Multiple choice

There are three urns A, B and C Urn A contains 4 white balls and 5 blue balls. Urn B contains 4 white balls and 3 blue balls. Urn C contains 2 white balls and 4 blue balls. One ball is drawn from each of these urns. What is the probability that out of these three balls drawn, 2 are white balls and 1 is blue ball?

  1. 3/7

  2. 2/7

  3. 1/189

  4. 64/189
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Urn A: P(W)=4/9, P(B)=5/9. Urn B: P(W)=4/7, P(B)=3/7. Urn C: P(W)=2/6=1/3, P(B)=4/6=2/3. We need 2 white and 1 blue. Possible cases: (W,W,B) = (4/9)(4/7)(2/3) = 32/189; (W,B,W) = (4/9)(3/7)(1/3) = 12/189; (B,W,W) = (5/9)(4/7)(1/3) = 20/189. Total = (32+12+20)/189 = 64/189.

AI explanation

Using the theorem of total probability for independent draws, calculate the probability of each valid permutation of drawing two white balls and one blue ball. The three possible scenarios are drawing (White, White, Blue), (White, Blue, White), and (Blue, White, White) from urns A, B, and C respectively. The sum of these probabilities is (4/9 x 4/7 x 4/6) + (4/9 x 3/7 x 2/6) + (5/9 x 4/7 x 2/6), which equals 64/189. The probability is 64/189.

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