# The urns

### Problem statement

There are 11 urns labeled by \( u \in \{0, 1, 2, ... 10\} \), each containing ten balls. Urn \( u \) contains \( u \) black balls and \( 10 -u \) white balls. Fred selects an urn \( u \) uniformly at random and draws \( N \) times with replacement from that urn, obtaining \( n_B \) blacks and \( N-n_B \) whites. Fred's friend, Bill, looks on. If after \( N=10 \) draws \( n_B = 3 \) blacks have been drawn, what is the probability that the urn Fred is using is urn \( u \) from Bill's point of view?

### Solution

Firstly, we know how to express the probability distribution of the number of blacks given the urn and the total ball count:

Where we define [ \( f_u := ? \) ].

Adding the \( u \) dimension creates the joint probability distribution of the random variables \( u \) and \( n_b \), \( P(u, n_B | N) \). It can be written as:

Notices how \( P(u) \) appears, the probability that Fred selects a given urn, which is uniform at \( \frac{1}{11} \) as per the question.

From here, we need Bayes Theorem, which is:

From the joint probability of \( u \) and \( n_B \) we can obtain the conditional distribution of \( u \) given \( n_B \) using Bayes Theorem:

We so far know 2 of the 3 expressions in that quotient. The third, the denominator, \( P(n_B|N) \), is the marginal probability of \( n_B \) which we can obtain using the sum rule:

So the conditional probability of \( u \) given \( n_B \) is:

We must calculate and sum all probabilities of the form (1) above, then p the sum by the probabilities from (1) for the given urn.

By the way, the sum for when \( n_B \) is 3 is: \( P(n_B = 3|N=10) = 0.083 \).

Todo: compare this to the bent coin.