The integers $r$ and $k$ are randomly selected, where $-3 < r < 6$ and $1 < k < 8$. What is the probability that the division $r \div k$ is an integer value? Express your answer as a common fraction.
The total number of possible outcomes is the number of ways to choose $r$ and $k$ from their respective ranges. There are 8 possible values for $r$, from $-2$ to $5$, and 6 possible values for $k$, from $2$ to $7$. So the total number of possible outcomes is $8 \times 6 = 48$.
[*} The favorable outcomes are those where $r \div k$ is an integer, which means that $k$ must be a factor of $r$. For $r = -2$, the factors are $\pm 1$ and $\pm 2$. Only $k = 2$ is in the range for $k$, so there is one favorable outcome. For $r = -1$, the only factor is $\pm 1$. There is no matching value of $k$, so there are no favorable outcomes. For $r = 0$, any value of $k$ is a factor, so there are 6 favorable outcomes. For $r = 1$, the only factor is $\pm 1$. There is no matching value of $k$, so there are no favorable outcomes. For $r = 2$, the factors are $\pm 1$ and $\pm 2$. Only $k = 2$ is in the range for $k$, so there is one favorable outcome. For $r = 3$, the factors are $\pm 1$ and $\pm 3$. Only $k = 3$ is in the range for $k$, so there is one favorable outcome. For $r = 4$, the factors are $\pm 1$, $\pm 2$, and $\pm 4$. Only $k = 2$ and $k = 4$ are in the range for $k$, so there are two favorable outcomes. For $r = 5$, the factors are $\pm 1$ and $\pm 5$. Only $k = 5$ is in the range for $k$, so there is one favorable outcome. Adding up all the favorable outcomes, I get $1 + 0 + 6 + 0 + 1 + 1 + 2 + 1 = 12$. So the probability that $r \div k$ is an integer is $\frac{12}{48} = \frac{1}{4}$. 