Consider 3 urns. Um A contains 2 white and 4 red balls, um B contains 8 white and 4 red balls, and um C contains 1 white and 3 red balls. If 1 ball is selected from each urn, what is the probability that the ball chosen from um A was white given that exactly 2 white balls were selected?
This problem is in the conditional probability unit, so I'm confident that you have to use P(E|F) = P(E∩F) / P(F)... but since E and F aren't independent I don't know how to calculate P(E∩F)...
Thanks
Either through experimentation or theory. Here you can evaluate from the model of drawing balls from urns without bias.
Here you have $F$, the event of obtaining exactly two white balls when drawing one ball from each of the three urns, and $E_A, E_B, E_C$, the respective events of obtaining a white ball when drawing from the indicated urn. Note: these events will be independent, and you can easily calculate $\mathsf P(E_A)$, $\mathsf P(E_B)$ and $\mathsf P(E_C)$
Then $F= (E_A\cap E_B\cap E_C^\complement)\cup(E_A\cap E_B^\complement\cap E_C)\cup(E_A^\complement\cap E_B\cap E_C)$
You seek $\mathsf P(E_A\mid F)$, and do know :
$$\begin{align}\mathsf P(E_A\mid F) &=\dfrac{\mathsf P(E_A\cap F)}{ \mathsf P(F)}\\[2ex]&=\dfrac{\mathsf P\big(E_A\cap((E_B\cap E_C^\complement)\cup(E_B^\complement\cap E_C))\big)}{\mathsf P(F)}\end{align}$$
I shall leave the rest to you.