Suppose that a store opens at 0pm and customers arrive according to a non-homogenous Poisson process ${N(t),t\geq0}$ with the intensity function $λ(t)=3t$ per hour.
a) Let $S_2$ denote the time at which the second customer arrives. Calculate $P[2<S_2\leq3|N(1)=1]$.
b) Find the probability that the total number of customers arriving at the store during the three intervals 1-1:15pm, 2-2:15pm and 3-3:30pm is at least 3.
Here is what I tried so far for a)
$P[2<S_2\leq3|N(1)=1]$
=$P[2<S_2|N(1)=1]$-$P[3\leq S_2|N(1)=1]$
=$P[N(2)\leq2|N(1)=1]$-$P[N(3)\leq2|N(1)=1]$
=$P[N(2)-N(1)\leq1]$-$P[N(3)-N(1)\leq1]$
=$e^{4.5}$-$e^{12}$+$4.5e^{4.5}$-$12e^{12}$
Can someone confirm if I'm on the right track ?