Suppose that the probability that a passenger will miss a flight is 0.0925. Airlines do not...

Suppose that the probability that a passenger will miss a flight is 0.0925. Airlines do not like flights with empty seats, but it is also not desirable to have overbooked flights because passengers must be "bumped" from the flight. Suppose that an airplane has a seating capacity of 53 passengers. Suppose that 57 tickets are sold. What is the probability that a passenger will have to be "bumped"?

(a) Suppose that 57 tickets are sold. What is the probability that a passenger will have to be "bumped"?

(b). For a plane with seating capacity of 51 passengers, how many tickets may be sold to keep the probability of a passenger being "bumped" below 55%?

For part a using binomcdf (59,0.9075,53)= .4676 1-.4676=0.5324. However when doing it by hand ₅₉nCr₅₃(0.9075)⁵³ (1-0.9075)⁶ =.1646 . Can someone tell me what I am doing wrong when using the formula to solve the question? Thank you

For part b not sure how to do it

Expert Solution

orchestra answered 1 year ago

Suppose that the probability that a passenger will miss a flight is 0.0925 Airlines do not...

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