Denote by 
the random variables representing the integer values 
, respectively. Then 
and 
, where 
denotes the discrete uniform distribution over the interval ![[a,b]](https://tex.z-dn.net/?f=%5Ba%2Cb%5D)
. So 
and 
have probability mass functions
We want to find 
, where 
is any integer.
We have six possible choices for :
(i) if 
, then 
is an integer when 
;
(ii) if 
, then is an integer when 
;
(iii) if 
, then is an integer when 
;
(iv) if 
, then is an integer when 
;
(v) if 
or 
, then is an integer only when 
in both cases.
If the selection of 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}\dfrac1{48}&\text{for }(r,k)\in[-2,5]\times[2,7]\\\\0&\text{otherwise}\end{cases}](https://tex.z-dn.net/?f=p_%7BR%2CK%7D%28r%2Ck%29%3Dp_R%28r%29%5Ccdot%20p_K%28k%29%3D%5Cbegin%7Bcases%7D%5Cdfrac1%7B48%7D%26%5Ctext%7Bfor%20%7D%28r%2Ck%29%5Cin%5B-2%2C5%5D%5Ctimes%5B2%2C7%5D%5C%5C%5C%5C0%26%5Ctext%7Botherwise%7D%5Cend%7Bcases%7D)
That is, there are 48 possible events in the sample space. We counted 12 possible outcomes in which is an integer, so the probability of this happening is 
.
