If I assume that $\Pr\{N(t) : t \ge 0\}$ is a Poisson process with intensity $\lambda$, I'm supposed to evaluate $\Pr\{N(s)=3\mid N(t)=8\}$ for $0 < s < t$.
Since $N(t)$ is poisson, would I just use this formula?:
$$\Pr\{X(s)=k\mid X(t)=n\}=\frac{n!}{k!(n-k)!} \left(\frac{s}{t}\right)^k \left(1-\frac{s}{t}\right)^{n-k}$$
Hence, since 0 < s < t, we know the distribution of arrivals in $[0,t]$ is Uniform. Since each occurrence is uniform and independent of all others, The formula you stated is correct.
Basically, view the events $A_i$ = {arrival i occurs in the first s minutes of $[0,t]$} for $1 ≤ i ≤ 8$
$A_i$ occurs with probability $\frac st$ by uniformity. Each $A_i$ also occurs independently. If we look at the $A_i$'s as independent Bernoulli trials, we can view the probability that 3 of the $A_i$ occur in $[0,s]$ as a binomial random variable with $n =8, k=3, p = \frac st$
Here's a good quick reference:
http://www.columbia.edu/~ww2040/3106F14/Concise.pdf
http://www.columbia.edu/~ww2040/3106F14/lec1014sols.pdf