Question

One bag contains 4 yellow and 5 red balls. Another bag contains 6 yellow and 3 red balls. A ball is transferred from the first bag to the second bag and then a ball is drawn from the second bag. Find the probability that ball drawn is yellow.

Answer 1

A yellow ball can be drawn in two mutually exclusive ways:

(I) By transferring a red ball from first to second bag, then drawing a yellow ball
(II) By transferring a yellow ball from first to second bag, then drawing a yellow ball

Let E₁, E₂ and A be the events as defined below:

E₁ = A red ball is transferred from first to second bag
E₂ = A yellow ball is transferred from first to second bag
A = A yellow ball is drawn

$$\begin{gathered} \therefore P\left(E_1\right)=\frac{5}{9} \\ P\left(E_2\right)=\frac{4}{9} \\ \mathrm{Now}, \\ P\left(A/E_1\right)=\frac{6}{10} \\ P\left(A/E_2\right)=\frac{7}{10} \\ \mathrm{Using}\mathrm{the}\mathrm{law}\mathrm{of}\mathrm{total}\mathrm{probability},\mathrm{we}\mathrm{get} \\ \mathrm{Required}\mathrm{probability}=P\left(A\right)=P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right) \\ =\frac{5}{9}\times \frac{6}{10}+\frac{4}{9}\times \frac{7}{10} \\ =\frac{30}{90}+\frac{28}{90} \\ =\frac{58}{90}=\frac{29}{45} \end{gathered}$$