One of three boxes is chosen: Box I with probability \(\displaystyle\frac{1}{2}\), Box II with probability \(\displaystyle\frac{1}{3}\) and Box III with probability \(\displaystyle\frac{1}{6}\). Box I contains 1 red and 3 blue balls, Box II contains 2 red and 2 blue balls, and Box III contains 3 red and 1 blue ball. A ball is drawn at random from the chosen box. Consider the following statements:
(A) The probability that the ball is red is \(\displaystyle\frac{5}{12}\).
(B) If the ball is red, the probability that it came from Box III is \(\displaystyle\frac{1}{2}\).
(C) If the ball is red, the probability that it came from Box II is \(\displaystyle\frac{2}{5}\).
(D) If the ball is blue, the probability that it came from Box III is \(\displaystyle\frac{1}{14}\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (C) and (D) only
Let \(E_1, E_2, E_3\) be the events that Box I, II, III is chosen, and R, B the events that the ball is red, blue. The chances of red are \(\frac{1}{4}\), \(\frac{2}{4} = \frac{1}{2}\) and \(\frac{3}{4}\).
\( \displaystyle P(E_1)P(R|E_1) = \frac{1}{2}\times\frac{1}{4} = \frac{3}{24} \)
\( \displaystyle P(E_2)P(R|E_2) = \frac{1}{3}\times\frac{1}{2} = \frac{4}{24} \)
\( \displaystyle P(E_3)P(R|E_3) = \frac{1}{6}\times\frac{3}{4} = \frac{3}{24} \)
By the theorem of total probability:
\( \displaystyle P(R) = \frac{3+4+3}{24} = \frac{10}{24} = \frac{5}{12},\quad P(B) = \frac{7}{12} \)
By Bayes’ theorem:
\( \displaystyle P(E_3|R) = \frac{3}{10},\quad P(E_2|R) = \frac{4}{10} = \frac{2}{5} \)
\( \displaystyle P(E_3|B) = \frac{\frac{1}{6}\times\frac{1}{4}}{\frac{7}{12}} = \frac{\frac{1}{24}}{\frac{14}{24}} = \frac{1}{14} \)
So P(red) = \(\frac{5}{12}\), P(Box III | red) = \(\frac{3}{10}\), P(Box II | red) = \(\frac{2}{5}\) and P(Box III | blue) = \(\frac{1}{14}\).
Checking the statements
(A) The probability that the ball is red is \(\frac{5}{12}\). — Correct. \(\frac{3}{24}+\frac{4}{24}+\frac{3}{24} = \frac{5}{12}\).
(B) If the ball is red, the probability that it came from Box III is \(\frac{1}{2}\). — Incorrect. It is \(\frac{3}{24}\div\frac{10}{24} = \frac{3}{10}\); Box III has the most red balls but is chosen least often.
(C) If the ball is red, the probability that it came from Box II is \(\frac{2}{5}\). — Correct. \(\frac{4}{24}\div\frac{10}{24} = \frac{2}{5}\).
(D) If the ball is blue, the probability that it came from Box III is \(\frac{1}{14}\). — Correct. \(\frac{1}{24}\div\frac{14}{24} = \frac{1}{14}\).