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Question
An urn contains m white and n black balls. A ball is drawn at random and is put back into the urn along with k additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Show that the probability of drawing a white ball now does not depend on k.
Solution
Let A be the event having m white and n black balls
E1 = {first ball drawn of white colour}
E2 = {first ball drawn of black colour}
E3 = {second ball drawn of white colour}
∴ P(E1) = `"m"/("m" + "n")` and P(E2) = `"n"/("m" + "n")`
`"P"("E"_3/"E"_1) = ("m" + "k")/("m" + "n" + "k")` and `"P"("E"_3/"E"_2) = "m"/("m" + "n" + "k")`
Now P(E3) = `"P"("E"_1) * "P"("E"_3/"E"_1) + "P"("E"_2)("E"_3/"E"_2)`
= `"m"/("m" + "n") xx ("m" + "k")/("m" + "n" + "k") + "n"/("m" + "n") xx "m"/("m" + "n" + "k")`
= `"m"/("m" + "n" + "k")[("m" + "k")/("m" + "n") + "n"/("m" + "n")]`
= `"m"/("m" + "n" + "k")[("m" + "n" + "k")/("m" + "n")]`
= `"m"/("m" + "n")`
Hence, the probability of drawing a white ball does not depend upon k.
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