Question

Coloured balls are distributed in four boxes as shown in the following table:

Box	Colour
	Black	White	Red	Blue
I II III IV	3 2 1 4	4 2 2 3	5 2 3 1	6 2 1 5

A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black, what is the probability that ball drawn is from the box III.

Answer 1

Let A, E₁, E₂, E₃ and  E₄ denote the events that the ball is black, box I selected, box II selected, box III is selected and box IV is selected, respectively.

\[\therefore P\left( E_1 \right) = \frac{1}{4}\]
\[ P\left( E_2 \right) = \frac{1}{4}\]
\[ P\left( E_3 \right) = \frac{1}{4} \]
\[ P\left( E_4 \right) = \frac{1}{4}\]
\[\text{Now }, \]
\[P\left( A/ E_1 \right) = \frac{3}{18}\]
\[P\left( A/ E_2 \right) = \frac{2}{8}\]
\[P\left( A/ E_3 \right) = \frac{1}{7}\]
\[P\left( A/ E_4 \right) = \frac{4}{13}\]
\[\text{ Using Bayes' theorem, we get} \]
\[\text{ Required probability } = P\left( E_3 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)}\]
\[ = \frac{\frac{1}{4} \times \frac{1}{7}}{\frac{1}{4} \times \frac{3}{18} + \frac{1}{4} \times \frac{2}{8} + \frac{1}{4} \times \frac{1}{7} + \frac{1}{4} \times \frac{4}{13}}\]
\[ = \frac{\frac{1}{7}}{\frac{1}{6} + \frac{1}{4} + \frac{1}{7} + \frac{4}{13}} = \frac{156}{947}\]