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.

Answer 1

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.