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 dice is cast, if the face 1 or 3 turns up, a ball is taken from the first bag and if any other face turns up, a ball is chosen from the second bag. The probability of choosing a black ball is .........

The probability of getting 1 or 3 when die is cast is $$\dfrac{2}{6}=\dfrac{1}{3}$$

The probability of selecting a black from the first bag is $$\dfrac{3}{3+4}=\dfrac{3}{7}$$

So, the probability of getting 1 or 3 and selecting a black ball is = $$\dfrac{1}{3}\times\dfrac{3}{7}=\dfrac{3}{21}$$

Similarly, the probability of getting 2, 4, 5 or 6 when die is cast is $$\dfrac{4}{6}=\dfrac{2}{3}$$

The probability of selecting a black from the second bag is $$\dfrac{4}{4+3}=\dfrac{4}{7}$$

So, the probability of getting 2, 4, 5 or 6 and selecting a black ball is = $$\dfrac{2}{3}\times\dfrac{4}{7}=\dfrac{8}{21}$$

Hence, the total probability of selecting a black ball is = $$\dfrac{3}{21}\ +\dfrac{8}{21}=\dfrac{11}{21}$$