Question

Air-USA has a policy of booking as many as 15 persons on an airplane that can seat only 14. (Past studies have revealed that only 83% of the booked passengers actually arrive for the flight.)

Find the probability that if Air-USA books 15 persons, not enough seats will be available.
prob =

Answer 1

To solve this problem, we will assume that the number of passengers that actually show up can be modeled using a binomial distribution. Specifically, let's denote the number of passengers that show up as \( X \).

The number of passengers booked (15) is \( n = 15 \), and the probability of any individual passenger showing up is \( p = 0.83 \). Thus, \( q = 1 - p = 0.17 \) is the probability of a passenger not showing up.

We want to find the probability that more than 14 passengers show up, i.e., \( P(X > 14) \).

Calculating \( P(X > 14) \) can be done by finding \( P(X = 15) \) and then:

\[
P(X > 14) = P(X = 15)
\]

Since \( X \) follows a binomial distribution \( B(n=15, p=0.83) \):

\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]

For \( k = 15 \):

\[
P(X = 15) = \binom{15}{15} (0.83)^{15} (0.17)^{0}
\]

Calculating \( P(X = 15) \):

\[
\binom{15}{15} = 1
\]

\[
(0.83)^{15} = 0.83^{15} \approx 0.228767 (approximately)
\]

Thus,

\[
P(X = 15) = 1 \cdot 0.228767 \cdot 1 = 0.228767
\]

Therefore,

\[
P(X > 14) \approx 0.228767
\]

Now, we can round this to a suitable number of significant digits. So:

**The probability that not enough seats will be available is approximately** \( 0.229 \) or **22.9%**.