poisson and uniform distributions

Question

I have an answer to this question from someone else but I do not think it is right. Here is the question:

Customers arrive at a bank at a Poisson rate lambda. Suppose two customers arrive during the first hour. What is the probability that both arrived during the first 20 minutes?

The answer I have says the arrivals are uniform (0,60). How can that be?

Answer 1

2 ways to obtain the same answer (1/9), 1 easier than the other. Both use the definition of conditional probability as the quotient of joint probability and probability of 1 RV, but the easier method considers number of customers as RVs, while the other considers arrival times.

1. Let $n(t)$ be an RV for the number of customers arriving during time $t$. Since $n(t)$ increments in disjoint intervals are Poisson and thus independent, the probability of $n(t_1)$ given $n(t_2)$ is

$$ \frac{P(n(t_1)\cap n(t_2))}{P_{Poi}(n(t_2);\lambda t_2)}=\frac{P_{Poi}(n(t_1);\lambda t_1)P_{Poi}(n(t_2)-n(t_1);\lambda(t_2-t_1))}{P_{Poi}(n(t_2);\lambda t_2)} $$

2. Let $\tau_i$ be an RV for the arrival time of customer i. Then $\Delta \tau_i$ are independent and exponentially distributed. The desired probability is the probability $\tau_i<t_1$ given $\tau_i<t_2$ and $\tau_{i+1}>t_2$, or

$$ \frac{P_t(\tau_i<t_1\cap\tau_i<t_2\cap\tau_{i+1}>t_2)}{P_t(\tau_i<t_2\cap\tau_{i+1}>t_2)}=\frac{P_t(\tau_i<t_1\cap\tau_{i+1}>t_2)}{P_t(\tau_i<t_2\cap\tau_{i+1}>t_2)} $$ $$ P_t(\tau_i<t\cap\tau_{i+1}>t_2)=\int_0^td\Delta\tau_1...\int_0^{t-\sum_{j=1}^{i-1}\Delta\tau_j}d\Delta\tau_i\int_{t_2-\sum_{j=1}^i\Delta\tau_j}^\infty d\Delta\tau_{i+1}\prod_{j=1}^{i+1}P_{exp}(\Delta\tau_j;\lambda) $$