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 if it shows up any other number, a ball is chosen from the second bag. Then, the probability of choosing a black ball is

• A. $\frac{10}{21}$
• B. $\frac{1}{21}$
• C. $\frac{5}{21}$
• D. $\frac{11}{21}$

Answer 1

Answer: (D)

Let $A$ : first bag is chosen.
Let $B$ : second bag is chosen.
Let $E$ denote the event of choosing a black ball.
$\because$ A die is thrown and if it shows up 1 or 3 , a ball is taken from the first bag
$\therefore P(A)=\frac{2}{6}=\frac{1}{3}$
Since, the second bag is chosen when the die shows any of the numbers $2,4,5,6$.
$\text{ Therefore, }P(B)=\frac{4}{6}=\frac{2}{3}$
Bag I contains 3 black and 4 white balls.
Therefore, $P\left(\frac{E}{A}\right)=\frac{3}{7}$
Similarly, bag Il contains 4 black and 3 white balls.
Therefore, $P\left(\frac{E}{B}\right)=\frac{4}{7}$
Now, probability of choosing a black ball is given by
$P(E)=P(A)\times P\left(\frac{E}{A}\right)+P(B)\times P\left(\frac{E}{B}\right)$
$=\frac{1}{3}\times \frac{3}{7}+\frac{2}{3}\times \frac{4}{7}$
$=\frac{3}{21}+\frac{8}{21}=\frac{11}{21}$