91Ó°ÊÓ

The airline sells 200 tickets on an airplane that has 198 seats.

Let X denote the number of people who do not appear for the flight.

The number of people who appear will have a seat if \(X \ge 2\).

X follows a binomial distribution with parameters \(n = 200\) and \(p = 0.01\).

A Poisson distribution with a mean can approximate the distribution

\(\begin{array}{c}\lambda = 200 \times 0.01\\ = 2\end{array}\).

The p.m.f of X is

\(\begin{array}{c}f\left( x \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x \ge 0\\ = \frac{{{e^{ - 2}}{2^x}}}{{x!}}\end{array}\)

The probability that everyone who appears on the flight will have a seat is \(P\left( {X \ge 2} \right)\).

\(\begin{array}{c}P\left( {X \ge 2} \right) = 1 - P\left( {X \le 1} \right)\\ = 1 - \left( {\frac{{{e^{ - 2}}{2^0}}}{{0!}} + \frac{{{e^{ - 2}}{2^1}}}{{1!}}} \right)\\ = 1 - 0.1353 - 0.2706\\ = 0.5941\end{array}\)

Hence, the probability that everyone who appears on the flight will have a seat is 0.5941.