There are two bags, one of which contains 3 black and 4 white balls while the other contains 4 black and 3 white balls. A die is thrown. If it shows up 1 or $3$, a ball is taken from the Ist bag but it shows up any other number, a ball is chosen from the II bag. Find the probability of choosing a black ball.

Since, Bag ${\rm{I}} = \{ 3$ black, 4 white balls $\}$,

Bag ${\rm{II}} =${4 black, 3 white balls}

Let ${E_1}$ be the event that bag ${\rm{I}}$ is selected and ${E_2}$ be the event that bag II is selected.

Let ${E_3}$ be the event that black ball is chosen.

$\therefore$ $\quad P\left( {{E_1}} \right) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}$

and $P\left( {{E_2}} \right) = 1 - \frac{1}{3} = \frac{2}{3}$

and $P\left( {{E_3}/{E_1}} \right) = \frac{3}{7}$

and $P\left( {{E_3}/{E_2}} \right) = \frac{4}{7}$

$\therefore$ $P\left( {{E_3}} \right) = P\left( {{E_1}} \right) \cdot P\left( {{E_3}/{E_1}} \right) + P\left( {{E_2}} \right) \cdot P\left( {{E_3}/{E_2}} \right)$

$= \frac{1}{3} \cdot \frac{3}{7} + \frac{2}{3} \cdot \frac{4}{7} = \frac{{11}}{{21}}$