An airline finds that 4 percent of the passengers that make reservations on a particular flight will not show up. Consequently, their policy is to sell 100 reserved seats on a plane that has only 98 seats. Find the probability that every person who shows up for the flight will find a seat available.

References

Doyle, Peter G. (2006). "Grinstead and Snell's Introduction to Probability" (PDF). Retrieved June 6, 2024.

Solution: D

The number of persons that show up for the flight equals [math]N[/math] and has a binomial distribution with [math]n=100, p = 0.96 [/math] and we are interested in the probability that [math]N \leq 98 [/math]. This equals

[[math]] 1-P(N=99)-P(N=100) = 1- 100 \cdot 0.96^{99} \cdot 0.04 - 0.96^{100} = 0.9128. [[/math]]