question_answer

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 does not depend on k.

Answer:

Let U = {m white balls, n black balls}      \[{{E}_{1}}=\] Event that first ball drawn of white colour      \[{{E}_{2}}=\]Event that first ball drawn of black colour and \[{{E}_{3}}=\] Event that second ball drawn of white colour Then, \[P({{E}_{1}})=\frac{m}{m+n}\,\,\text{and}\,\,P({{E}_{2}})=\frac{n}{m+n}\] Also,\[P\left( \frac{{{E}_{3}}}{{{E}_{1}}} \right)=\frac{m+k}{m+n+k}\]and \[P\left( \frac{{{E}_{3}}}{{{E}_{2}}} \right)=\frac{m}{m+n+k}\] \[\therefore \] \[P({{E}_{3}})=P({{E}_{1}})\cdot P\left( \frac{{{E}_{3}}}{{{E}_{1}}} \right)+P({{E}_{2}})\cdot P\left( \frac{{{E}_{3}}}{{{E}_{2}}} \right)\] \[=\frac{m}{m+n}\cdot \frac{m+k}{m+n+k}+\frac{n}{m+n}\cdot \frac{m}{m+n+k}\] \[=\frac{m(m+k)+nm}{(m+n+k)(m+n)}\] \[=\frac{m(m+k+n)}{(m+n+k)(m+n)}=\frac{m}{m+n}\] Hence the probability of drawing a white ball does not depend on k.