Question

There are $3$ bags $A, B$ and $C$. Bag $A$ contains $2$ white and $3$ black balls, bag $B$ contains $4$ white and $2$ black balls and Bag $C$ contains $3$ white and $2$ black balls. If a ball is drawn at random from a randomly chosen bag. then the probability that the ball drawn is black, is

• A. $\frac{2}{3}$
• B. $\frac{4}{9}$
• C. $\frac{5}{9}$
• D. $\frac{1}{9}$

Answer 1

Answer: (B)

$\frac{4}{9}$

Answer 2

Given that $A$ has $2$ white, $3$ black balls, $B$ has $4$ white and $2$ black balls and $C$ has $3$ white and $2$ black balls.
Let event of drawing black ball from bags $A, B$ and $C=\frac{1}{3}$
$\therefore$ Required probability
$=$ Probability of black ball from bag $A$ + Probability of black ball from bag $B$ + Probability of black ball from bag $C$
$=\frac{1}{3} \times \frac{3}{5}+\frac{1}{3} \times \frac{2}{6}+\frac{1}{3} \times \frac{2}{5}=\frac{3}{15}+\frac{2}{18}+\frac{2}{15}$
$=\frac{1}{5}+\frac{1}{9}+\frac{2}{15}=\frac{9+5+6}{45}$
$=\frac{20}{45}=\frac{4}{9}$