Question

The sales of Lexus automobiles in the Detroit area follow a Poisson distribution with a mean of 5 per day. (i) What is the probability that no Lexus is sold on a particular day? (ii) What is the probability that at least two Lexus is sold on consecutive 3 days? 

Answer 1

(i) Let X be the number of cars sold on a day, then X~Pois(5)

$P(X=0)={\frac{5^0} {0!}}e^{-5}=e^{-5}\approx0.0067$

(ii) Let X be the number of cars sold on three days, then X~Pois(3*5)=Pois(15)

$P(X≥2)=1-P(X<2)=1-P(X=0)-P(X=1)=1-{\frac{15^0} {0!}}e^{-15}-{\frac{15^1} {1!}}e^{-15}\approx1$