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

$$\begin{gathered} \therefore P\left(E_1\right)=\frac{{}^3{C}_2}{{}^{13}{C}_2}=\frac{3}{78} \\ P\left(E_2\right)=\frac{{}^{10}{C}_2}{{}^{13}{C}_2}=\frac{45}{78} \\ P\left(E_3\right)=\frac{{}^{10}{C}_1{\times }^3{C}_1}{{}^{13}{C}_2}=\frac{30}{78} \\ \mathrm{Now}, \\ P\left(A/E_1\right)=\frac{3}{10} \\ P\left(A/E_2\right)=\frac{5}{10} \\ P\left(A/E_3\right)=\frac{4}{10} \\ \mathrm{Using}\mathrm{the}\mathrm{law}\mathrm{of}\mathrm{total}\mathrm{probability},\mathrm{we}\mathrm{get} \\ \mathrm{Required}\mathrm{probability}=P\left(A\right)=P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)+P\left(E_3\right)P\left(A/E_3\right) \\ =\frac{3}{78}\times \frac{3}{10}+\frac{45}{78}\times \frac{5}{10}+\frac{30}{78}\times \frac{4}{10} \\ =\frac{9}{780}+\frac{225}{780}+\frac{120}{780} \\ =\frac{354}{780}=\frac{59}{130} \end{gathered}$$