A bag contains 3 balls, each of which is either white or black. All the possible numbers of white balls in the bag (0, 1, 2 or 3) are equally likely. A ball is drawn at random from the bag and is found to be white. The probability that all three balls in the bag are white is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{2}\)
Let \(E_0, E_1, E_2, E_3\) be the events that the bag has 0, 1, 2, 3 white balls, and W the event that the ball drawn is white.
\( \displaystyle P(E_0) = P(E_1) = P(E_2) = P(E_3) = \frac{1}{4} \)
\( \displaystyle P(W|E_0) = 0,\quad P(W|E_1) = \frac{1}{3},\quad P(W|E_2) = \frac{2}{3},\quad P(W|E_3) = 1 \)
By Bayes’ theorem, the common factor \(\frac{1}{4}\) cancels:
\( \displaystyle P(E_3|W) = \frac{1}{0+\frac{1}{3}+\frac{2}{3}+1} \)
\( \displaystyle = \frac{1}{2} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{4}\) — Incorrect. This is \(P(E_3)\), the chance before the white ball is seen.
Option (2) \(\displaystyle\frac{1}{3}\) — Incorrect. This is \(P(E_2|W) = \frac{2/3}{2}\), the probability that the bag has exactly 2 white balls.
Option (3) \(\displaystyle\frac{2}{3}\) — Incorrect. This is \(P(W|E_2)\), the chance of drawing white from a bag with 2 white balls.