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 $i = 1, 2, 3$. If a white ball is selected, what is the probability that it came from Bag 2?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{2}{11}$ ##
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.
$∴P(F) = P(E_1) \cdot P(F | E_1) + P(E_2) \cdot P(F | E_2) + P(E_3) \cdot P(F | E_3)$
$= \left( \frac{1}{6} \times 0 \right) + \left( \frac{2}{6} \times \frac{1}{3} \right) + \left( \frac{3}{6} \times 1 \right) = \frac{1}{9} + \frac{3}{6} = \frac{11}{18}$
Using Bay's theorem, we have
$P(E_2 | F) = \frac{P(E_2) \cdot P(F | E_2)}{P(E_1) \cdot P(F | E_1) + P(E_2) \cdot P(F | E_2) + P(E_3) \cdot P(F | E_3)}$
$= \frac{\frac{2}{6} \times \frac{1}{3}}{\frac{1}{6} \times 0 + \frac{2}{6} \times \frac{1}{3} + \frac{3}{6} \times 1} = \frac{\frac{2}{18}}{\frac{2}{18} + \frac{3}{6}}$
$= \frac{2/18}{\frac{2 + 9}{18}} = \frac{2}{11}$