SOLUTION: Two hundred and fifty passenger have mad reservations for airplane flight. If the probability that a passenger who has a reservation show up is 0.98. what is the exactly five that

Algebra.Com
Question 1163265: Two hundred and fifty passenger have mad reservations for airplane flight. If the probability that a passenger who has a reservation show up is 0.98. what is the exactly five that will not show up?
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

It is   =  = 0.1772.    ANSWER


It is a binomial distribution problem 

with the number of trials n = 250 and the number of success trials of 245 = 250-5.

Solved.

------------------

To see many other similar solved problems, look into the lessons
    - Simple and simplest probability problems on Binomial distribution
    - Typical binomial distribution probability problems
in this site.



RELATED QUESTIONS

Musquodoboit World Airways operates a fleet of small passenger planes. Like most major... (answered by stanbon)
Suppose that the probability that a passenger will miss a flight is 0.0999. Airlines do... (answered by ikleyn)
The previous answer provided was not correct. Please help me figure out this difficult... (answered by stanbon)
A small regional carrier accepted 16 reservations for a particular flight with 12 seats.... (answered by Boreal)
small regional carrier accepted 17 reservations for a particular flight with 13 seats. 11 (answered by math_tutor2020)
An airport limousine can accommodate up to 4 passengers on any trip. The company will... (answered by donrys)
Because not all air line passengers show up for their reserved seat, an airline sells 125 (answered by rothauserc,Theo)
Consider a low cost airline operating a 120-seat plane. The company typically sells 125... (answered by stanbon)
A small feeder airline knows that the probability is .10 that a reservation holder will... (answered by stanbon)