From the question given
Past studies have shown that only about 84% of the booked passengers show up
This can be taken as the probability of success (P)
Since we are told to find the probability that if Air-USA books 21 people, not enough people will show up
Then
In this case
[tex]\begin{gathered} P=84\text{\%=}\frac{84}{100}=0.84 \\ n=21 \end{gathered}[/tex]We will use the formula below to find the probability that if Air-USA books 21 people, not enough people will show up
[tex]P(x>20)=P^n[/tex]since we have the values of P and n
[tex]P(x>20)=0.84^{21}=0.0257[/tex][tex]\begin{gathered} \text{Hence,} \\ P(x>20)=0.0257 \end{gathered}[/tex]For the B part of the question
let us convert the answer obtained to percentage so we can compare it with 5% probability
[tex]0.0257=2.57\text{ \%}[/tex]Thus,
yes, the probability is low enough so that overbooking is low enough to be a concern.
For the part C of the question
since we have already converted to a percentage, then we will compare to 10%
Since
[tex]\begin{gathered} 2.57\text{ \% is less than 10\%} \\ \text{Then} \end{gathered}[/tex]yes, the probability is low enough so that overbooking is low enough to be a concern.