Given three customers arrive in the first hour, find probability third arrives in second half

Question

Assume customers arrive according to a Poisson process with rate $\lambda = 3$.

Given three customers arrive in the first hour, find the probability that the third customer arrives in the second half of the hour.

So that is if $X_3$ is the third arrival time: $P(X_3 \in [0.5,1] \mid N(1) = 3)$

and we have $X_3 \sim$ Gamma$(3,3)$.

To solve this I have cases : $\bigg \{X_1,X_2 \in [0,.5]$ and $X_3 \in [.5,1]\bigg\}$ or

$\bigg \{X_1 \in [0,.5]$ and $X_2,X_3 \in [.5,1]\bigg\}$ or $\bigg \{X_1,X_2, X_3 \in [.5,1] \bigg \}$

However, isn't there a way to solve this using the Uniform distribution?

Answer 1

Given that exactly three events were observed in the first hour, the distribution of their event times are independent and uniform over that hour. So the probability the third event occurs in the second half hour is equal to the complement of the probability that all three events occur in the first half hour. But the probability of all three events occurring in the first half hour is simply $(1/2)^3 = 1/8$; therefore, the desired answer is $1 - 1/8 = 7/8$.

Answer 2

I think the key to solving this problem is to identify the following relationship: $$\{N(1)=3\}=\{X_3\leq 1,X_4>1\}$$ This way we can say $$P\bigg(\frac{1}{2}\leq X_3 \leq 1\bigg|N(1)=3\bigg)=\frac{P\bigg(\frac{1}{2}\leq X_3 \leq 1,X_4>1\bigg)}{P(N(1)=3)}$$ Using the total law of probability, $$P\bigg(\frac{1}{2}\leq X_3 \leq 1,X_4>1\bigg)=\int_{1/2}^1P\bigg(\frac{1}{2}\leq X_3 \leq 1,X_4>1\bigg|X_3=t\bigg)f_{X_3}(t)dt$$ Since inter$-$arrival times are independent and $\sim \exp(3)$ we have $$P\bigg(\frac{1}{2}\leq X_3 \leq 1,X_4>1\bigg|X_3=t\bigg)=\int_{1-t}^{\infty}3e^{-3t}dt=e^{3t-3}$$ Using the fact that $X_3 \sim \text{Erlang}(3,3)$ and $N(1)\sim \text{Poisson}(3)$ we get $$P\bigg(\frac{1}{2}\leq X_3 \leq 1,X_4>1\bigg)=\int_{1/2}^1e^{3t-3}f_{X_3}(t)dt=\frac{63}{16e^3}$$ $$P(N(1)=3)=\frac{27}{6e^3}$$ Dividing these two gives us our desired probability: $$P\bigg(\frac{1}{2}\leq X_3 \leq 1\bigg|N(1)=3\bigg)=\frac{7}{8}$$