How to generate pairwise independence

Select a prime p, $n\le p\le 2n$. Such a prime exists because of the density of primes. For each vertex u, choose an integer a_u such that $\frac{a_u}{p}\approx \frac 1{2d(u)}$, and an interval A_u, |A_u|=a_u in Z_p.

Let X(u) be the random variable from uniformly choosing $x,y\in Z_p$ and setting

$$X(u)= \begin{cases} 1&\text{ if $xu+y\in A_u$, and}\ 0& \text{ otherwise.} \end{cases}$$ (13)

This means that $\Pr(X(u)=1)=\frac {a_u}p\approx \frac 1{2d(u)}$ because
$$\begin{multline} \Pr_{x,y}(X(u)=1)=\Pr(\exists a\in A_u: xu+y=a)\ =\sum_{a\in A_u}\Pr(x=(a-y)u^{-1}) = \sum_{a\in A_u}\frac 1p=\frac {a_u}p.\end{multline}$$
For pairwise independence, we would like to conclude that

$$\Pr(X(u)=1, X(v)=1)=\Pr(X(u)=1)\Pr(X(v)=1)=\frac{a_ua_v}{p^2}.$$ (14)

This is also true because of the following.

$$\begin{split} \Pr(X(u)=1, X(v)=1) &= \Pr(\exists a\in A_u,... ...b)\in A_u\times A_v}\frac1{p^2}=\frac{a_ua_v}{p^2} \end{split}$$ (15)