a) Consider a consulting firm in a particular city where past records show that on average 6 customers arrive daily to seek consulting services. The demand for consulting services in the firm is distributed according to a Poisson distribution. Calculate the probability of exactly 0, 1, 2 and 3 customers arriving for the consulting services.

Let $X=$ the number of customers arriving for the consulting services: $X\sim Po(\lambda)$

Given $\lambda=6$

$P(X=0)=\dfrac{e^{-\lambda}\cdot \lambda^0}{0!}=\dfrac{e^{-6}}{1}\approx0.00247875$

$P(X=1)=\dfrac{e^{-\lambda}\cdot \lambda^1}{1!}=\dfrac{6e^{-6}}{1}\approx0.014873$

$P(X=2)=\dfrac{e^{-\lambda}\cdot \lambda^2}{2!}=\dfrac{36e^{-6}}{2}\approx0.044618$

$P(X=3)=\dfrac{e^{-\lambda}\cdot \lambda^3}{3!}=\dfrac{216e^{-6}}{6}\approx0.089235$