Consider a collection of $N+1$ urns, numbered $0,1,2,..N$ , each containing a total of $N$ red and white balls.The urn number $K$ has $K$ red and $N-K$ white balls .$(K=0,1,...N)$.An urn is chosen at random and $n<N$ random drawings are made from it by without replacement .What is the conditional probability that all the $n$ balls drawn are taken from the $N$th urn given that they are all red?

I am defining $3$ events: $E_1$: AN urn is chosen at random and $n$ drawings are made

$E$:All balls come out to be red

$E_2$:The $N$th urn is chosen.

We need $P(E_2|E)$.

But I can't evaluate this.

  • $\begingroup$ Have you heard about Bayes' formula? $\endgroup$ – Arthur Dec 7 '16 at 9:20
  • $\begingroup$ @Arthur: This problem is a bit deeper than simply applying Bayes formula... $\endgroup$ – Richard Ambler Dec 7 '16 at 10:20
  • $\begingroup$ @RichardAmbler I agree that $P(E)$ is a bit of work to calculate, but once you have that, $\frac{P(E\mid E_2)P(E_2)}{P(E)}$ is relatively straight-forward. $\endgroup$ – Arthur Dec 7 '16 at 10:26

The $k$th urn contains $k$ red balls and $N-k$ white balls. Since the $n$ balls are taken without replacement, the number of red balls taken, given the urn, follows a hypergeometric distribution. Thus, the probability of selecting $n$ red balls given the $k$th urn is given by:

$$P(n \text{ red balls}|\text{urn } k) = \frac {\displaystyle{k \choose n}{N-k \choose 0}}{\displaystyle {N \choose n}}=\frac {\displaystyle {k \choose n}}{\displaystyle {N \choose n}}$$

Since the urn is chosen at random, our implicit prior probability that the $k$th urn is selected is $\displaystyle \frac 1 N$, i.e. each urn is equally likely to be chosen.

Thus, "reversing the condition" (and simplifying) we get:

$$P(\text{urn } k|n \text{ red balls})=\frac{\displaystyle{k \choose n}}{\displaystyle\sum_{j=0}^{N}{j\choose n}}$$


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