Conditional Probability with a Ball Chosen From a Random Urn I have 10 Urns; 9 contain 2 White and 2 Black Balls, one urn contains 5 White and one Black ball. A ball chosen from a random urn is white. What is the probability that it came from the urn with the five white balls? 
I know I have to use Bayes Theorem here, but I am not quite sure how to use it in this given case. Any help is highly appreciated. 
Thanks, Daniel 
 A: Bayes theorem says that:
$$
P(A|B)P(B)=P(B|A)P(A)
$$
let $A$ be the event of the urn you draw from having 5 white balls, and $B$ be the event 'draw a white ball'. Now, you want $P(A|B)$, and this can be expressed in the other three probabilities, which are easier to calculate.
A: To elaborate a little on the previous answer. Number your urns 1 through 10, with the tenth urn being the one with 5 white and 1 black ball, and let P(10|W) denote the conditional probability that the ball came from the 10th urn, given that it is white.
From the definition of conditional probability, you know: 
P(10|W) = P(W and 10th)/P(W) 
You can compute the probability that it came from the tenth urn and is white, since this is just (1/10)(5/6). So the question is what is P(W)? But we can decompose this event as the sum of the disjoint events: 
P(W) = P(W and 1) + P(W and 2) + .... P(W and 10) 
Note that if n is not equal to 10 (i.e. n=1,2,...,9), then P(n and W) = (1/10)*(1/2), so 
P(W) = (1/10)(9/2 + 5/6) = 32/60 = 8/15
