For $0\leq t \leq 3$
Type A customers arrive at a rate $\lambda_a=0.7\frac{-t^2+8t+84}{10}$
Type B customers arrive at a rate $\lambda_b=6$
What is the probability of $40$ Type A arrivals until either a single Type B arrival or time $t=3$
I already simplified the problem to this, using super-positioning.
If someone could show me how to do such probabilities, it would be great.
I know the tower property for Expectation, but not Probabilities: $E(X)=E(E(X|Y))$
I also know the probability of $40$ A arrivals before $t=3$
$$\Lambda=\int_0^3 0.7\frac{-t^2+8t+84}{10} dt$$
Then $\mathbb P(40)=e^{-\Lambda}\Lambda^{40}/40!$
Let $A_n$ and $B_n$ the arrival times of customers of type $A$ and $B$, respectively. Then $B_1\sim{\mathsf{Exp}}(6)$. Let $\tau=3\wedge B_1$, then $$\mathbb P(\tau=3)=\mathbb P(B_1>3) = e^{-18} $$ and for $0<t<3$ we have $$\mathbb P(B_1\leqslant t)=1-e^{-6t}. $$ It follows that $\tau$ has distribution $$F_\tau(t)=(1-e^{-6t})\mathsf 1_{(0,3)}(t) + e^{-18}\mathsf 1_{[3,\infty)}(t). $$ Let $$N_A(t)=\sup\{n: A_n\leqslant t\}$$ be the number of type $A$ arrivals in $(0,t]$. For $t>0$ we have $$\Lambda(t) = \int_0^t \frac7{100}(6+s)(14-s)\ \mathsf ds = \frac 7{300}t(252+12t -t^2),$$ and so \begin{align} \mathbb P(A_{40}>t) &=\mathbb P(N_A(t)\leqslant 39)\\ &= \sum_{j=0}^{39}\frac{\Lambda(t)^j}{j!}e^{-\Lambda(t)}\\ & = \mathbb P(G\geqslant\Lambda(t)), \end{align} where $G\sim\mathsf{Gamma}(40,1)$. Since $$\lim_{t\to\infty}\Lambda(t)=-\infty, $$ it follows that $A_{40}$ has density $$f_{A_{40}}(t) = \frac{\Lambda(t)^{39}}{39!}e^{-\Lambda(t)}\lambda(t)\mathsf 1_{(0,\infty)}(t).$$ The desired probability may then be computed by \begin{align}\mathbb P(A_{40}\leqslant \beta) &= \mathbb P(A_{40}\leqslant B_1,B_1<3) + \mathbb P(A_{40}\leqslant 3,B_1\geqslant 3)\\ &= \mathbb P(A_{40}\leqslant B_1)\mathbb P(B_1<3) + \mathbb P(A_{40}\leqslant 3)\mathbb P(B_1\geqslant 3)\\ &= \mathbb P(A_{40}\leqslant B_1)(1-e^{-18}) + \mathbb P(A_{40}\leqslant 3)e^{-18}. \end{align}
Here I will pause, because there is a technical problem - the intensity function of a nonhomogeneous Poisson process is normally assumed to be strictly positive, which is not the case here...