Answer: 0.9826
Step-by-step explanation:
Given : Mean : [tex]\lambda =3\text{ calls per minute}[/tex]
For two minutes period the new mean would be :
[tex]\lambda_1=2\times3=6\text{ calls per two minutes}[/tex]
The formula to calculate the Poisson distribution is given by :_
[tex]P(X=x)=\dfrac{e^{-\lambda_1}\lambda_1^x}{x!}[/tex]
Then ,the required probability is given by :-
[tex]P(X\geq2)=1-(P(X\leq1))\\\\=1-(P(0)+P(1))\\\\=1-(\dfrac{e^{-6}6^0}{0!}+\dfrac{e^{-6}6^1}{1!})\\\\=1-0.0173512652367\\\\=0.982648734763\approx0.9826[/tex]
Hence, the probability that at least two calls are received in a given two-minute period is 0.9826.
