Three bags contain a number of red and white balls as follows: Bag 1 : 3 red balls, Bag 2 : 2 red balls and 1 white ball and Bag 3 : 3 white balls. The probability that bag $i$ will be chosen and a ball is selected from it is $\frac{i}{6}$, where $i = 1, 2, 3$. What is the probability that a white ball is selected?

$\frac{11}{18}$

The correct answer is Option (2) → $\frac{11}{18}$ ##

Bag 1 : 3 red balls and 0 white ball.

Bag 2 : 2 red balls and 1 white ball.

Bag 3 : 0 red ball and 3 white balls.

Let $E_1, E_2$ and $E_3$ be the events that bag 1, bag 2 and bag 3 is selected respectively and a ball is chosen from it.

$P(E_1) = \frac{1}{6}, P(E_2) = \frac{2}{6} \text{ and } P(E_3) = \frac{3}{6}$

Let $F$ be the event that a white ball is selected.

$∴P(F) = P(E_1) \cdot P(F | E_1) + P(E_2) \cdot P(F | E_2) + P(E_3) \cdot P(F | E_3)$

$= \left( \frac{1}{6} \times 0 \right) + \left( \frac{2}{6} \times \frac{1}{3} \right) + \left( \frac{3}{6} \times 1 \right) = \frac{1}{9} + \frac{3}{6} = \frac{11}{18}$