Question

Suppose 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.

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

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

Answer 1

n=5,p=0.5

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

$P(X=4)+P(X=5)= \binom{5}{4}(0.5)^4(1-0.5)^{5-4}+\binom{5}{5}(0.5)^5(1-0.5)^{5-5}= (5+1)(0.5)^5=0.1875$

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

$P(X=0)+P(X=1)+P(X=2)=\binom{5}{0}(0.5)^0(1-0.5)^{5-0}+\binom{5}{1}(0.5)^1(1-0.5)^{5-1}+\binom{5}{2}(0.5)^2(1-0.5)^{5-2}=(1+5+10)(0.5)^5=0.5$