An urn contains 3 red balls and 6 blue balls. Two balls are drawn without replacement and the second is found to be red. What is the probability that the first ball was also red?
I thought it would simply be 1/3 because that was the initial chance of drawing a red. Then I tried multiplying 1/3 by 1/4 to get a 1/12 but that was wrong too. The answer is 1/4. Would someone mind explaining how to do this problem? :/ I've looked at other urn problems, but I don't think any others had this question except this person. But it had no answers (Edit: it does now.)