A coin is tossed. If it shows a head, a ball is drawn at random from Bag I, which contains 2 red and 3 green balls; if it shows a tail, a ball is drawn at random from Bag II, which contains 4 red and 1 green ball. Consider the following statements:
(A) The probability that the ball drawn is red is \(\displaystyle\frac{3}{5}\).
(B) The probability that the ball drawn is green is \(\displaystyle\frac{1}{2}\).
(C) If the ball drawn is red, the probability that it came from Bag II is \(\displaystyle\frac{2}{3}\).
(D) If the ball drawn is green, the probability that it came from Bag I is \(\displaystyle\frac{1}{4}\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) and (C) only
Let \(E_1\) and \(E_2\) be the events that Bag I and Bag II are used. Each has probability \(\frac{1}{2}\), since the coin decides.
\( \displaystyle P(R|E_1) = \frac{2}{5},\quad P(R|E_2) = \frac{4}{5} \)
By the theorem of total probability:
\( \displaystyle P(R) = \frac{1}{2}\times\frac{2}{5} + \frac{1}{2}\times\frac{4}{5} = \frac{3}{5} \)
\( \displaystyle P(G) = 1 - \frac{3}{5} = \frac{2}{5} \)
By Bayes’ theorem:
\( \displaystyle P(E_2|R) = \frac{\frac{1}{2}\times\frac{4}{5}}{\frac{3}{5}} = \frac{2}{3} \)
\( \displaystyle P(E_1|G) = \frac{\frac{1}{2}\times\frac{3}{5}}{\frac{2}{5}} = \frac{3}{4} \)
So P(red) = \(\frac{3}{5}\), P(green) = \(\frac{2}{5}\), P(Bag II | red) = \(\frac{2}{3}\) and P(Bag I | green) = \(\frac{3}{4}\).
Checking the statements
(A) The probability that the ball drawn is red is \(\frac{3}{5}\). — Correct. \(\frac{1}{2}\times\frac{2}{5}+\frac{1}{2}\times\frac{4}{5} = \frac{3}{5}\).
(B) The probability that the ball drawn is green is \(\frac{1}{2}\). — Incorrect. It is \(1-\frac{3}{5} = \frac{2}{5}\).
(C) If the ball drawn is red, the probability that it came from Bag II is \(\frac{2}{3}\). — Correct. \(\frac{2}{5}\div\frac{3}{5} = \frac{2}{3}\).
(D) If the ball drawn is green, the probability that it came from Bag I is \(\frac{1}{4}\). — Incorrect. It is \(\frac{3}{10}\div\frac{2}{5} = \frac{3}{4}\); \(\frac{1}{4}\) is the chance for Bag II.