Suppose the number of cars arriving at a gas station per hour follows a Poisson Process with $\lambda = 3$, Suppose $6$ cars arrive after $2$ hours, what is the probability only $1$ arrived in the first hour?
So I know that $P(X_t=k)=e^{\lambda t}\frac{(\lambda t)^k}{k!}$ represents the probability the process is at k at time t.
So because the states are independent, I should just have the $P(X_1=1)P(X_1=5)$? Since the probability of $5$ cars arriving in the 2nd hour should be the same as the probability $5$ arrive in the first hour?
Which gives me $P(X_1=1)P(X_1=5)=(3e^3)(\frac{3^5e^3}{5!})$?
Or does independence mean knowing the total after $2$ hours doesn't matter? And I only need to compute $P(X_1=1)$
Best Answer
Paul gave some good hints — I’ll flesh it out a bit if that helps.
So we know that over 2 hours we got 6 cars and want to know if only 1 of those 6 arrived in the first hour.
Let $X,Y \sim \text{Poi(3)}$ and $Z=X+Y \implies Z\sim \text{Poi(6)}$
Then $P(X=1|Z=6) = \frac{P(X=1,Y=5)}{P(Z=6)}$
The numerator is essentially one of the many ways you can get 6 cars in two hours.
So we have, by independence:
$$P(X=1,Y=5) = \left(e^{-3} \frac{3^1}{1!} \right) \left(e^{-3} \frac{3^5}{5!} \right)= e^{-6}\frac{3^6}{5!} $$
Denominator is $P(Y=6)= e^{-6}\frac{6^6}{6!}$
So we get
$$P(X=1|Z=6) = 2^{-6}$$