Given $4$ arrivals of a Poisson process, find the distribution of the time of the $3$rd arrival.

Let $\{N_t\}$ be a Poisson process of rate $\mu>0.$ If at time $t=10$ we have $N_{10}=4$ arrivals, what is the distribution of the time $T$ of the $3$rd arrival?

Attempt. If we were not given the total number of arrivals, the distribution of time $T$ would be $Gamma(3,\mu).$ Given the total number $N$ of arrivals though, the total times are uniformly equidistributed on interval $[0,T]$. Does this lead me somewhere?

Thank you in advance for the help.

• What you say is correct, so you are asking for the third order statistic of four independent uniform random variables on the interval $[0,10]$. – angryavian Jul 4 '18 at 18:03

Let $U_{(i)},~i=1,2,3,4$ be the uniformly equidistributed arrival times on $[0,10]$ ($U_{(i)}$ means the $i$-th arrival time $U_i,~i=1,2,3,4$ with respect to the function ordering). Then $U_{(3)}$ takes values on $[0,10]$ and $U_{(3)}\leq u$ if and only if:
either $U_1,~U_2,~U_3,~U_4\leq u$, with probability $(u/10)^4$,
either exactly one of the $U_i'$s is $>u$, with probability $4\,(u/10)^3\,(1-u/10)$.
So the desired distribution is given as: $$F_{(3)}(u)=P(U_{(3)}\leq u)=(u/10)^4+4\,(u/10)^3\,(1-u/10)$$ for $u\in [0,1]]$, as $0$ for $u<0$ and as $1$ for $u>10.$