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
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.
$$P(U_1|E) = \frac{P(U_1 \cap E)}{P(E)} = \frac{P(U_1)P(E|U_1)}{\sum_{i=1}^3 P(U_i)P(E|U_i)}.$$
The computation in the denominator uses what is often called the Law of Total Probability.