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DISCLAIMER: This is school related, however it's not homework. This is from an exercise document one of the previous students posted. Sadly, there are no solutions, so I can't verify if I'm thinking about the problem the right way.

The problem:

There are ten boxes. In 9 of the boxes, there are 2 white balls and 2 black balls. In the tenth box there is 5 white and 1 black ball. By random chance, a white ball was picked. What's the probability the white ball was inside the tenth box?

To illustrate my understanding of the problem:


The way I am thinking about solving the problem

  1. $H_{1...9}$: Picked from inside a box containing 2b & 2w balls
  2. $H_{10}$: Picked from inside a box containing 1b & 5w balls;
  3. Picking a box: $\Pr(H_{1...10})=\frac{1}{10}$
  4. A white ball is picked: $A$.
  5. The white ball was picked from one of the nine boxes: $\Pr(A|H_{1...9})=\frac{2}{4}$

  6. The white ball was picked from the tenth box: $\Pr(A|H_{10})=\frac{5}{6}$

  7. Calculate that the probability that the white ball was picked from within the tenth box: $$ \Pr(H_{10}|A) = \frac{\Pr(H_i)\Pr(A|H_i)}{\Pr(A)} $$

    Where $\Pr(A)$ is:

    $$\Pr(A) = \sum_{n=1}^{10}\Pr(H_n)\Pr(A|H_n)$$

    So, in the end I'm left with:

$$ \Pr(H_{10}|A) = \frac{\Pr(H_i)\Pr(A|H_i)} {9*[\Pr(H_{1...10})\Pr(A|H_{1...9})]+ \Pr(H_{1...10})\Pr(A|H_{10})} $$

Can someone point me in the right direction if this is wrong? Thanks! Also, sorry for creating noise, in the unlikely scenario that this is actually correct.

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You are absolutely correct with the approach and I believe in your last step, the denominator should be $9[Pr(H_{1...9})Pr(A|H_{1...9})]+Pr(H_{10})Pr(A|H_{10})$. Just pointing out a notation error. A classic example of Bayes !! The worst thing that someone could do to this problem is provide some prior probabilities about picking boxes. Then you just would have had to play around with $P(H_{i})$ too. – Sudarsan Sep 23 '13 at 0:56
@Sudarsan Oh, thanks! I used $\Pr(H_{1...10})$ because it's same for all of the ten hypotheses: $\frac{1}{10}$, but it does look neater with your notation! Thanks :-) – omninonsense Sep 23 '13 at 1:06
Also I hope you could see the difference between the following 2 questions here: a. A ball is drawn at random; what is the probability that it is a white ball and it is from the 10th box? (A slightly different question) b. A ball is drawn at random and it is white; what is the probability that it came from the 10th box? (Your question). As an exercise on probability it's always good to think about more questions as there are always subtle differences. – Sudarsan Sep 23 '13 at 1:24
@Sudarsan Oh yes, I see the difference. – omninonsense Sep 23 '13 at 1:45
up vote 3 down vote accepted

What you’ve done is correct, though you can shorten it a little by thinking of the first nine boxes as one box that you pick with probability $0.9$. Then the probability of drawing a white ball is $$\frac9{10}\cdot\frac12+\frac1{10}\cdot\frac56=\frac{32}{60}\;,$$ to which the last box contributes $\frac5{60}$, so the desired probability is $$\frac{5/60}{32/60}=\frac5{32}\;.$$

Or you can be really slick and notice that the probabilities aren’t changed if we add one white and one black ball to each of the first nine boxes. Now, however, all boxes contain the same number of balls, so picking a box at random and then drawing a ball from the box is equivalent to picking a ball at random: each ball has one chance in $60$ of being chosen. And since there are now $32$ white balls, $5$ of which are in the last box, we get the desired result immediately.

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Oh, this is just brilliant, but I'm afraid this way of thinking is way too elegant and advanced for me! I can just marvel at your logic. Thanks! – omninonsense Sep 23 '13 at 1:11
@destielstarship: You’re welcome! I have to admit that I didn’t notice the slick solution until I’d already done the other calculation and was looking for a nice way to check it. – Brian M. Scott Sep 23 '13 at 1:13

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