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Urn 1 contains 5 white balls and 7 black balls. Urn 2 contains 3 whites and 12 black. A fair coin is flipped; if it is Heads, a ball is drawn from Urn 1, and if it is Tails, a ball is drawn from Urn 2. Suppose that this experiment is done and you learn that a white ball was selected. What is the probability of choosing a white ball?

I thought P(W) = 8/27 or 0.29 since P(H or T) = 0.5

BUT P(W) = P(W|T) P(T) + P(W|T') P(T') = 3/15 x 1/2 + 5/12 x 1/2 = 1/10 + 5/24 = 37/120 (= 0.302)

I understand the even though getting H or T is same, the frequency of white balls in each urn is different. But I still think probability is 8/27 (of course I am wrong but dont know how to change my opinion). Could someone better explain whats happening and/or point me to other examples? I always get such questions wrong and want to train myself instinctively for such weighted probabilities. Need to train myself to spot them.

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Now, imagine the same experiment with the first urn containing 1000 white balls plus 1000 black balls and the second urn containing only one white ball. Another case would consist in having only 2000 black balls in the first urn and only one white in the second. You see that in those two examples, your first reasoning is obviously wrong.

Bottom line: always check the extreme cases to make sure your solution is correct.

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But wouldnt P(W) still be 1001/2001 OR 1/2000? – user18601 Jan 12 at 14:12
Of course not, if you have 1000 black balls in the first urn and only one white ball in the second urn, you see that the probability of picking a white ball is equivalent to the probability of picking the second urn ($=\frac{1}{2}$), that is having Tail when the coin is flipped. Try to picture the experiment in your head and count the number of times you pick a white ball as opposed to a black ball. – ThePawn Jan 12 at 14:19
this clarified everything. thanks a lot!!! – user18601 Jan 12 at 23:09
Looking at extremes is indeed useful general advice for checking one's thinking--as well as one's calculations. (+1) – whuber♦ Mar 13 at 20:43
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Suppose that you are in a probability course in which the professor begins every class by doing the experiment you described. However, you are late to class every day and come in just as the ball is being withdrawn. In the rush to get to your seat, you fail to observe that on some days Urn 1 is being used while on other days Urn 2 is being used. You keep track of whether the ball is white. Over 120 days (what can I say, it is a long course), on 60 days it was Urn 1 that was used from which 25 white and 35 black balls white drawn while on the other 60 days it was Urn 2 from which 12 white and 48 black balls were drawn. So, as far as you can tell, the probability of drawing a white ball is (25+12)/120 = 37/120. Your more punctual classmates, on the other hand, have more data and know that the probability of drawing a white ball depends on which urn is chosen, but over 120 days, and ignoring the fine details of the data that they have gathered, they too work out the relative frequency of white balls as 37/120, same as you, and this is indeed the maximum-likelihood estimate for the probability that a white ball is drawn at the beginning of class in your course.

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(+1) I would like to suggest--and emphasize--that the value of this contribution lies in the principle it uses: we can almost always reduce probability calculations to counting. People tend to reason inaccurately about probabilities, yet very accurately about counts. In fact, people whose job it is to communicate probabilities (such as those who talk about risk in human endeavors) have learned to rephrase probabilities as counts or relative frequencies. – whuber♦ Mar 13 at 20:45

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