Question

The number of patients entering the intensive care unit of Great Care Hospital (GCH) on any one day has a Poisson probability distribution with a mean equal to 5 patient per day.
What is the probability that the number of patient entering the intensive care unit on a particular day is greater that two.

Answer 1

$p(x)=\frac{e^{-\lambda}\lambda^{x}}{x!}, where\ x=0,1,2,3...\\ \lambda=5\\ p(x>2)=1-p(x\le2)\\ p(x>2)=1-[\frac{e^{-5}5^{0}}{0!}+\frac{e^{-5}5^{1}}{1!}+\frac{e^{-5}5^{2}}{2!}]\\ p(x>2)=1-[0.0067+0.0337+0.0842]\\ p(x>2)=1-0.1247\\ p(x>2)=0.8753$