Conditional probability/

Each of $n$ urn contains four white and six black balls,while another contains five white and five black balls. An urn is chosen at random from the $n+1$ urns and two balls are drawn for it, both being black. The probability that five white an three black balls remain in the chosen urn is $\frac{1}{7}$. Find $n$. My attempt: Let S be the event that we chose one of the first n urns, and H the event that five white and three black balls remain. $$p(H)=p(H \cap S)+p(H \cap S^{c})=p(S)p(H|S)+p(S^{c})p(H|S^{c})=0+\frac{1}{n+1}$$ . I think something went wrong in here. (I thought that the event H is the certain event if we chose $S^{c}$ and was the empty event if we chose an $S$). Any suggestions/hints?

• You seem to have taken no account of the fact an urn with fewer black balls is less likely to deliver two black balls – Henry Oct 21 '15 at 0:09

1 Answer

The events you should investigate are: $L$ the event that you choose the last urn, and $B$ the event of drawing two black balls.

The information you are given is, in essence, that : $\mathsf P(L) = \frac 1 {n+1} \\ \mathsf P(B\mid L) = \tfrac 2 9 \\ \mathsf P(B\mid L^\complement) = \tfrac {3}{9} \\ \mathsf P(L\mid B) =\tfrac 1 7$

Use this to find $n$.