Answer to Question #206593 in Statistics and Probability for Pavankumar

Question

A travel agency had 2 cars which it hires daily. The number of demands for a car on each

day is distributed as a Poisson cariate with mean 1.5. Find the probability that on a particular

day (i) there was no demand (ii) a demand is refused?

Answer 1

$P(X=0)=\frac{1.5^0e^{-1.5}}{0!}=0.2231.$

$P(X>2)=1-P(X=0)-P(X=1)-P(X=2)=$

$=1-\frac{1.5^0e^{-1.5}}{0!}-\frac{1.5^1e^{-1.5}}{1!}-\frac{1.5^2e^{-1.5}}{2!}=0.1912.$