Urn I contains 1 white and 4 black balls, Urn II contains 3 white and 2 black balls and Urn III contains 4 white and 1 black ball. An urn is chosen at random and a ball is drawn at random from it. If the ball drawn is white, then the probability that it was drawn from Urn III is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{2}\)
Let \(E_1, E_2, E_3\) be the events that Urn I, II, III is chosen, and W the event that the ball is white. Each urn is equally likely:
\( \displaystyle P(E_1) = P(E_2) = P(E_3) = \frac{1}{3} \)
Each urn has 5 balls, so
\( \displaystyle P(W|E_1) = \frac{1}{5},\quad P(W|E_2) = \frac{3}{5},\quad P(W|E_3) = \frac{4}{5} \)
By Bayes’ theorem:
\( \displaystyle P(E_3|W) = \frac{\frac{1}{3}\times\frac{4}{5}}{\frac{1}{3}\times\frac{1}{5}+\frac{1}{3}\times\frac{3}{5}+\frac{1}{3}\times\frac{4}{5}} \)
\( \displaystyle = \frac{4}{1+3+4} = \frac{4}{8} = \frac{1}{2} \)
Shortcut: When the urns are equally likely and hold the same number of balls, the answer is simply white balls in Urn III over all white balls: \(\frac{4}{1+3+4} = \frac{1}{2}\).