Bag I contains 3 red and 2 black balls, and Bag II contains 2 red and 4 black balls. One ball is drawn at random from Bag I and put into Bag II without its colour being noted. A ball is then drawn at random from Bag II. The probability that this ball is red is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{13}{35}\)
Let \(E_1\) and \(E_2\) be the events that the transferred ball is red and black, and let A be the event that the ball drawn from Bag II is red. After the transfer Bag II has 7 balls.
\( \displaystyle P(E_1) = \frac{3}{5},\quad P(A|E_1) = \frac{3}{7} \) (Bag II then has 3 red, 4 black)
\( \displaystyle P(E_2) = \frac{2}{5},\quad P(A|E_2) = \frac{2}{7} \) (Bag II then has 2 red, 5 black)
By the theorem of total probability:
\( \displaystyle P(A) = \frac{3}{5}\times\frac{3}{7} + \frac{2}{5}\times\frac{2}{7} \)
\( \displaystyle P(A) = \frac{9}{35}+\frac{4}{35} = \frac{13}{35} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{3}{7}\) — Incorrect. This assumes the transferred ball was red.
Option (3) \(\displaystyle\frac{1}{3}\) — Incorrect. This is \(\frac{2}{6}\), the chance of red from Bag II as it was before the transfer.
Option (4) \(\displaystyle\frac{9}{35}\) — Incorrect. This counts only the case in which a red ball is transferred and leaves out the black case.