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.

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