# How to use bayes theorem to find the probability of the following

I'm not sure what the math would be for this. If you can explain that, it would be greatly appreciated!

There are three urns that each contain $10$ balls. The first contains $2$ white and 8 red balls, the second $8$ white and $2$ red, and the third all 10 white. Each urn has an equal probability of being selected. An urn is selected and a ball is drawn and replaced, and then another ball is drawn from the same urn. Suppose both are white.

What is the probability of first urn being selected? What is the probability of second urn being selected?

Would I use the following formula?: P(E)=P(F)(E|F) + P(F^C)P(E|F^C)

E being the event is the ball is white and F being the event if the urn was either the first or second

• Good evening Kevin Diez! Please provide your attempt(s)/thought(s) about a proposed solution. Thanks! – Rustyn Sep 28 '17 at 2:07
• I'm not quite sure how to follow through with it. My lecture for the class talked about @Rustyn – Kevin Diez Sep 28 '17 at 2:09
• P(E)=P(F) P(E|F) + P(F^C)P(E|F^C), would that be the formula to use? – Kevin Diez Sep 28 '17 at 2:11
• @KevinDiez Please specify what E and F are in this problem, and I do believe you need a little more than that formula, but I believe you are on the right track. – Isaac Browne Sep 28 '17 at 3:09
• Did you mean to ask what is the probability that the first urn is selected given that the two selected balls are white? – N. F. Taussig Sep 29 '17 at 10:38

Outline: Let $U_i$ be the event that Urn $i$ is chosen. $P(U_1) = P(U_2) = P(U_3) = 1/3.$
Denote the event $E = \{\text{Draw 2 white}\}$. Conditional on each of the three urns (separately) find $P(E|U_i).$ For example, $P(E|U_3) = 1.$
Then use Bayes' Theorem with three 'partition sets', the $U_i$'s.
$$P(U_1|E) = \frac{P(U_1 \cap E)}{P(E)} = \frac{P(U_1)P(E|U_1)}{\sum_{i=1}^3 P(U_i)P(E|U_i)}.$$