Question

Suppose that an airline accepted 12 reservations for a commuter plane with 10 seats.
They know that 7 reservations went to regular commuters who will show up for sure.
The other 5 passengers will show up with a 50% chance, independently of each other.
Find the probability that the
ight will be overbooked.

Answer 1

We are looking at what happens after 7 people have been counted. Hence there are 3 spaces left for 3 people.

$n=12-7=5, p=0.5$

(a) Find the probability that the flight will be overbooked. 

$$P(X=4)+P(X=5)=$$

$$=\binom{5}{4}(0.5)^4(1-0.5)^{5-4}+\binom{5}{5}(0.5)^5(1-0.5)^{5-5}=$$

$$=(5+1)(0.5)^5={3\over 16}=0.1875$$

(b) Find the probability that there will be empty seats.

$$P(X=0)+P(X=1)+P(X=2)=$$

$$=\binom{5}{0}(0.5)^0(1-0.5)^{5-0}+\binom{5}{1}(0.5)^1(1-0.5)^{5-1}+$$

$$+\binom{5}{2}(0.5)^2(1-0.5)^{5-2}=(1+5+10)(0.5)^5=$$

$$={1\over 2}=0.5$$