Question

There are two bags, one of which contains three black and four white balls while the other 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. Find the probability of choosing a black ball.

• A. $\frac{1}{3}$
• B. $\frac{11}{21}$
• C. $\frac{2}{3}$
• D. $\frac{10}{21}$

Answer 1

Answer: (B)

$\frac{11}{21}$

Let $E_1$ be the event that a ball is drawn from first bag.

$E_2$ be the event that a ball is drawn form the second bag.

$E$ be the event a black ball is chosen.

$P\left(E_1\right)=\frac{2}{6}=\frac{1}{3}$

$P\left(E_2\right)=\frac{4}{6}=\frac{2}{3}$

Now, $P\left(E\right)=P\left(E_1\right)P\left(E/E_1\right)+P\left(E_2\right)P\left(E/E_2\right)$

$=\frac{1}{3}.\frac{3}{7}+\frac{2}{5}.\frac{4}{7}$

$=\frac{11}{21}.$