A small regional carrier accepted 19 reservations for a particular flight with 17 seats. 16 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 44% chance, independently of each other.

Question

(Report answers accurate to 4 decimal places.)

Find the probability that overbooking occurs.

Find the probability that the flight has empty seats.

Answer 1

To solve this problem, we need to analyze the situation involving the one remaining reservation.

Given:

• Total reservations = 19
• Seats available = 17
• Regular customers arriving = 16 (will definitely show up)
• One additional reservation has a 44% chance of showing up

Let \( X \) be a random variable representing the number of additional passengers that show up. \( X \) can take values 0 or 1, and follows a binomial distribution:

\[ X \sim \text{Binomial}(n=1, p=0.44) \]

Step 1: Calculate the probability that overbooking occurs

Overbooking occurs when the total number of passengers who show up exceeds the number of seats available (17). Since 16 regular customers will definitely show up, we need to find the cases where:

\[ 16 + X > 17 \]

This simplifies to:

\[ X > 1 \]

Given that \( X \) can only be 0 or 1 (since it represents the one additional passenger), we find:

\[ P(X > 1) = 0 \]

Step 2: Probability of \( X = 0 \)

Calculating the probability that 0 additional passengers show up:

\[ P(X = 0) = (1 - p)^n = (1 - 0.44)^1 = 0.56 \]

Step 3: Calculate the probability that the flight has empty seats

The flight has empty seats when the number of passengers showing up is less than the number of seats available (17). Thus, this condition translates to:

\[ 16 + X < 17 \]

This simplifies to:

\[ X < 1 \]

Thus, we need to consider the case when \( X = 0 \):

\[ P(X = 0) = 0.56 \]

Conclusion

1. Probability that overbooking occurs:

\[ P(\text{overbooking occurs}) = P(X > 1) = 0 \]

2. Probability that the flight has empty seats:

\[ P(\text{empty seats}) = P(X < 1) = P(X = 0) = 0.56 \]

Final answers are:

• Probability that overbooking occurs: \( 0.0000 \)
• Probability that the flight has empty seats: \( 0.5600 \)