A bag holds four balls. One of them is white. Each of the others is either black or white at equal chance. You randomly draw out two balls and discover they are both white. If you then randomly draw a third ball, what is the chance that it is white?
I've seen answers varying from 0.27 through 0.58 to 0.75. What is the correct answer?
The probabilities that we have $(1,2,3,4)$ white balls in the bag are $$\left({1\over8},{3\over8},{3\over8},{1\over8}\right)\ .$$ Given that we have $(1,2,3,4)$ white balls in the bag the probabilities that we draw two white balls are $$\left(0,{1\over6}, {1\over2},1\right)\ .$$ The overall probability $p(2w)$ to draw two white balls then is given by $$p(2w)={1\over8}\cdot0+{3\over8}\cdot{1\over6}+{3\over8}\cdot{1\over2}+{1\over8}\cdot1={3\over8}\ .$$ Finally the probability that all three balls are white computes to $$p(3w)={1\over8}\cdot0+{3\over8}\cdot0+{3\over8}\cdot{1\over4}+{1\over8}\cdot1={7\over32}\ .$$ It follows that the conditional probability $p(3w|2w)$ is given by $$p(3w|2w)={p(3w\wedge 2w)\over p(2w)}={p(3w)\over p(2w)}={7\over12}=0.58333\ldots\ .$$