Poisson process So this is a problem from the Essentials of Stochastic Processes,by Richard Durrett.

A policewoman on the evening shift writes a Poisson mean $6$ number of tickets per hour. Two-third’s of these are for speeding and cost $\$100$. One-third’s of these are for DWI and cost ${$}400$. 



*

*(a) Find the mean and standard deviation for the total revenue from the tickets she writes in an hour. 

*(b) What is the probability that between $2$ and $3$ AM she writes five tickets for speeding and one for DWI. 

*(c) Let $A$ be the event that she writes no tickets between $1$ and $1:30$ AM, and $N$ be the number of tickets she writes between $1$ and $2$ AM. Which is larger $P(A)$ or $P(A\mid N = 5)?$ 


My try:


*

*(a) \begin{align}
E(X) & = 100\cdot(2/3)+400\cdot(1/3)=200\\[0.2cm] 
E(X^2) & = 100^2\cdot(2/3)+400^2\cdot(1/3)=60000 \\[0.2cm]
Var(X) & = E(X^2) - E(X)^2 = 20000\end{align} 

*(b) $X_S \sim \text{Pois}(6\cdot(2/3)) = \text{Pois}(4)$ and $X_D\sim \text{Pois}(6\cdot(1/3)) = \text{Pois}(2)$.


$\begin{align}P(X_D=1,X_S=5)~=~&\frac{e^{-\lambda P_{D}t} (\lambda P_Dt)^{X_D}}{X_D!}\frac{e^{-\lambda P_{S}t} (\lambda P_St)^{X_S}}{X_S!} \\~=~& \frac{e^{-2}(2)^1}{1!}\frac{e^{-4}(4)^5}{5!}\\~=~&2e^{-2}(128/15)e^{-4}\\~=~&256e^{-6}/15\end{align}$
But I am stuck on (c), so how to compare those two possibilities?

Comments on (c):
Without restriction the average number of tickets in the first half hour
will be 3, and the average number in the whole hour will be 6.
So if there are 5 tickets in the whole hour, intuitively it seems
less likely there were no tickets in the first half hour.
Here is a simulation using R statistical software, of a million performances of the experiment.
The first approximation should be accurate to almost three places;
the conditional approximation is based on only about 160,000 iterations and
so may be less accurate. Use @nicola's suggestion to see.
n1 = rpois(10^6, 3);  n2 = rpois(10^6, 3) # nr in 1st and 2nd half hr
n = n1 + n2    # nr in whole hour
mean(n1==0)
## 0.050041    # aprx P(A)
dpois(0, 3)
## 0.04978707  # exact P(A)
mean(n1[n==5]==0)
## 0.03132917  # aprx P(A|N=5)

