Question

A bag X contains 2 white and 3 black balls and another bag Y contains 4 white and 2 black balls. One bag is selected at random and a ball is drawn from it. Then, the probability chosen to be white is
(a) 2/15
(b) 7/15
(c) 8/15
(d) 14/15

Answer 1

(c) 8/15

A white ball can be drawn in two mutually exclusive ways:
(I) Selecting bag X and then drawing a white ball from it.
(II) Selecting bag Y and then drawing a white ball from it.

Let E₁, E₂ and A be three events as defined below:
E₁ = Selecting bag X
E₂ = Selecting bag Y
A = Drawing a white ball

We know that one bag is selected randomly.

$$\begin{gathered} \therefore \mathrm{P}\left(E_1\right)=\frac{1}{2} \\ \mathrm{P}\left(E_2\right)=\frac{1}{2} \\ P\left(A/E_1\right)=\frac{2}{5} \\ P\left(A/E_2\right)=\frac{4}{6}=\frac{2}{3} \\ \mathrm{Using}\mathrm{the}\mathrm{law}\mathrm{of}\mathrm{total}\mathrm{probability},\mathrm{we}\mathrm{get} \\ \mathrm{Required}\mathrm{probability}=P\left(A\right)=P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right) \\ =\frac{1}{2}\times \frac{2}{5}+\frac{1}{2}\times \frac{2}{3} \\ =\frac{1}{5}+\frac{1}{3} \\ =\frac{3+5}{15}=\frac{8}{15} \end{gathered}$$