An urn has 9 balls, numbered with the numbers from 1 to 9. José and Maria simultaneously draw a ball from the urn. With the balls drawn, they form a two-digit number, where the number written on José's ball is the tens digit and the number written on Maria's ball is the units digit. What is the probability that this number is even?
Let $a$ and $b$ be the numbers written on the balls drawn by José and Maria, respectively. There are then 9 possibilities for $a$ and 8 possibilities for $b$. Thus, there are $9 \times 8=72$ possibilities for the number $a b$.

On the other hand, to count how many of these numbers are even, we need to analyze two cases separately:

- the numbers $a$ and $b$ are even;
- the number $a$ is odd and the number $b$ is even.

In the first case, where $a$ and $b$ are even, there are 4 possibilities for $a$ and 3 possibilities for $b$. Thus, there are $4 \times 3=12$ possibilities.

In the second case, where $a$ is odd and $b$ is even, there are 5 possibilities for $a$ and 4 possibilities for $b$. Thus, there are $5 \times 4=20$ possibilities.

Therefore, the probability that the number $a b$ is even is $\frac{12+20}{72}=\frac{32}{72}=\frac{4}{9}$.