Bag I contains 3 white and 3 red balls, and Bag II contains 1 white and 5 red balls. One of the bags is chosen at random and a ball is drawn from it. If the ball drawn is white, then the probability that it was drawn from Bag I is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{3}{4}\)
Let \(E_1\) and \(E_2\) be the events that Bag I and Bag II are chosen, and let W be the event that the ball is white.
\( \displaystyle P(E_1) = P(E_2) = \frac{1}{2} \)
\( \displaystyle P(W|E_1) = \frac{3}{6} = \frac{1}{2},\quad P(W|E_2) = \frac{1}{6} \)
By Bayes’ theorem:
\( \displaystyle P(E_1|W) = \frac{P(E_1)P(W|E_1)}{P(E_1)P(W|E_1)+P(E_2)P(W|E_2)} \)
\( \displaystyle P(E_1|W) = \frac{\frac{1}{2}\times\frac{1}{2}}{\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{6}} \)
\( \displaystyle P(E_1|W) = \frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{12}} = \frac{\frac{3}{12}}{\frac{4}{12}} = \frac{3}{4} \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{1}{4}\) — Incorrect. This is \(P(E_2|W)\), the chance that the white ball came from Bag II.
Option (3) \(\displaystyle\frac{1}{2}\) — Incorrect. This is \(P(W|E_1)\), the chance of white when Bag I is known; the condition is the wrong way round.
Option (4) \(\displaystyle\frac{1}{3}\) — Incorrect. This is P(W), the overall chance of a white ball, \(\frac{1}{4}+\frac{1}{12}\).