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 1st bag; but it shows up any other number, a ball is chosen from the second bag. Find the probability of choosing a black ball.

• A. $\frac{5}{21}$
• B. $\frac{6}{21}$
• C. $\frac{10}{21}$
• D. $\frac{11}{21}$

Answer 1

The correct option is D $\frac{11}{21}$

Bag 1: {$3$ black , $4$ white}
Bag 2: {$4$ black, $3$ white }
Let $E_1$ be the event bag $1$ is selected and $E_2$ be the event bag $2$ is selected.
Let $E$ be the event black ball is chosen
$P\left(E_1\right)=\frac{2}{6},P\left(E_2\right)=\frac{4}{6}$
$P\left(\frac{E}{E_1}\right)=\frac{3}{7}$ and $P\left(\frac{E}{E_2}\right)=\frac{4}{7}$
Using Total Theorem of probablity,
$P\left(E\right)=P\left(E_1\right).P\left(E/E_1\right)+P\left(E_2\right).P\left(E/E_2\right)$
$P\left(E\right)=\frac{2}{6}\times \frac{3}{7}+\frac{4}{6}\times \frac{4}{7}=\frac{22}{42}=\frac{11}{21}$