Conditional Poisson Question b)What is the probability that 4 customers arrive between 8:30 and 9:00 a.m., if exactly 8
customers arrived between 8:00 and 9:00 a.m.?
In the main body of the question, I'm given that the interarrival process is exponentially distributed with mean of 5 minutes. so I found rate ƛ = 1/5 arrivals per min. But for the question, how do I use conditional probability while accounting for the overlap in time overlap? I have found the individual probabilities of 4 customers arriving between 8:30 and 9 and the probability that 8 arrive between 8:00 and 9:00. How would I go about solving the rest? Would I need to adjust the rate of 1/5 that I was using based on the conditional statement or would I completely disregard the condition based on the memoryless property of the Poisson process?
Thanks
 A: I’ll assume that you intended to ask about the probability that exactly $4$ (not at least $4$) customers arrived during that half of the hour.
Given the condition, each of the $8$ arrivals has the same independent probability to have arrived in either of the two halves of the hour. Thus the probability that $4$ arrived in the latter half is
$$
\binom84\left(\frac12\right)^4\left(\frac12\right)^4=\binom84\left(\frac12\right)^8=\frac{35}{128}\;.
$$
A: Let $X$ and $Y$ be the random variables corresponding to the number of arrivals from 8:00 to 8:30 and from 8:30 to 9:00 respectively. Let $Z=X+Y$, so $Z$ has a Poisson distribution with $\lambda=12$.
Raikov's theorem states that we can decompose $Z$ into two independent Poisson distributions, each with $\lambda=6$ – which must be $X$ and $Y$. Then the desired probability is
$$\frac{P(X=Y=4)}{P(Z=8)}=\frac{P(X=4)P(Y=4)}{P(Z=8)}=\frac{(6^4e^{-6})^2}{12^8e^{-12}}\binom84=\frac{35}{128}$$
A: Theorem 5.2 from Introduction to Probability Models, Ninth Edition by Sheldon Ross, referring to a Poisson process $\{N(t), t \ge 0\}$, states:

Given that $N(t) = n$, the $n$ arrival times $S_1, \dots , S_n$ have
the same distribution as the order statistics corresponding to $n$
independent random variables uniformly distributed on the interval
$(0,t)$.

So the probability that exactly $4$ out of $8$ Uniform(8:00, 9:00)-distributed random variables will lie in the interval from 8:30 to 9:00 is
$$\binom{8}{4} \left( \frac{1}{2} \right)^4 \left( \frac{1}{2} \right)^4$$
