Question

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.

Answer 1

Denote by
`R,K` the random variables representing the integer values
`r,k`, respectively. Then
`R\sim\mathrm{Unif}(-2,5)` and
`K\sim\mathrm{Unif}(2,7)`, where
`\mathrm{Unif}(a,b)` denotes the discrete uniform distribution over the interval
`[a,b]`. So
`R` and
`K` have probability mass functions

`p_R(r)=\begin{cases}\frac18&\text{for }r\in\{-2,-1,\ldots,5\}\\\\0&\text{otherwise}\end{cases}`

`p_K(k)=\begin{cases}\frac16&\text{for }k\in\{2,3,\ldots7\}\\\\0&\text{otherwise}\end{cases}`

We want to find
`P\left(\frac RK\equiv0\pmod n\right)`, where
`n` is any integer.

We have six possible choices for
`K`:

(i) if
`K=2`, then
`\frac RK` is an integer when
`R=\pm2,0,4`;

(ii) if
`K=3`, then
`\frac RK` is an integer when
`R=0,3`;

(iii) if
`K=4`, then
`\frac RK` is an integer when
`R=0,4`;

(iv) if
`K=5`, then
`\frac RK` is an integer when
`R=0,5`;

(v) if
`K=6` or
`K=7`, then
`\frac RK` is an integer only when
`R=0` in both cases.

If the selection of
`R,K` are made independently, then the joint distribution is the product of the marginal distribution, i.e.

`p_(R,K)(r,k)=p_R(r)\cdot p_K(k)=\begin{cases}\frac1{48}&\text{for }(r,k)\in[-2,5]*[2,7]\\\\0&\text{otherwise}\end{cases}`

That is, there are 48 possible events in the sample space. We counted 12 possible outcomes in which
`\frac RK` is an integer, so the probability of this happening is
`(12)/(48)=\frac14`.