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

Answer:

See the explanation.

Step-by-step 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 $5\times(\frac{50}{100} )^{5} = 5\times(\frac{1}{2} )^{5} = \frac{5}{32}$.

The probability that all of the 5 passengers will come is $\frac{1}{32}$.

Hence, the required probability is $\frac{5}{32} + \frac{1}{32} = \frac{6}{32} = \frac{3}{16}$.

(b)

The probability that there will be exactly 10 passengers on the flight is $^5C_3\times (\frac{1}{2} )^5 = \frac{20}{2} \times\frac{1}{32} = \frac{10}{32} = \frac{5}{16}$.

The probability that there will be no empty seats in the plane is $\frac{3}{16} + \frac{5}{16} = \frac{1}{2}$.

Hence, probability of having empty seats is $1 - \frac{1}{2} = \frac{1}{2}$.