There are 3 bags $A, B$ and $C$. Bag $A$ contains 2 white and 3 black balls, bag $B$ contains 4 white and 2…
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
$\frac{2}{3}$
$\frac{4}{9}$
$\frac{5}{9}$
$\frac{1}{9}$
Solution
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 $C$
$
\begin{aligned}
& =\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}
\end{aligned}
$