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Question
There are two bags, one of which contains 3 black and 4 white balls while the other contains 4 black and 3 white balls. A die is thrown. If it shows up 1 or 3, a ball is taken from the Ist bag; but it shows up any other number, a ball is chosen from the second bag. Find the probability of choosing a black ball.
Solution
Let E1 be the event of selecting Bag I
And E2 be the event of selecting Bag II
Let E3 be the event that black ball is selected
∴ P(E1) = `2/6 = 1/3` and P(E2) = `1 - 1/3 = 2/3`
`"P"("E"_3/"E"_1) = 3/7` and `"P"("E"_3/"E"_2) = 4/7`
∴ P(E3) = `"P"("E"_1) * "P"("E"_3/"E"_1) + "P"("E"_2) * "P"("E"_3/"E"_2)`
= `1/3*3/7 + 2/3*4/7`
= `(3 + 8)/21`
= `11/21`
Hence, the required probability is `11/21`.
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