To solve this problem, we can model the number of passengers who actually show up as a binomial random variable. Here’s the breakdown of the problem:

• Let $n = 54$: This is the number of tickets sold.
• Let $p = 0.9066$: This is the probability that a passenger will show up for the flight, given that the probability of a passenger missing the flight is 0.0934.
• The seating capacity is 52, so the flight is overbooked if 53 or 54 passengers show up.

We’re looking for $P(X = 53 \text{ or } X = 54)$, where $X$ is the number of passengers who show up.

1. Define $X$: $X \sim \text{Binomial}(n=54, p=0.9066)$.

2. Calculate $P(X = 53)$:

   Using the binomial probability formula: $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$ where $\binom{n}{k}$ is the binomial coefficient.

3. Calculate $P(X = 54)$:

   We use the same formula to find the probability that exactly 54 passengers show up.

4. Sum of Probabilities:

   Then, the probability of an overbooked flight is: $P(X = 53 \text{ or } X = 54) = P(X = 53) + P(X = 54)$

Step-by-Step Calculation (using a calculator or software):

Let me know if you’d like a detailed calculation using exact values for each probability.

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Would you like to see the detailed calculations? Here are some additional questions to deepen your understanding:

1. How would the probability change if the probability of missing the flight were higher?
2. What if only 53 tickets were sold instead of 54—how would the overbooking probability be affected?
3. How could we approximate this probability if $n$ were much larger?
4. What if we wanted to find the probability that at most 52 passengers show up?
5. How could we use the normal approximation to the binomial distribution to solve this problem?

Tip: The binomial distribution is particularly useful in probability problems involving repeated independent events with only two outcomes (e.g., show up or not show up).