# Posterior probability or Prior probability

I have an arguement with my friends on a probability question.

Question: There are lots of stone balls in a big barrel A, where 60% are black and 40% are white, black and white ones are identical, except the color.

First, John, blindfolded, takes 110 balls into a bowl B; afterwards, Jane, blindfolded also, from bowl B takes 10 balls into cup C -- and find all 10 balls in C are white.

Now, what's the expectation of black balls in bowl B?

It seems there are 3 answers

Answer 1 My friend thinks the probability of black stones in bowl B is still 60%, or 60% * 100 = 60 black balls expected in bowl B.

Answer 2 However, I think 60% is just prior probability; with 10 balls all white from bowl B, we shall calculate the posterior probability. Denote $B_k$ as the black ball numbers in bowl B, and $C_{w}=10$ as the event that 10 balls in cup C are all white.

$$E(B_B | C_w=10) = \sum_{x=10}^{110}\left[x P(B_k = x | C_w=10)\right] = \sum_{x=10}^{110} \left[x \frac{P(B_k = x \text{ and } C_w=10)}{P(C_w=10)}\right] = \sum_{x=10}^{110} \left[x \frac{P(B_k = x) P( C_w=10 | B_k=x)}{P(C_w=10)}\right]$$

, where $$P(C_w=10) = \sum_{x=10}^{110} \left[ P(B_k = x) P(C_w = 10 | B_k=x) ) \right]$$

, and according to binomial distribution $$P(B_k=x) = {110 \choose x} 0.6^x 0.4^{110-x}$$ , and $$P(C_w = 10 | B_k=x) = \frac{1}{(110-x)(110-x-1)\cdots (110-x-9)}$$

Answer 3 This is from Stefanos below: You can ignore the step that John takes 110 balls into a bowl B, this does not affect the answer. The expected percentages in bowl $B$ are again $60\%$ and $40\%$ percent, i.e. 66 black balls and 44 white balls. Now, after Jane has drawn 10 white balls obviously the posterior probability changes, since the expected number of balls is now 66 and 34. So, you are correct.

I sort of don't agree with Stefanos that, the black ball distribution from bowl B could vary a lot from barrel A, as the sampling distribution could be different from the universe distribution.

in other words, if Janes draws a lot balls and all are white, e.g. 50 balls are all white, I fancy it's reasonable to suspect that bowl B does not have a 60%-40% black-white distribution.

You can ignore the step that John takes 110 balls into a bowl B, this does not affect the answer. The expected percentages in bowl $B$ are again $60\%$ and $40\%$ percent, i.e. 66 black balls and 44 white balls. Now, after Jane has drawn 10 white balls obviously the posterior probability changes, since the expected number of balls is now 66 and 34. So, you are correct.