Check out some similar questions!

Adamcustom · Aug 24, 2009, 03:14 AM

there is a 2% probability that an urn with my name in it contains 10 names only, and a 98% probability that it contains a thousand. These " prior" probabilities are my personal estimates, before any name are drawn. If my name is among the first three drawn, then the posterior probability of there having been only ten names is..
How should calculate it using Bayes' rule. I know the answer is ar.67%, but how to arrive at it?

Adamcustom

ebaines · Aug 24, 2009, 06:55 AM

Hello Adamcustom. You're going to have to be a bit clearer in wording this problem, because I don't understand what you'e gettng at. You have an urn that may only have 10 slips of paper with names on them, or it may have 1000 slips of papre with 1000 names? Or do you mean that the urn is known to have 1000 slips of paper and they may have just 10 names repeated over and over, or they may all be individual names? Is there some additional condition having to do with the presence of your name on one of the slips of paper? If this is a homework question, please copy the question exactly as it is written.

galactus · Aug 24, 2009, 11:06 AM

Let the urn with 10 names be A and the urn with 1000 names be B.

The prob of choosing urn A is .02 and of choosing urn B is .98.

$P(\text{urn A}\wedge \text{your name})=(.1)(.02)=.002$

$P(\text{your name})=(.02)(.1)+(.001)(.98)=.00298$

$\frac{.002}{.00298}=.67$

This is assuming I am interpreting correctly.