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%.