One of two boxes, A and B, is chosen and a ball is drawn at random from it. The probability of drawing a red ball is 0.6 from Box A and 0.2 from Box B. The probability that the ball drawn is red is 0.44. If the ball drawn is red, then the probability that it was drawn from Box A is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{9}{11}\)
The chance of choosing Box A is not given, so let it be p. Then Box B is chosen with probability 1 − p. Let R be the event that the ball is red.
By the theorem of total probability:
\(P(R) = 0.6p + 0.2(1-p)\)
\(0.44 = 0.2 + 0.4p\)
\(p = 0.6\)
By Bayes’ theorem:
\( \displaystyle P(A|R) = \frac{0.6\times 0.6}{0.44} = \frac{0.36}{0.44} = \frac{9}{11} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{3}{5}\) — Incorrect. This is p = 0.6, the chance of choosing Box A before the colour is seen.
Option (3) \(\displaystyle\frac{3}{4}\) — Incorrect. This is \(\frac{0.6}{0.6+0.2}\), which treats the two boxes as equally likely.
Option (4) \(\displaystyle\frac{2}{11}\) — Incorrect. This is \(\frac{0.4\times 0.2}{0.44}\), the chance that the red ball came from Box B.