Chapter 3: Q.3.6 (page 109)

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 $b / (b + r + c)$ .

Expert verified

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

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

Step by step solution

Events:

$B_{1}$ - first drawn ball is black.

$R_{2}$ - second ball drawn is red.

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

P (B_{1}) = \frac{b}{b + r} \Rightarrow P (B_{1}^{c}) = \frac{r}{b + r}

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

$P (R_{2} ∣ B_{1}) = \frac{r}{r + b + c}$

$P (R_{2} ∣ B_{1}^{c}) = \frac{r + c}{r + b + c}$

$B_{1}$ and $B_{1}^{c}$ are mutually exclusive events whose union is the whole outcome space.

That is when we can use the Bayes formula:

$P (B_{1} ∣ R_{2}) = \frac{P (R_{2} ∣ B_{1}) P (B_{1})}{P (R_{2} ∣ B_{1}) P (B_{1}) + P (R_{2} ∣ B_{1}^{c}) P (B_{1}^{c})}$ .

All that is left is to substitute the known values:

P (B_{1} ∣ R_{2}) = \frac{\frac{r}{r + b + c} \cdot \frac{b}{b + r}}{\frac{r}{r + b + c} \cdot \frac{b}{b + r} + \frac{r + c}{r + b + c} \cdot \frac{r}{b + r}} = \frac{b}{r + b + c}

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

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

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