Question

There are $3$ bags A, B and C. Bag A contains $1$ Red and $2$ Green balls, bag B contains $2$ Red and $1$ Green balls and bag C contains only one Green ball. One ball is drawn from bag A and put into bag B then one ball is drawn from bag B and put into bag C and finally one ball is drawn from bag C and put into bag A. When this operation is completed, probability that bag A contains $2$ Red and $1$ Green balls, is -

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{1}{6}$

Answer 1

The correct option is D $\frac{1}{6}$


So Probability $=\frac{2}{3}\times \frac{2}{4}\times \frac{1}{2}=\frac{1}{6}$