An urn has six balls numbered from 1 to 6. If two balls are drawn, what is the probability that the difference between the numbers on these two balls is equal to 1?
Solution 1: The pairs

$$
\begin{aligned}
& \{1,2\},\{1,3\},\{1,4\},\{1,5\},\{1,6\} \\
& \{2,3\},\{2,4\},\{2,5\},\{2,6\},\{3,4\} \\
& \{3,5\},\{3,6\}, \quad\{4,5\}, \quad\{4,6\}, \quad\{5,6\}
\end{aligned}
$$

are the only pairs of different balls that can be drawn from the urn. Therefore, there are $5+4+3+2+1=15$ different pairs of balls that can be drawn from the urn. Among these, there are only 5 pairs of balls numbered with numbers that differ by one unit, namely, $\{1,2\},\{2,3\},\{3,4\},\{4,5\}$, and $\{5,6\}$. Thus, the probability that one of these pairs is drawn is $\frac{5}{15}=\frac{1}{3}$.

Solution 2: Observe that if we draw the first ball with a number between 2 and 5, then among the five remaining balls in the urn, there are two possible balls that meet the condition of the problem. Therefore, in this case, the probability that the second ball meets the condition is $\frac{2}{5}$, and the probability that the first ball has a number between 2 and 5 is $\frac{4}{6}$. On the other hand, if the first ball drawn is 1 or 6, there is only one ball in the urn that meets the condition, so in this case, the probability for the second ball is $\frac{1}{5}$, and the probability that the first ball is 1 or 6 is $\frac{2}{6}$. Thus, the probability that the balls are consecutive is

$$
\frac{4}{6} \times \frac{2}{5}+\frac{2}{6} \times \frac{1}{5}=\frac{5}{15}=\frac{1}{3}
$$