There are three boxes. First box has 2 white, 3 black and 4 red balls. Second…
2022
There are three boxes. First box has 2 white, 3 black and 4 red balls. Second box has 3 white, 2 black and 2 red balls. Third box has 4 white, 1 black and 3 red balls. A box is chosen at random and 2 balls are drawn out of which 1 is white, and 1 is red. What is the probability that the balls came from first box?
Answer: A. 0.237 — Key idea: use Bayes' theorem. Compute the probability of drawing one white and one red from each box, then combine with equal priors. First box (2 white, 4…
- A.
0.237
- B.
0.723
- C.
0.18
- D.
0.452
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Correct answer: A
Key idea: use Bayes' theorem. Compute the probability of drawing one white and one red from each box, then combine with equal priors.
First box (2 white, 4 red, total 9): P = (2/9)*(4/8) + (4/9)*(2/8) = 2/9 ≈ 0.2222
Second box (3 white, 2 red, total 7): P = (3/7)*(2/6) + (2/7)*(3/6) = 2/7 ≈ 0.2857
Third box (4 white, 3 red, total 8): P = (4/8)*(3/7) + (3/8)*(4/7) = 3/7 ≈ 0.4286
Apply Bayes' theorem with equal priors (1/3 each): P(first box | one white & one red) = (2/9) / (2/9 + 2/7 + 3/7).
Answer: (2/9) / (2/9 + 2/7 + 3/7) = (2/9) / (59/63) = 14/59 ≈ 0.2373.
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