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.
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.
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