An urn has nine balls, numbered from 1 to 9. José and Maria each draw a ball from the urn simultaneously. With the balls drawn, they form a two-digit number, where the number on José's ball is the tens digit and the number on Maria's ball is the units digit. What is the probability that this number is even?
Solution 1: Let $a$ and $b$ be the numbers written on the balls drawn by José and Maria, respectively. There are, then, nine possibilities for $a$ and eight possibilities for $b$. Thus, there are $9 \times 8=72$ possibilities for the number $a b$. To count how many of these numbers $a b$ are even, we need to analyze two cases separately, as follows.

- Both 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 four possibilities for $a$ and three possibilities for $b$. Thus, there are $4 \times 3=12$ possibilities in total.

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

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

Solution 2: The parity of the number to be formed depends on the parity of the number written on the ball to be drawn by Maria. Among the integers from 1 to 9, there are five odd numbers, $1,3,5,7$ and 9, and four even numbers, $2,4,6$ and 8. Therefore, the probability that the number to be formed is even is $\frac{4}{5+4}=\frac{4}{9}$.