CMA Foundation · Fundamentals of Business Mathematics and Statistics · Probability
Box 1 holds 3 red and 2 black balls. Box 2 holds 1 red and 4 black balls. A box is chosen with probability 1/3 for Box 1 and 2/3 for Box 2. Two balls are then drawn with replacement from the chosen box, and both are red. What is the probability that the box chosen was Box 1?
The probability is 9/11. With replacement, two reds have probability 9/25 from Box 1 and 1/25 from Box 2. Weighting by priors 1/3 and 2/3 gives 9/75 and 2/75. Box 1's share is 9/(9+2) = 9/11 by Bayes' theorem.
- A9/11Correct
- B3/5
- C2/11
- D1/3
Explanation
P(RR | Box 1) = (3/5)^2 = 9/25 and P(RR | Box 2) = (1/5)^2 = 1/25. Joint: (1/3)(9/25) = 3/25 = 9/75 and (2/3)(1/25) = 2/75. Posterior = 9/(9+2) = 9/11. The option 3/5 results from using only one red draw instead of two.
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