To solve these probability problems related to flight attendance and overbooking, we can use the binomial distribution model. Let’s go through each part step by step.

Part 1: Probability of Overbooking with 53 Tickets Sold

Given:

• Probability of missing a flight $p = 0.0957$
• Probability of showing up $q = 1 - p = 0.9043$
• Total passengers who bought tickets $n = 53$
• We want to find the probability that 52 or 53 passengers show up.

This can be calculated using the binomial probability formula: $P(X = k) = \binom{n}{k} q^k p^{n-k}$ where $\binom{n}{k}$ is the binomial coefficient.

We need to calculate $P(X = 52) + P(X = 53)$.

1. Calculate $P(X = 52)$: $P(X = 52) = \binom{53}{52} (0.9043)^{52} (0.0957)^{1}$

2. Calculate $P(X = 53)$: $P(X = 53) = \binom{53}{53} (0.9043)^{53} (0.0957)^{0}$

3. Total Probability of Overbooking: $P(\text{overbooked}) = P(X = 52) + P(X = 53)$

Calculating these:

1. $P(X = 52) = 53 \cdot (0.9043)^{52} \cdot (0.0957)$
2. $P(X = 53) = (0.9043)^{53}$

Let's compute these values.

Part 2: Probability of Bumping with 57 Tickets Sold

Given:

• Total passengers who bought tickets $n = 57$

We need to find the probability that more than 51 show up, which means we want: $P(X > 51) = 1 - P(X \leq 51)$

This can be computed as: $P(X \leq 51) = P(X = 51) + P(X = 50) + ... + P(X = 0)$

Use the binomial formula for each value from 0 to 51 and sum those probabilities.

Part 3: Maximum Tickets for 290 Passengers with Bumping < 1%

To find the maximum number of tickets $n$ that can be sold such that: $P(X > 290) < 0.01$

We can approach this by trying different values of $n$ and calculating the cumulative distribution until the cumulative probability of bumping exceeds 1%.

Let's calculate these probabilities step-by-step.

Calculations:

I'll do the calculations for Parts 1 and 2, then we can look at Part 3.

Part 1: Calculation of Overbooking Probability
Part 2: Calculation of Bumping Probability
Part 3: Finding Maximum Tickets

Let me calculate these probabilities.Here are the calculated probabilities and results for your questions:

Part 1: Probability of Overbooking with 53 Tickets Sold

• The probability that 52 or 53 passengers show up (resulting in an overbooked flight) is approximately: $P(\text{overbooked}) \approx 0.0320$

Part 2: Probability of Bumping with 57 Tickets Sold

• The probability that at least one passenger will have to be bumped (more than 51 passengers show up) is approximately: $P(\text{bump}) \approx 0.5328$

Part 3: Maximum Tickets for 290 Passengers with Bumping < 1%

• The largest number of tickets that can be sold while keeping the probability of a passenger being bumped below 1% is: $290$

If you have any questions or would like further details on any part, let me know! Here are some related questions for further exploration:

1. How does increasing the number of tickets sold affect the probability of overbooking?
2. What strategies can airlines use to minimize the impact of overbooking?
3. How would the results change if the probability of missing a flight was lower?
4. Can the airlines use simulations to predict overbooking outcomes? How?
5. What are the legal implications of overbooking flights for airlines?

Tip: Understanding the binomial distribution and its properties can help in analyzing similar scenarios in different contexts, such as inventory management or event attendance.