Given, we have \[3\] urns:
Urn \[1\] = \[2\] white and \[3\] black balls
Urn \[2\] = \[3\] white and 2 black balls
Urn \[3\] = \[4\] white and \[1\] black balls
Now, the probabilities of choosing either of the urns are
\[P({{U}_{1}})\text{ }=\text{ }P({{U}_{2}})\text{ }=\text{ }P({{U}_{3}})\text{ }=\text{ }1/3\]
Let H be the event of drawing white ball from the chosen urn.
So,
\[P(H/{{U}_{1}})\text{ }=\text{ }2/5,\text{ }P(H/{{U}_{2}})\text{ }=\text{ }3/5\]and \[P(H/{{U}_{3}})\text{ }=\text{ }4/5\]
By using Baye’s Theorem, we have
Therefore, the required probability is \[1/3\].