Problem Setup:

• Customers arrive following a Poisson process with rate $\lambda = 4$ customers per hour.
• The shop opens at 10:00 am.
• Given condition: The second customer arrives at 10:40 am. Let this time be $T_{(2)} = 40$ minutes after opening.
• Question: Find the probability that no customer arrived before 10:30 am. Let the time of the first arrival be $T_{(1)}$. We need to find $P(T_{(1)} > 30 \mid T_{(2)} = 40)$.

Rate Conversion:

The rate $\lambda = 4$ customers/hour needs to be converted to customers per minute:

$\lambda = \frac{4 \text{ customers}}{60 \text{ minutes}} = \frac{1}{15}$ customers/minute.

Using Poisson Process Properties:

A key property of Poisson processes states that if $N(t)=n$ (i.e., $n$ events occur in the interval $[0, t]$), then the times of these $n$ events are distributed as the order statistics of $n$ independent random variables uniformly distributed on $[0, t]$.

In this problem, the interval is $[0, 40]$ minutes (from 10:00 am to 10:40 am). We are given that the second arrival occurs exactly at $t=40$ minutes ($T_{(2)} = 40$). This implies that $N(40)=2$.

We can model the two arrival times, $T_{(1)}$ and $T_{(2)}$, as the order statistics of two independent random variables, $U_1$ and $U_2$, drawn from a Uniform distribution on $[0, 40]$. That is, $T_{(1)} = \min(U_1, U_2)$ and $T_{(2)} = \max(U_1, U_2)$.

Applying the Condition:

We are given $T_{(2)} = \max(U_1, U_2) = 40$. This implies that at least one of $U_1$ or $U_2$ must be 40. Since $U_1$ and $U_2$ are drawn from $U[0, 40]$, this condition is met.

Specifically, if $\max(U_1, U_2) = 40$, then one of the variables must be 40 (say $U_2=40$), and the other variable $U_1$ is drawn from $U[0, 40]$.

Calculating the Probability:

We want to find the probability that the first arrival occurred after 10:30 am, meaning $T_{(1)} > 30$. This translates to $\min(U_1, U_2) > 30$.

Given $U_2 = 40$, the condition becomes $\min(U_1, 40) > 30$. Since $40 > 30$, this inequality simplifies to $U_1 > 30$.

Therefore, we need to find the probability $P(U_1 > 30)$, where $U_1$ is drawn uniformly from $[0, 40]$.

$ P(U_1 > 30) = \frac{\text{Length of interval } (30, 40]}{\text{Length of interval } [0, 40]} $

$ P(U_1 > 30) = \frac{40 - 30}{40 - 0} = \frac{10}{40} = \frac{1}{4} $

Conclusion:

The probability that no customer arrived before 10:30 am, given that the second customer arrived at 10:40 am, is $\frac{1}{4}$.