Correct Answer - Option 2 : 0.39
Concept:
For the Poisson process of rate λ, and for any t > 0, the Probability mass function for N(t) (i.e., the number of arrivals in (0,t]) is given by the Poisson PMF
\({P_{N\left( t \right)}}\left( n \right) = \frac{{{{\left( {\lambda t} \right)}^n}\exp \left( { - \lambda t} \right)}}{{n!}}\)
If the arrivals of a Poisson process are split into two new arrival processes, each new process is independent.
Calculation:
Given:
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
