Customers arrive at a shop according to the Poisson distribution with a mean of 10 customers/hour. The manager notes that no customer arrives for the first 3 minutes after the shop opens. The probability that a customer arrives within the next 3 minutes is

The correct option is 0.39.

Step-by-step Explanation:

1. Understanding the Memoryless Property of the Poisson Process:
The arrival of customers at the shop is modeled as a Poisson process. An important characteristic of a Poisson process is that it is memoryless. This means that the probability of an event occurring in a future time interval depends only on the length of that interval, and is completely independent of what occurred in the past.
In this problem, we are given that no customer arrived during the first 3 minutes. Due to the memoryless property, this past history does not affect the probability of arrivals in the subsequent period. Thus, the probability of a customer arriving within the next 3 minutes is exactly the same as the probability of a customer arriving within any 3-minute interval starting from scratch.

2. Finding the Arrival Rate:
The mean arrival rate is given as 10 customers per hour. We first convert this rate into minutes:

$\lambda =\frac{10\text{ customers}}{60\text{ minutes}}=\frac{1}{6}\text{ customers/minute}$

3. Calculating the Mean for a 3-Minute Interval:
Let $t$ be the duration of the interval, which is 3 minutes. The average number of arrivals (mean $\mu$) in this 3-minute interval is:

$\mu =\lambda \times t=\frac{1}{6}\times 3=0.5$

4. Computing the Probability of at Least One Arrival:
Let $X$ be the random variable representing the number of customers arriving in a 3-minute interval. The probability distribution of $X$ follows the Poisson formula:

$P(X=k)=\frac{e^{-\mu }{\mu }^k}{k!}$

We want to find the probability that at least one customer arrives (i.e., $X\ge 1$). This can be calculated using the complement probability:

$P(X\ge 1)=1-P(X=0)$

First, find $P(X=0)$ with $\mu =0.5$:

$P(X=0)=\frac{e^{-0.5}{0.5}^0}{0!}=e^{-0.5}\approx 0.6065$

Now, substitute this back to find the final probability of at least one arrival:

$P(X\ge 1)=1-0.6065=0.3935\approx 0.39$