A small regional carrier accepted 18 reservations for a particular flight with 17 seats. 14 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 51% chance, independently of each other.

Question

(Report answers accurate to 4 decimal places.)

1. Find the probability that overbooking occurs.
Answer here:______

2. Find the probability that the flight has empty seats.
Answer here: ______

Answer 1

To address this problem, we will denote the following:

• Let \( X \) be the number of passengers (out of the 4 who are not regular customers) that actually show up for the flight.
• \( X \) follows a binomial distribution, specifically \( X \sim \text{Binomial}(n=4, p=0.51) \).

1. Find the probability that overbooking occurs.

Overbooking occurs if the total number of passengers showing up exceeds the number of available seats (which is 17). Since there are already 14 regular customers who will arrive, we need to find the probability of more than 3 additional passengers showing up (i.e., \( X > 3 \)).

To calculate the probability of overbooking, we need \( P(X > 3) \):

\[ P(X > 3) = P(X = 4) \]

Now we calculate \( P(X = 4) \):

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

For \( X = 4 \):

\[ P(X = 4) = \binom{4}{4} (0.51)^4 (0.49)^0 \]

\[ P(X = 4) = 1 \cdot (0.51)^4 \cdot 1 = (0.51)^4 \]

Now we compute \( (0.51)^4 \):

\[ (0.51)^4 = 0.0679 \quad (\text{rounded to 4 decimal places}) \]

Thus, the probability that overbooking occurs is:

\[ \text{Answer 1: } 0.0679 \]

2. Find the probability that the flight has empty seats.

Empty seats occur if fewer than 17 passengers show up. Since there are 14 regular passengers, we need to determine the probability that no more than 2 of the additional passengers show up (i.e., \( X \leq 2 \)):

\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]

Calculating each of these:

1. For \( P(X = 0) \): \[ P(X = 0) = \binom{4}{0} (0.51)^0 (0.49)^4 = 1 \cdot 1 \cdot (0.49)^4 \approx 0.0576 \]

2. For \( P(X = 1) \): \[ P(X = 1) = \binom{4}{1} (0.51)^1 (0.49)^3 = 4 \cdot (0.51) \cdot (0.49)^3 \approx 4 \cdot 0.51 \cdot 0.1176 \approx 0.2406 \]

3. For \( P(X = 2) \): \[ P(X = 2) = \binom{4}{2} (0.51)^2 (0.49)^2 = 6 \cdot (0.51)^2 \cdot (0.49)^2 \approx 6 \cdot 0.2601 \cdot 0.2401 \approx 0.3762 \]

Now sum these probabilities:

\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \approx 0.0576 + 0.2406 + 0.3762 = 0.6744 \]

Thus, the probability that the flight has empty seats is:

\[ \text{Answer 2: } 0.6744 \]

Final Answers:

1. The probability that overbooking occurs: 0.0679
2. The probability that the flight has empty seats: 0.6744