An urn I contains 3 white and 4 blue balls, while urn II contains 5 white and 6 blue balls. One ball is drawn at random from one of the urns and it is found to be white. The probability that it was drawn from urn II is

$\frac{35}{68}$

The correct answer is Option (4) → $\frac{35}{68}$

$P(\text{Urn I})=\frac{1}{2},\;P(\text{Urn II})=\frac{1}{2}$

$P(W|\text{I})=\frac{3}{7},\;P(W|\text{II})=\frac{5}{11}$

$P(W)=\frac12\cdot\frac37+\frac12\cdot\frac5{11} =\frac{33+35}{154} =\frac{68}{154} =\frac{34}{77}$

$P(\text{Urn II}|W)=\frac{P(W|\text{II})P(\text{II})}{P(W)} =\frac{\frac12\cdot\frac5{11}}{\frac{34}{77}} =\frac{5}{22}\cdot\frac{77}{34} =\frac{35}{68}$

The probability that the ball was drawn from Urn II is $\frac{35}{68}$.