**Question:**

**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 second bag. Find the probability of choosing a black ball.**

**Solution:**

Let E1 be the event of selecting Bag 1 and E2 be the event of selecting Bag 2.

Also, let E3 be the event that black ball is selected

Now,

P(E1) = 2/6 = 1/3 and P(E2) = 1 – 1/3 = 2/3

P(E3/E1) = 3/7 and P(E3/E2) = 4/7

So,

P(E3) = P(E1). P(E3/E1) + P(E2). P(E3/E2)

= 1/3. 3/7 + 2/3. 4/7 = (3 + 8)/ 21 = 11/21

Therefore, the required probability is 11/21.