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There are two bags in which first bag contains three black and four white balls, second bag contains four black and three white balls. A die 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 (assume balls of the same colour are distinct in size)

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$\frac{9}{21}$

$\frac{11}{21}$

$\frac{12}{21}$

$\frac{10}{21}$

answer is B.

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Detailed Solution

Let $E_{1}$ be the event that a ball is drawn from first bag, $E_{2}$ the event that a ball is drawn from the second bag and E the event a black ball is chosen. Then, we have

$P (E) = P (E_{1}) P (E / E_{1}) + P (E_{2}) P (E / E_{2}) = \frac{1}{3} . \frac{3}{7} + \frac{2}{3} . \frac{4}{7} = \frac{11}{21} .$

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