Simple Conditional Probability in a Poisson Process

Question

Buses arrive at a station in a Poisson process at a rate of 3 per hour.
Suppose 4 buses arrive in the first hour, what is the probability that a total of 6 buses arrive in the first two hours? Suppose 6 buses arrive in the first two hours, what is the probability that a 4 buses arrived in the first hour?

Just looking for a check on my reasoning here.

For the first question, if we are given that $X_1 = 4$ then the only way that $X_1 + X_2 = 6$ is to have $X_2 = 2$. Thus, the probability is simply $P(X_2 = 2)$.

For the second question, there are 7 ways to have non-negative $X_1 + X_2 = 6$. Only one involves $X_1 = 4$. So here $P = 1/7$.

Answer 1

The first problem is correctly done, but the answer should be more explicit.

For the second question, let $A$ be the event $4$ buses arrive in the first hour, and $B$ the event there are $6$ buses in the $2$ hours. We want $\Pr(A\mid B)$, which is $\Pr(A\cap B)/\Pr(B)$.

Compute. We have $\Pr(B)=e^{-6}\frac{6^6}{6!}$.

The probability of $A\cap B$ is the probability of $4$ buses in the first hour and $2$ in the second. This is $e^{-3}\frac{3^4}{4!}e^{-3}\frac{3^2}{2!}$.

Divide. Our conditional probability is $\frac{6!}{4!2!}\frac{3^2 3^4}{6^6}$, which is more nicely expressed as $\binom{6}{4}\cdot \frac{1}{2^6}$.

A similar calculation shows that in general, the conditional distribution of the number of buses in the first $a$ hours, given that there were $n$ buses in the first $a+b$ hours, is binomial, parameters $p=\frac{a}{a+b}$, and $n$.

Answer 2

It is not clear what you mean by $X_2$.

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Part 1. In other words, you want two arrivals in in the interval $(1,2]$. Since disjoint blocks of time yield independent Poisson distributions, then you want $$P(X_{(1,2)} = 2)$$ where $X_{(1,2)}\sim\text{Pois}(t\lambda = (2-1) 3)$.

Part 2. No. The quick way is to notice that you are dealing with a Binomial(6, 1/2). Otherwise, expand by Bayes' rule $$P(X_{(0,1)} = 4|X_{(0,2)} = 6) = \frac{P(X_{(0,1)}= 4, X_{(0,2)} = 6)}{P(X_{(0,2)} = 6)}$$