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  • 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 bal

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.

Answers (1)

Solution

Given-

An urn contains m white and n black balls.

Let E1 be the first ball drawn of white colour
E2 be the first ball drawn in black colour
And E3 be the second ball drawn in white colour

\begin{aligned} &P=\left(E_{1}\right)=\frac{m}{m+n} \text { and } P\left(E_{2}\right)=\frac{n}{m+n}\\ &\text { And, }\\ &\mathrm{P}\left(\mathrm{E}_{3} \mid \mathrm{E}_{1}\right)=\frac{\mathrm{m}+\mathrm{k}}{\mathrm{m}+\mathrm{n}+\mathrm{k}} \text { and } \mathrm{P}\left(\mathrm{E}_{3} \mid \mathrm{E}_{2}\right)=\frac{\mathrm{m}}{\mathrm{m}+\mathrm{n}+\mathrm{k}} \end{aligned}
Using the probability theorem, we get,
\\ \therefore P\left(E_{3}\right)=P\left(E_{1}\right) \cdot P\left(E_{3} \mid E_{1)}+P\left(E_{2}\right) \cdot P\left(E_{3} \mid E_{2}\right)\right. \\\\=\frac{\mathrm{m}}{\mathrm{m}+\mathrm{n}} \times \frac{\mathrm{m}+\mathrm{k}}{\mathrm{m}+\mathrm{n}+\mathrm{k}}+\frac{\mathrm{n}}{\mathrm{m}+\mathrm{n}} \times \frac{\mathrm{m}}{\mathrm{m}+\mathrm{n}+\mathrm{k}} \\\\=\frac{m(m+k)+n m}{(m+n+k)(m+n)}=\frac{m^{2}+m k+n m}{(m+n+k)(m+n)} \\\\=\frac{m(m+k+n)}{(m+n+k)(m+n)} \\\\=\frac{\mathrm{m}}{\mathrm{m}+\mathrm{n}}

Hence, the probability of drawing a white ball does not depend on k.

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infoexpert22

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