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