To solve this problem, we can use the binomial distribution, where we want to find the probability of having more than 14 passengers show up for a flight when 15 have been booked, and the probability that a booked passenger shows up is 83% (or 0.83).
Let \( X \) be the random variable representing the number of passengers that show up. Since there are 15 persons booked, \( X \) follows a binomial distribution given by:
\[ X \sim \text{Binomial}(n = 15, p = 0.83) \]
We are interested in the situation where there are not enough seats available, which occurs when 15 or more passengers show up. Since it’s impossible to have more than 15 passengers when only 15 were booked, we want to find the probability where \( X \geq 15 \).
To find \( P(X \geq 15) \), we can use the complement rule. This means that we will calculate \( P(X < 15) \) and subtract from 1:
\[ P(X \geq 15) = 1 - P(X < 15) = 1 - P(X \leq 14) \]
To calculate \( P(X \leq 14) \), we can sum the probabilities for \( k = 0 \) to \( k = 14 \):
\[ P(X \leq 14) = \sum_{k=0}^{14} P(X = k) \]
where \( P(X = k) \) is given by the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \]
However, it's important to note that:
\[ P(X = k) = \binom{15}{k} (0.83)^k (0.17)^{15-k} \]
We can calculate \( P(X = 15) \) since we only need that.
Thus,
\[ P(X = 15) = \binom{15}{15} (0.83)^{15} (0.17)^{0} = (0.83)^{15} \]
Now, let's compute \( (0.83)^{15} \):
\[ (0.83)^{15} \approx 0.243 \]
So,
\[ P(X \geq 15) = P(X = 15) = (0.83)^{15} \approx 0.243 \]
Thus, the probability that not enough seats will be available (meaning 15 or more passengers show up) is approximately:
\[ \text{prob} \approx 0.243 \]
The probability that if Air-USA books 15 persons, not enough seats will be available is approximately 0.243 or 24.3%.