Question

An urn contains $b$ black balls and r red balls. One of the balls is drawn at random, but when it is put back in the urn, $c$ additional balls of the same color are put in with it. Now, suppose that we draw another ball. Show that the probability that the first ball was black, given that the second ball drawn was red, is$b/(b+r+c)$.

Short answer

Use the Bayes formula, depending on which color was drawn first

$\frac{b}{r+b+c}$.

Step-by-step solution

Given Information

Events:

$B_1$- first drawn ball is black.

$R_2$- second ball drawn is red.

Explanation

As there are $r$ red and $b$ black balls to begin with:

$P\left(B_1\right)=\frac{b}{b+r}\Rightarrow P\left(B_1^c\right)=\frac{r}{b+r}$

As $c$balls of the corresponding color are added the conditional distribution of balls is such that:

$P\left(R_2\mid B_1\right)=\frac{r}{r+b+c}$

$P\left(R_2\mid B_1^c\right)=\frac{r+c}{r+b+c}$

Explanation

$B_1$and $B_1^c$are mutually exclusive events whose union is the whole outcome space.

That is when we can use the Bayes formula:

$P\left(B_1\mid R_2\right)=\frac{P\left(R_2\mid B_1\right)P\left(B_1\right)}{P\left(R_2\mid B_1\right)P\left(B_1\right)+P\left(R_2\mid B_1^c\right)P\left(B_1^c\right)}$.

All that is left is to substitute the known values:

$P\left(B_1\mid R_2\right)=\frac{\frac{r}{r+b+c}·\frac{b}{b+r}}{\frac{r}{r+b+c}·\frac{b}{b+r}+\frac{r+c}{r+b+c}·\frac{r}{b+r}}=\frac{b}{r+b+c}$.

Final Answer

Use the Bayes formula, depending on which color was drawn first

$\frac{b}{r+b+c}$.