Given integers $a,b,c,d\in[2^n,2^m]$ with $m>n>1$, how many primes $p$ are there in $[n^\alpha,n^\beta]$ for some $1<\alpha<\beta$ such that $$0<a\bmod p<n^{\alpha/k}$$ $$0<b\bmod p<n^{\alpha/k}$$ $$0<c\bmod p<n^{\alpha/2}$$ $$0<d\bmod p<n^{\alpha/2}$$ holds where $k>2$ is fixed?

Assume $n,m,\alpha,\beta,k$ are fixed.

Heuristically we have $$|\{p:a\bmod p<n^{\alpha/k}\}|\approx|\{p:b\bmod p<n^{\alpha/k}\}|\approx\frac{n^{\alpha/k}}{(n^{\beta}-n^\alpha)/\log(n^{\beta}-n^\alpha)}$$ $$|\{p:c\bmod p<n^{\alpha/2}\}|\approx|\{p:d\bmod p<n^{\alpha/2}\}|\approx\frac{n^{\alpha/2}}{(n^{\beta}-n^\alpha)/\log(n^{\beta}-n^\alpha)}$$

So $$\mathsf{Prob}(a\bmod p<n^{\alpha/k},b\bmod p<n^{\alpha/k},c\bmod p<n^{\alpha/2},d\bmod p<n^{\alpha/2}\underbrace{\approx}_{\mathsf{assuming}\mbox{ }\mathsf{independence}\mbox{ }\mathsf{of}\mbox{ }a,b,c,d}\frac1{n^{2\alpha(1/k+1/2)}}$$

So I think we should have approximately $$\frac{n^{\beta}-n^\alpha}{n^{2\alpha(1/k+1/2)}\log(n^{\beta}-n^\alpha)}$$ such prime numbers.

Am I correct?