# conditional probability density of arrival times

I am trying to figure out how to solve problem:

Let $N(t)$ be a Poisson process of rate $λ$. Given that $N(t) = 3$, determine the conditional probability density functions of each of the three arrival times $S_1$, $S_2$, $S_3$.

I know that $S_1$ and $S_2$ are uniformly distributed at $[0,1]$ and $S_3$ seems to be independent of them, so I can find its pdf from joint density of $S_1$, $S_2$ and $S_3$ which is equal to $3!/t^3$. Then pdfs of $S_1$ and $S_2$ are $1/t$ and pdf of $S_3 = 6/t$. Or maybe they are all the same $(2/t)$?

$S_1$ and $S_2$ are not uniformly distributed, and just where you get $[0,1]$ you don't tell us. The support of the distribution of each of the three random variables $S_1$, $S_2$, $S_3$ is $[0,t]$, not $[0,1]$.
In fact $S_1,S_2,S_3$ have the same distribution as the order statistics from an i.i.d. sample of size $3$ from a uniform distribution. That implies that none of them is uniformly distributed.
Unconditionally, or marginally (if you prefer that word), we have $\Pr(S_1<s) = 1 - e^{-\lambda s}$ for $s>0.$ We want $\Pr(S_1<s\mid N(t)=3)$.
This is the c.d.f. on the interval $[0,t]$. Differentiating it with respect to $s$ gives you the p.d.f. on that interval. If you differentiate this with respect to $s$, using the product rule, you'll get five terms, and all but one of them will cancel.
$S_2$ and $S_3$ can be handled similarly.