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 IInd bag. Find the probability of choosing a black ball.

$\frac{11}{21}$

The correct answer is Option (2) → $\frac{11}{21}$ ##

Since, Bag I = {3 black balls, 4 white balls}, Bag II = {4 black balls, 3 white balls}

Let $E_1$ be the event that bag 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.

$∴P(E_1) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}$

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

$\text{and } P(E_3 | E_1) = \frac{3}{7}$

$\text{and } P(E_3 | E_2) = \frac{4}{7}$

$∴P(E_3) = P(E_1) \cdot P(E_3 | E_1) + P(E_2) \cdot P(E_3 | E_2)$

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