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

Urn I contains 3 white and 4 blue balls (7 total).

Urn II contains 5 white and 6 blue balls (11 total).

One urn is chosen randomly and a white ball is drawn.

The probability of drawing white from Urn I:

$P(\text{White from Urn I}) = \frac{3}{7}$

The probability of drawing white from Urn II:

$P(\text{White from Urn II}) = \frac{5}{11}$

Since each urn is chosen with probability $\frac{1}{2}$:

$P(\text{Urn I and White}) = \frac{1}{2} \times \frac{3}{7}$

$= \frac{3}{14}$

$P(\text{Urn II and White}) = \frac{1}{2} \times \frac{5}{11}$

$= \frac{5}{22}$

The total probability of drawing a white ball:

$P(\text{White}) = \frac{3}{14} + \frac{5}{22}$

$= \frac{33}{154} + \frac{35}{154}$

$= \frac{68}{154}$

Using Bayes' theorem:

$P(\text{Urn II | White}) = \frac{P(\text{Urn II and White})}{P(\text{White})}$

$= \frac{5/22}{68/154}$

$= \frac{5}{22} \times \frac{154}{68}$

$= \frac{770}{1496}$

$= \frac{35}{68}$

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