Bag I contains 3 white and 1 black ball, and Bag II contains 1 white and 3 black balls. One of the bags is chosen at random, and two balls are drawn from it one after the other, each ball being put back before the next draw. Both balls are white. A third ball is now drawn from the same bag in the same way. The probability that this third ball is white is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{7}{10}\)
Let \(E_1\) and \(E_2\) be the events that Bag I and Bag II are chosen, and let A be the event that the first two balls are white. The ball is put back each time, so
\( \displaystyle P(A|E_1) = \frac{3}{4}\times\frac{3}{4} = \frac{9}{16},\quad P(A|E_2) = \frac{1}{4}\times\frac{1}{4} = \frac{1}{16} \)
By Bayes’ theorem, after two white balls:
\( \displaystyle P(E_1|A) = \frac{\frac{1}{2}\times\frac{9}{16}}{\frac{1}{2}\times\frac{9}{16}+\frac{1}{2}\times\frac{1}{16}} = \frac{9}{10} \)
\( \displaystyle P(E_2|A) = \frac{1}{10} \)
For the third draw these are the new chances of the two bags. By the theorem of total probability:
\( \displaystyle P(\text{third white}) = \frac{9}{10}\times\frac{3}{4}+\frac{1}{10}\times\frac{1}{4} \)
\( \displaystyle P(\text{third white}) = \frac{27}{40}+\frac{1}{40} = \frac{28}{40} = \frac{7}{10} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{3}{4}\) — Incorrect. This assumes the bag is certainly Bag I; after two white balls Bag II is still possible, with chance \(\frac{1}{10}\).
Option (3) \(\displaystyle\frac{1}{2}\) — Incorrect. This is the chance of white before any ball is seen, \(\frac{1}{2}\times\frac{3}{4}+\frac{1}{2}\times\frac{1}{4}\); it ignores the two white balls already drawn.
Option (4) \(\displaystyle\frac{9}{10}\) — Incorrect. This is the chance that the bag is Bag I, not the chance that the third ball is white.