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The contents of three given urns I, II and III are as follows

  • 1 white, 2 black and 3 red
  • 2 white, 1 black and 1 red
  • 4 white, 5 black and 3 red

One urn is chosen at random and two balls are drawn. They happen to be white and red. What is the probability they come from urns I, II or III?

My attempt::

Event = choose white and red ball

Total probability = P(urn 1) * P(event / urn 1) + P(urn 2) * P(event / urn 2) + P(urn 3) * P(event / urn 3)

where P(A/B) is the conditional probability. Is this the correct approach?

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  • $\begingroup$ That probability is one if I take your wording serious. Is the question not more about calculating the probability for the rw event for each urn? $\endgroup$ – mvw Jul 4 '14 at 13:38
  • $\begingroup$ This question is similar, except it has only two "urns" (raining or not-raining) instead of three: math.stackexchange.com/questions/165831/conditional-probability $\endgroup$ – David K Jul 4 '14 at 16:59
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Yes, that's right, and then the probability it was from Urn 1 is

$$P(urn 1)*P(event|urn1)/Total Probability$$

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Using the conditional probability is a good idea. Let A be the event a red and white ball are drawn. The law of the total probability can be used to prove Bayes' theorem. Using Bayes' theorem gives: $$P(I \mid A) = \frac{P(A \mid \text{I})P(\text{I})}{P(A \mid \text{I})P(\text{I}) + P(A \mid \text{II})P(\text{II}) + P(A \mid \text{III})P(\text{III})}.$$

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