Bag I contains 4 white and 2 black balls, and Bag II contains 3 white and 5 black balls. One of the bags is chosen at random and two balls are drawn from it together. If both balls are white, then the probability that they were drawn from Bag I is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{56}{71}\)
Let \(E_1\) and \(E_2\) be the events that Bag I and Bag II are chosen, and let A be the event that both balls are white. Each bag is chosen with probability \(\frac{1}{2}\).
Two balls drawn together are a selection of 2 balls, so
\( \displaystyle P(A|E_1) = \frac{{}^{4}C_{2}}{{}^{6}C_{2}} = \frac{6}{15} = \frac{2}{5} \)
\( \displaystyle P(A|E_2) = \frac{{}^{3}C_{2}}{{}^{8}C_{2}} = \frac{3}{28} \)
By Bayes’ theorem (the factors \(\frac{1}{2}\) cancel):
\( \displaystyle P(E_1|A) = \frac{\frac{2}{5}}{\frac{2}{5}+\frac{3}{28}} \)
\( \displaystyle P(E_1|A) = \frac{\frac{56}{140}}{\frac{56}{140}+\frac{15}{140}} = \frac{56}{71} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{15}{71}\) — Incorrect. This is \(P(E_2|A)\), the chance that the two white balls came from Bag II.
Option (3) \(\displaystyle\frac{2}{5}\) — Incorrect. This is \(P(A|E_1)\), the chance of two white balls when Bag I is known; the condition is the wrong way round.
Option (4) \(\displaystyle\frac{16}{25}\) — Incorrect. This is the answer for one white ball, \(\frac{4}{6}\div\left(\frac{4}{6}+\frac{3}{8}\right)\), not for two.