A small regional carrier accepted 22 reservations for a particular flight with 20 seats. 14 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 42% chance, independently of each other.
a. Find the probability that overbooking occurs.
b. Find the probability that the flight has empty seats.

There are 8 reservations for passengers that may go or not, such that the probability of arriving to the flight is P = 0.42, and the probability of not arriving is Q = 0.58

a) There will be overbooking if 7 or more passengers go to the flight, the probability that the 8 attend is:

P(8) = (0.42)^8

The probability that 7 attend is:

P(7) = 8*(0.42)^7*(0.58)

(where the factor 8 is because there are 8 permutations)

The total probability is:

P = P(8) + P(7) = 0.012

b) Will be empty seats if less than 6 people go.

The probability that exactly 6 of these people go to the flight is:

P(6) = (28)*(0.42)^6*(0.58)^2

Where there are 28 permutations, that is why we used that factor.

Then, the probability of having empty seats is:

P = 1 - P(6) - P(7) - P(8) = 0.937

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