A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls.

Let E1 and E2 be the events of selecting first bag and second bag respectively

$\therefore \mathrm{P}\left(\mathrm{E}_{1}\right)=\mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{1}{2}$

Let A be the event of getting a red ball.

$\Rightarrow \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)=\mathrm{P}($ drawing a red ball from first bag $)=\frac{4}{8}=\frac{1}{2}$

$\Rightarrow \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)=\mathrm{P}($ drawing a red ball from second bag $)=\frac{2}{8}=\frac{1}{4}$

The probability of drawing a ball from the first bag, given that it is red, is given by $P\left(E_{2} \mid A\right)$.

By using Bayes’ theorem, we obtain

$P\left(E_{1} \mid A\right)=\frac{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A \mid E_{2}\right)}$

$=\frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2} \cdot \frac{1}{2}+\frac{1}{2} \cdot \frac{1}{4}}$

$=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{8}}$

$=\frac{\frac{1}{4}}{\frac{3}{8}}$

$=\frac{2}{3}$