Question

The following table shows the probability of selecting the boxes $A, B$ and $C$ and the number of balls of different colours contained in them

Box	Number of balls			Probability
A	1	2	3	$\frac{1}{2}$
B	2	3	1	$\frac{1}{3}$
C	3	1	2	$\frac{1}{6}$

A box is selected at random and a ball is drawn from it. If it is given that the ball drawn is green, then the probability that it has come from box C is

• A. $\frac{1}{13}$
• B. $\frac{6}{13}$
• C. $\frac{5}{13}$
• D. $\frac{7}{13}$

Answer 1

Answer: (A)

Probability of green ball $P(G)$
$=\frac{1}{2} \times \frac{2}{6}+\frac{1}{3} \times \frac{3}{6}+\frac{1}{6} \times \frac{1}{6}=\frac{1}{6}+\frac{1}{6}+\frac{1}{36} $
$=\frac{6+6+1}{36}=\frac{13}{36}$
Let probability of drawn ball is green comes from bag $C$ is $P\left(\frac{C}{G}\right)$, then
$P\left(\frac{C}{G}\right)=\frac{P\left(\frac{G}{C}\right) \times P(C)}{P(G)}$
$=\frac{\frac{1}{6} \times \frac{1}{6}}{\frac{13}{36}}$
$=\frac{1}{36} \times \frac{36}{13}=\frac{1}{13}$