Answer
a)$p(X\geq 5)=0.9961$
a)$p(X\gt5)=0.9886$
Work Step by Step
Let X be the random variable of the number of passengers who do not show up for flight. then X is distributed as binomial distribution with $n=125$, and $p=0.1$
this means that
$p(X=x)=\frac{125!}{x!(125-x)!}(0.1)^x(0.9)^{125-x},x=0,1,...,125$
a) the probability that every passenger who shows up
can take the flight
$p(X\geq 5)=1-p(X\leq4)=1-(p(X=0)+p(X=1)+p(X=2)+p(X=3)+p(X=4))\approx 0.9961$
b)the probability that the flight departs with empty
seats
$p(X\gt 5)=1-p(X\leq5)=1-(p(X=0)+p(X=1)+p(X=2)+p(X=3)+p(X=4)+p(X=5))\approx 0.9886$
