A small regional carrier accepted 22 reservations for a particular flight with 19 seats. 17 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 43% chance, independently of each other.
Hint: Use the binomial distribution with p = 0.43.

(Report answers accurate to 4 decimal places.)

Find the probability that overbooking occurs. 
  
Find the probability that the flight has empty seats. 
  

We are given a small regional carrier flight with 19 seats, where 22 reservations have been made. Among these, 17 reservations are for regular customers who will definitely show up. The remaining 5 reservations are for passengers who each have a 43% chance of arriving, independently of each other. The task is to compute:

1. The probability that overbooking occurs (i.e., more than 19 passengers show up).
2. The probability that the flight has empty seats (i.e., fewer than 19 passengers show up).

We can approach this using the binomial distribution for the 5 passengers who may or may not arrive. The binomial distribution formula is:

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

Where:
- \( n = 5 \) (the 5 passengers who may or may not show up),
- \( p = 0.43 \) (the probability that each of these passengers shows up),
- \( k \) is the number of passengers that actually show up.

### 1. Probability of Overbooking

Overbooking occurs when more than 19 passengers show up. Since 17 passengers are guaranteed to arrive, overbooking happens if 3 or more of the remaining 5 passengers also show up. We need to compute the probability that the number of additional passengers, \( X \), is at least 3.

\[
P(\text{Overbooking}) = P(X \geq 3) = 1 - P(X \leq 2)
\]

We will first compute \( P(X \leq 2) \), which is the probability that 0, 1, or 2 of the 5 passengers show up. Then we subtract that from 1 to get the probability of overbooking.

### 2. Probability of Empty Seats

For the flight to have empty seats, fewer than 19 passengers must show up. This happens if the number of additional passengers, \( X \), is less than 2 (since 17 are guaranteed to arrive).

\[
P(\text{Empty Seats}) = P(X < 2) = P(X \leq 1)
\]

Let's calculate these probabilities using the binomial distribution.The probabilities are as follows:

1. The probability that overbooking occurs (more than 19 passengers show up) is approximately **0.3705**.
2. The probability that the flight has empty seats (fewer than 19 passengers show up) is approximately **0.2871**.

Would you like more details on these calculations, or have any other questions?

Here are 5 related questions to explore further:
1. What is the expected number of passengers that will show up for the flight?
2. How does the overbooking probability change if the probability of a passenger showing up increases?
3. What is the probability that exactly 19 passengers show up?
4. How would these probabilities change if there were more seats on the flight?
5. What would the variance be for the number of passengers showing up?

**Tip**: In binomial distribution problems, the "success" probability \( p \) can significantly alter the outcome; adjusting \( p \) even slightly can shift probabilities.

A small regional carrier accepted 22 reservations for a particular flight with 19 seats. 17 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 43% chance, independently of each other. Find the probability that overbooking occurs. Find the probability that the flight has empty seats.

Calculate the probability of overbooking and empty seats for a flight with 22 reservations and 19 seats. 17 passengers are guaranteed to arrive, while the rest have a 43% chance. Explore the solution using binomial distribution and key probability concepts. Suitable for Grades 10-12.

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