Question

An urn contains 10 white and 3 black balls. Another urn contains 3 white and 5 black balls. Two are drawn from first urn and put into the second urn and then a ball is drawn from the latter. Find the probability that its is a white ball.

Answer 1

A white ball can be drawn in three mutually exclusive ways:
(I) By transferring two black balls from first to second urn, then drawing a white ball
(II) By transferring two white balls from first to second urn, then drawing a white ball
(III) By transferring a white and a black ball from first to second urn, then drawing a white ball
Let E₁, E₂, E₃ and A be the events as defined below:
E₁ = Two black balls are transferred from first to second bag
E₂ = Two white balls are transferred from first to second bag
E₂ = A white and a black ball is transferred from first to second bag
A = A white ball is drawn

\[\therefore P\left( E_1 \right) = \frac{{}^3 C_2}{{}^{13} C_2} = \frac{3}{78}\]
\[ P\left( E_2 \right) = \frac{{}^{10} C_2}{{}^{13} C_2} = \frac{45}{78}\]
\[ P\left( E_3 \right) = \frac{{}^{10} C_1 \times^3 C_1}{{}^{13} C_2} = \frac{30}{78}\]
\[\text{ Now }, \]
\[P\left( A/ E_1 \right) = \frac{3}{10}\]
\[P\left( A/ E_2 \right) = \frac{5}{10}\]
\[P\left( A/ E_3 \right) = \frac{4}{10}\]
\[\text{ Using the law of total probability, we get } \]
\[\text{ Required probability } = P\left( A \right) = P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + P\left( E_3 \right)P\left( A/ E_3 \right)\]
\[ = \frac{3}{78} \times \frac{3}{10} + \frac{45}{78} \times \frac{5}{10} + \frac{30}{78} \times \frac{4}{10}\]
\[ = \frac{9}{780} + \frac{225}{780} + \frac{120}{780}\]
\[ = \frac{354}{780} = \frac{59}{130}\]