Question

There are three different Urns, Urn-I, Urn-II and Urn-III containing 1 Blue, 2 Green, 2 Blue, 1 Green, 3 Blue, 3 Green balls respectively. If two Urns are randomly selected and a ball is drawn from each Urn and if the drawn balls are of different colours then the probability that chosen Urn was Urn-I and Urn-II is

• A. $\frac{1}{7}$
• B. $\frac{5}{13}$
• C. $\frac{5}{14}$
• D. $\frac{5}{7}$

Answer 1

Answer: (C)

$\frac{5}{14}$

Required probability$=\frac{\frac{1}{3}(\frac{1}{3}.\frac{1}{3}+\frac{2}{3}.\frac{2}{3})}{\frac{1}{3}(\frac{1}{3}.\frac{2}{3}.\frac{2}{3})+\frac{1}{3}(\frac{2}{3}.\frac{3}{6}+\frac{1}{3}.\frac{3}{6})+\frac{1}{3}(\frac{3}{6}.\frac{2}{3}+\frac{3}{6}.\frac{1}{3})}$

$=\frac{\frac{5}{9}}{\frac{5}{9}+\frac{9}{18}+\frac{9}{18}}=\frac{5}{14}$