Bag I contains 2 white and 3 black balls, and Bag II contains 4 white and 5 black balls. One ball is drawn at random from Bag I and put into Bag II without its colour being seen. A ball is then drawn at random from Bag II and is found to be white. The probability that the ball moved from Bag I to Bag II was white is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{5}{11}\)
Let \(E_1\) be the event that the ball moved is white, \(E_2\) the event that it is black, and W the event that the ball drawn from Bag II is white.
\( \displaystyle P(E_1) = \frac{2}{5},\quad P(E_2) = \frac{3}{5} \)
After a white ball is moved, Bag II has 5 white and 5 black balls; after a black ball is moved, it has 4 white and 6 black balls:
\( \displaystyle P(W|E_1) = \frac{5}{10},\quad P(W|E_2) = \frac{4}{10} \)
By Bayes’ theorem:
\( \displaystyle P(E_1|W) = \frac{P(E_1)P(W|E_1)}{P(E_1)P(W|E_1)+P(E_2)P(W|E_2)} \)
\( \displaystyle P(E_1|W) = \frac{\frac{2}{5}\times\frac{5}{10}}{\frac{2}{5}\times\frac{5}{10}+\frac{3}{5}\times\frac{4}{10}} \)
\( \displaystyle P(E_1|W) = \frac{\frac{10}{50}}{\frac{10}{50}+\frac{12}{50}} = \frac{10}{22} = \frac{5}{11} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{2}{5}\) — Incorrect. This is the chance that the moved ball is white before the second draw is known.
Option (2) \(\displaystyle\frac{11}{25}\) — Incorrect. This is \(\frac{22}{50}\), the overall chance that the ball drawn from Bag II is white.
Option (3) \(\displaystyle\frac{6}{11}\) — Incorrect. This is the chance that the moved ball was black.