Suppose there are three balls, each either black or white with presumed equal probability, in a covered container. Based on this premise, the probability of all three balls being white is 1/8. Now suppose we blindly reach in and pull out a ball - it is white. We return that ball to the container and make another blind selection - it again is white. We return the ball and make another pick - white again. So, we have made three random selections and come out with three white balls. What, now, is the probability that all three balls are white? Bayes Theorem is the easiest way of computing the answer. For any two events X and Y, the probability of both X and Y being true is clearly equal to the probability of {X is true} times the probability of {Y is true given that X is true}. This can be written as Pr{X and Y} = Pr{X} Pr{Y|X}, where the symbol "|" is read as "given". From this it follows that Pr{X and Y} Pr{Y|X} = ------------- (1) Pr{X} So if we define Y as the event {All three balls are white} and X as the event {All three of our selections are white}, then obviously the probability that all 3 balls are white AND all 3 selections are white is just equal to the probability that all 3 balls are white, which is 1/8. So we only need to determine Pr{X}, the probability that all 3 selections are white. As already noted, the unconditional probability that all 3 balls are white is 1/8, and we are certain to have all 3 selections white in that case. The unconditional probability that exactly 2 balls are white is 3/8, in which case we have (2/3)^3 probability of all 3 selections being white. The unconditional probability of exactly 1 white ball is 3/8, in which case we have (1/3)^3 probability of all 3 selections being white. Therefore, the overall probability of all 3 selections being white is (1/8)*(1) + (3/8)*(2/3)^3 + (3/8)*(1/3)^3 = 1/4 So, we've determined that, of all possible outcomes, 1/4 result in all 3 selections being white, and 1/8 have all 3 balls white AND all 3 selections white. It follows that of the cases when all 3 selections are white, exactly half are cases with all 3 balls being white. Hence the conditional probability of all three balls being white, given that all three of our selections was white, is 1/2. Now, sometimes students think that the probability of all the balls being white should be 2/9, based on the notion that it equals the probability of selecting a different ball on each of our three trials, which equals (1)(2/3)(1/3)= 2/9. It's true that if we select a different ball each time, and if all 3 selections are white, then all three balls are white, but the converse is not true. If we do NOT select a different ball each time, it's still possible that all 3 balls are white. So, to work the problem this way, we need to consider all the possibilities: We have 2/9 chance of selecting 3 different balls, and the probability of all 3 selections being white is 1/8. We have 6/9 chance of selecting one ball twice and another ball once, and in this case our chances of all 3 selections being white are 1/4. We have 1/9 chance of selecting the same ball all three times, and in this case our chance is 1/2 of all three selections being white. Combining these, the total probability of all 3 selections being white is (2/9)(1/8) + (6/9)(1/4) + (1/9)(1/2) = 1/4 so again the probability of all three balls being white, given that all 3 selections are white, is (1/8)/(1/4) = 1/2. For a related problem, suppose a box contains six white balls and an unknown number of blue balls, but not more than six. Three balls are removed successively without replacement, and all are blue. If a 4th ball is removed, what is the probability that it is white? This problem, again, calls for some Bayesian analysis, to infer the probability of a prior condition from some subsequent results. In this case we are trying to figure out the probabilities that the original number of blue balls was 0,1,2,..,or 6, given the fact that we drew 3 blue balls in a row. Again we use a form of Bayes theorem Pr{A} Pr{A|B} = Pr{B|A} ----- Pr{B} which is really just a consequence of the intuitive fact that Pr{A and B} = Pr{A}Pr{B|A}, and by symmetry we also have Pr{A and B} = Pr{B}Pr{A|B}, so we can equate these two to give the above version of Bayes' theorem. The probability that there were originally 5 blue balls, given that we drew 3 blue balls in a row, is equal to [the probability that we would draw 3 in a row GIVEN that there were originally 5 blue balls] times [the probability that there were originally 5 blue balls] divided by [the probability that we would draw 3 in a row]. Notice that if we assume the "a priori" probability of 5 blue balls originally is the same as 4 or 3 or 2 or 1 or 6, then that factor is the same for all the possibilities. Also, the overall probability of drawing 3 blue balls (unconditionally) is the same for all cases. Thus, the probabilities are PROPORTIONAL to the respective values of Pr{B|A}, i.e., the probability that we would draw 3 blue in a row GIVEN that there were originally k blue balls (k=3,4,5,6). The results are 3 2 1 4 3 2 5 4 3 6 5 4 p3 = - - - p4 = -- - - p5 = -- -- - p6 = -- -- -- 9 8 7 10 9 8 11 10 9 12 11 10 Letting Pt = p3 + p4 + p5 + p6, the overall result is found by multiplying the probability of drawing a white on the next ball given each of these four possible priors, so we have p3 p4 p5 p6 677 -- (1) + -- (6/7) + -- (6/8) + -- (6/9) = --- pt pt pt pt 909 which is about 74.5%. Incidentally, suppose we re-worded the problem like this: A box contains from 9 to 12 balls, 6 of which are white, and the remainder red. If someone randomly draws four balls (without replacement) and the first three of those four are red, what is the probability that the fourth one is white? The answer is the same as with the original wording. We can make the original problem explicitly symmetrical over all seven of the equally likely prior conditions (which consist of there being 0, 1, 2,..., or 6 blue balls at the start) by letting pk denote the probability that we would draw 3 blue balls in a row GIVEN that there were originally k blue balls. Then define pt = p0+p1+p2+p3+p4+p5+p6 to give the "weights" (pk/pt) of the seven possible conditions. Of course, it turns out that p0 = p1 = p2 = 0, so the solution reduces to the same form as given originally. In other words, the problem has the same answer with the original wording as with the alternate wording. The only assumption that's needed (in either case) is that the prior probabilities of having 3, 4, 5, or 6 blue balls are all equal. It doesn't matter if these priors are each 1/4 (meaning they are are only four possible priors) or each 1/7 (meaning there could have been 0, 1, or 2 blue balls also). The reason is that the subsequent condition of drawing 3 blue balls in a row automatically sets the "post" probabilities of the 0,1,2 cases to zero. The priors of 1/7 or 1/4 are just used to determine the PROPORTIONALITY between the resulting non-zero weights. As long as their priors are all the same (all 1/4 or all 1/7) the final answer will be the same.

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