Consider an urn that contains one black ball, four white balls, and some blue balls. A ball is randomly drawn from this urn, its color is observed, and the ball is returned to the urn. Then, another ball is randomly drawn from this urn. For which quantities of blue balls is the probability that the two balls drawn have the same color equal to $1 / 2$?
Let's call n the number of blue balls in the box. When we draw the two balls, they can be:

- Two black balls. The probability is

$$
\frac{1}{n+5} \times \frac{1}{n+5}=\left(\frac{1}{n+5}\right)^{2}
$$

- Two white balls. The probability is

$$
\frac{4}{n+5} \times \frac{4}{n+5}=\left(\frac{4}{n+5}\right)^{2}
$$

- Two blue balls. The probability is

$$
\frac{n}{n+5} \times \frac{n}{n+5}=\left(\frac{n}{n+5}\right)^{2}
$$

Therefore, the probability that the two balls are of the same color is the sum of the individual probabilities:

$$
\left(\frac{1}{n+5}\right)^{2}+\left(\frac{4}{n+5}\right)^{2}+\left(\frac{n}{n+5}\right)^{2}=\frac{1+16+n^{2}}{(n+5)^{2}}=\frac{1}{2}
$$

Simplifying the equation, we get $n^{2}-10 n+9=0$, from which $n$ is equal to 1 or 9.