Bag I contains 3 black and 2 white balls, Bag II contains 2 black and 4 white balls. A bag is selected at random and then a ball is drawn from it. The probability that the ball drawn is black is:

$\frac{7}{15}$

The correct answer is Option (1) → $\frac{7}{15}$

Bag I: 3 black, 2 white → total 5 balls

Bag II: 2 black, 4 white → total 6 balls

Probability of choosing each bag = $\frac{1}{2}$

Probability of drawing a black ball from Bag I = $\frac{3}{5}$

Probability of drawing a black ball from Bag II = $\frac{2}{6} = \frac{1}{3}$

Total probability (by law of total probability):

$P(\text{black}) = \frac{1}{2}\cdot\frac{3}{5} + \frac{1}{2}\cdot\frac{1}{3}$

$= \frac{1}{2}\left(\frac{9 + 5}{15}\right) = \frac{1}{2}\cdot\frac{14}{15} = \frac{7}{15}$

Required probability = $\frac{7}{15}$