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.)

  • $\begingroup$ Did you try applying Bayes' Rule? $\endgroup$ – Fred Apr 23 '14 at 23:22
  • $\begingroup$ Hint: $2!=4$, and I'm considering $2!$ since $3-1=2$. $\endgroup$ – user122283 Apr 23 '14 at 23:23

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 $\frac{1}{4}$.


Let $R_1$ and $R_2$ be the event of picking a red ball on the first or second draw, respectively. Then: $$P(R_1 \mid R_2) = \frac{2}{8} = \frac{1}{4}$$

Reason: If you've picked 1 red ball then $2$ of the remaining $8$ balls are red. The order of doing this does not matter, so if you know the second ball is red then there is a $\frac14$ chance of the first ball was also red.

Or take the long way, using Bayes theorem and total probability:

We know: $\mathrm{P}(R_1)=\frac{3}{9}, \mathrm{P}(\overline{R_1})=\frac{6}{9}, \mathrm{P}(R_2\mid R_1)=\frac{2}{8}, \mathrm{P}(R_2\mid \overline{R_1})=\frac{3}{8}$

So:$$\begin{array}{l}\mathrm{P}(R_1 \mid R_2) \\ = \dfrac{\mathrm{P}(R_1\cap R_2)}{\mathrm{P}(R_2)} \\ = \dfrac{\mathrm{P}(R_1)\cdot\mathrm{P}(R_2\mid R_1)}{\mathrm{P}(R_1)\cdot\mathrm{P}(R_2\mid R_1)+\mathrm{P}(\overline{R_1})\cdot\mathrm{P}(R_2\mid\overline{R_1})} \\ = \dfrac{\frac{3}{9}\cdot\frac{2}{8}}{\frac{3}{9}\cdot\frac{2}{8}+\frac{6}{9}\cdot\frac{3}{8}} \\ = \dfrac{1}{4}\end{array}$$


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