= `(1/4)/(10/24)`
= `1/4 xx 24/10`
= `3/5`
Thus, probability that the red ball comes out form Box-II is `3/5`.
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