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.)
There are two ways for the second ball to be red. Either blue then red, or red then red. If a blue is picked first, then the chance of a red is $\frac{3}{8}$. If a red is picked first, the odds are $\frac{2}{8}$. For the first pick, red happens $\frac{1}{3}$ of the time and blue $\frac{2}{3}$. The overall odds of blue then red is $\frac{2}{3}$ times $\frac{3}{8}$ or $\frac{6}{24}$ . The overal odds of it being red then red is $\frac{1}{3}$ times $\frac{2}{8}$ or $\frac{2}{24}$. Together the odds are $\frac{8}{24}$ that the second one is red, and out of these $\frac{8}{24}$ only $\frac{2}{24}$ have a red first. $\frac{2}{24}$ divided by $\frac{8}{24}$ is $\frac{1}{4}$.