Answer :

Solution :

Customers arrive at a shop according to the Poisson distribution with a mean of 10 customers/hour.

⇒ λ = 10 customers/60 min = 1 customer/6 min = 1/6 customers per minute;

The manager notes that no customer arrives for the first 3 minutes after the shop opens,

⇒ t = 3 min and n = 0;

From the Poisson’s PMF,

The probability of zero customers in 3 minutes will be

$P_{N\left(3\right)}\left(0\right)=\frac{{\left(\frac{1}{6}\cdot 3\right)}^0\exp \left(-\frac{1}{6}\cdot 3\right)}{0!}=e^{-0.5}=0.606$

Now,

The probability that the customer arrives in the next 3 minutes = 1 - P(0)

∴ The probability that the customer arrives in the next 3 minutes = 1 - 0.606

∴ The probability that the customer arrives in the next 3 minutes = 0.39