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 is not 0.228767

Answer 1

To find the probability that not enough seats will be available when Air-USA books 15 persons, we can model the situation using a binomial distribution.

Given:
- Number of passengers booked, \( n = 15 \)
- Probability that a booked passenger shows up, \( p = 0.83 \)
- Number of seats available, \( k = 14 \)

We want to find the probability that more than 14 passengers show up, which can be expressed as:

\[
P(X > 14) = 1 - P(X \leq 14)
\]

Where \( X \) is the number of passengers that show up. Since \( X \) follows a binomial distribution, \( X \sim \text{Binomial}(n=15, p=0.83) \).

The probability of \( X = k \) in a binomial distribution is given by the formula:

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

We'll first use this formula to compute \( P(X \leq 14) \) as:

\[
P(X \leq 14) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14)
\]

However, for efficiency, we can also find \( P(X > 14) \) directly since the maximum that can show up is 15, so we will only calculate \( P(X = 15) \):

\[
P(X = 15) = \binom{15}{15} p^{15} (1-p)^{0}
\]

Calculating this, we have:

\[
P(X = 15) = 1 \cdot (0.83)^{15} \cdot (0.17)^{0} = (0.83)^{15}
\]

Now, calculating \( (0.83)^{15} \):

\[
(0.83)^{15} \approx 0.23977
\]

Thus,

\[
P(X > 14) = P(X = 15) \approx 0.23977
\]

Finally, the probability that not enough seats will be available (meaning more than 14 passengers show up) is approximately:

\[
P(X > 14) \approx 0.23977
\]

So the final answer is:

\[
\text{prob} \approx 0.23977
\]