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 the pdfs of $S_1$ and $S_2$ are $1/t$ and pdf of $S_3 = 6/t$. Or maybe they are all the same $(2/t)$?
 A: $S_1$ 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 $\Pr(S_1<s\mid N(t)=3)$.
\begin{align}
& \Pr(S_1<s \mid N(t) = 3) = \frac{\Pr(S_1<s\ \&\ N(t)=3)}{\Pr(N(t)=3)} \\[10pt]
= {} & \Pr(N(s)\ge 1 \mid N(t)= 3) \\[10pt]
= {} & \frac{\Pr\left( \begin{array}{r} \big(N(s)=1\ \&\ N(t)-N(s) = 2\big) \\
\text{ or } \big( N(s)=2\ \&\ N(t)-N(s)=1 \big) \\
\text{ or } \big( N(s)=3\ \&\ N(t)-N(s)=0 \big) \end{array} \right)}{\dfrac{\lambda^3 e^{-\lambda t}}{3!}} \\[10pt]
= {} & \frac{ \left( \begin{array}{r} \dfrac{(\lambda s)^1 e^{-\lambda s}}{1!} \cdot\dfrac{(\lambda(t-s))^2 e^{-\lambda (t-s)}}{2!} \\[10pt]
{} + \dfrac{(\lambda s)^2 e^{-\lambda s}}{2!} \cdot \dfrac{(\lambda(t-s))^1 e^{-\lambda (t-s)}}{1!} \\[10pt]
{} + \dfrac{(\lambda s)^3 e^{-\lambda s}}{3!} \cdot\dfrac{(\lambda(t-s))^0 e^{-\lambda (t-s)}}{0!} \end{array} \right) }{\dfrac{\lambda^3 e^{-\lambda t}}{3!}} \\[10pt]
= {} & \frac{\dfrac{s^1 (t-s)^2}{1!2!} + \dfrac{s^2(t-s)}{2!1!} + \dfrac{s^3}{3!}}{1/(3!)} \\[10pt]
= {} & \binom 3 1 s^1 (t-s)^2 + \binom 3 2 s^2 (t-s)^1 + \binom 3 3 s^3.
\end{align}
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.
