Question

Suppose an airline accepted 12 reservations for a commuter plane with 10seats. They know that 7 reservations went to regular commuters who will show up forsure. The other 5 passengers will show up with a 50% chance, independently of eachother.(a) Find the probability that the flight will be overbooked.(b) Find the probability that there will be empty seats.

Answer 1

See the explanation.

Explanation:

(a)

7 persons will show up for sure, hence there is not any kind of probability here.

The other 5 persons have 50% chance to show up.

The flight will be overbooked, if either 4 will come or 5 will come.

From the total 5 persons, 4 can be chosen in
`^5C_4 = 5` ways.

The probability that any four among the 5 will come is
`5*((50)/(100) )^(5) = 5*((1)/(2) )^(5) = (5)/(32)`.

The probability that all of the 5 passengers will come is
`(1)/(32)`.

Hence, the required probability is
`(5)/(32) + (1)/(32) = (6)/(32) = (3)/(16)`.

(b)

The probability that there will be exactly 10 passengers on the flight is
`^5C_3* ((1)/(2) )^5 = (20)/(2) *(1)/(32) = (10)/(32) = (5)/(16)`.

The probability that there will be no empty seats in the plane is
`(3)/(16) + (5)/(16) = (1)/(2)`.

Hence, probability of having empty seats is
`1 - (1)/(2) = (1)/(2)`.