Step 1:

Let $E_1$ be the event of choosing the bag I

Let $E_2$ be the event of choosing the bag II

Let A be the event of drawing a black ball

$\therefore P(E_1)=P(E_2)=\large\frac{1}{2}$

Step 2:

Now $P(\large\frac{A}{E_1})$=P(drawing a black ball from bag I)

$\Rightarrow \large\frac{3}{5}$

$P(\large\frac{A}{E_2})$=P(drawing a black ball from bag II)

$\Rightarrow \large\frac{2}{6}$

$\Rightarrow \large\frac{1}{3}$

Step 3:

$\therefore$ Probability of selecting a black ball is $\large\frac{1}{2}\times \frac{3}{5}+\frac{1}{2}\times \frac{1}{3}$

$\Rightarrow \large\frac{3}{10}+\frac{1}{6}$

$\Rightarrow \large\frac{7}{15}$