One of two boxes, A and B, is chosen, Box A with probability \(\displaystyle\frac{1}{3}\) and Box B with probability \(\displaystyle\frac{2}{3}\), and a ball is drawn at random from the chosen box. The probability of drawing a red ball from Box B is 0.3. If the probability that a red ball drawn came from Box A is \(\displaystyle\frac{1}{2}\), then the probability of drawing a red ball from Box A is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.6
Let \(E_1\) and \(E_2\) be the events that Box A and Box B are chosen, R the event that the ball is red, and \(P(R|E_1) = x\). By Bayes’ theorem:
\( \displaystyle P(E_1|R) = \frac{\frac{1}{3}x}{\frac{1}{3}x+\frac{2}{3}\times 0.3} \)
This equals \(\frac{1}{2}\) exactly when the two terms of the denominator are equal:
\( \displaystyle \frac{1}{3}x = \frac{2}{3}\times 0.3 \)
\( \displaystyle \frac{1}{3}x = 0.2 \)
\( x = 0.6 \)
Why the other options are incorrect
Option (1) 0.3 — Incorrect. Then \(P(E_1|R) = \frac{0.1}{0.1+0.2} = \frac{1}{3}\), not \(\frac{1}{2}\).
Option (2) 0.15 — Incorrect. Then \(P(E_1|R) = \frac{0.05}{0.05+0.2} = \frac{1}{5}\), not \(\frac{1}{2}\).
Option (4) 0.45 — Incorrect. Then \(P(E_1|R) = \frac{0.15}{0.15+0.2} = \frac{3}{7}\), not \(\frac{1}{2}\).