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Three urns; both balls from Urn B

Short answer

Using total probability and Bayes' theorem, P(both from Urn B | one red and one black) = (1/3 × 8/36) / 0.19630 = 20/53 ≈ 0.377.

Write down the difference between the sample space and sample points, dependent and independent events. Urn A contains 2 white, 1 black and 3 red balls. Urn B contains 3 white, 2 black and 4 red balls. Urn C contains 4 white, 3 black and 2 red balls. One Urn is choosen at random and 2 balls are drawn. They happen to be red and black. What is the probability that both come from Urn B.

Sample space and sample point

  • Sample space (S): the set of all possible outcomes of an experiment. For a die, S = {1, 2, 3, 4, 5, 6}.
  • Sample point: a single outcome in the sample space. For a die, "3" is one sample point.

Dependent and independent events

  • Independent: the occurrence of one does not change the probability of the other, so P(A ∩ B) = P(A) P(B). Example: two tosses of a coin.
  • Dependent: the occurrence of one changes the probability of the other, so P(A ∩ B) = P(A) P(B|A). Example: drawing two balls without replacement.

Problem

The question is read as: given that the two balls drawn are one red and one black, find the probability that they came from Urn B.

Let R·B be the event "one red and one black". Each urn is chosen with probability 1/3.

UrnContents (W, B, R)TotalP(one red, one black)
A2, 1, 36(3 × 1) / C(6,2) = 3/15 = 0.2000
B3, 2, 49(4 × 2) / C(9,2) = 8/36 = 0.2222
C4, 3, 29(2 × 3) / C(9,2) = 6/36 = 0.1667

Total probability:

P(RB) = (1/3)(0.2000 + 0.2222 + 0.1667) = 0.19630

Bayes' theorem:

P(B | RB) = (1/3)(8/36) / P(RB)
          = 0.07407 / 0.19630
          = 20/53
          = 0.377

Answer: the probability that the two balls came from Urn B is 20/53 ≈ 0.377.

✓ Answer checked by Claude Sonnet 5.5

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