91Ó°ÊÓ

An airport limousine can accommodate up to four passengers on any one trip. The company will accept a maximum of six reservations for a trip, and a passenger must have a reservation. From previous records, \(20 \%\) of all those making reservations do not appear for the trip. Answer the following questions, assuming independence wherever appropriate. a. If six reservations are made, what is the probability that at least one individual with a reservation cannot be accommodated on the trip? b. If six reservations are made, what is the expected number of available places when the limousine departs? c. Suppose the probability distribution of the number of reservations made is given in the accompanying table. $$ \begin{array}{l|lllll} \text { Number of reservations } & 3 & 4 & 5 & 6 \\ \hline \text { Probability } & .1 & .2 & .3 & .4 \end{array} $$ Let \(X\) denote the number of passengers on a randomly selected trip. Obtain the probability mass function of \(X\).

Step by step solution

01

Identifying the Probability Distribution

For each passenger with a reservation, there is an 80% chance they will appear for the trip (since 20% do not appear). Therefore, the number of "showing up" passengers out of six reservations follows a binomial distribution with parameters: number of trials \( n = 6 \) and success probability \( p = 0.8 \).

02

Calculating the Probability of Overbooking

To find the probability that at least one individual cannot be accommodated, calculate the probability that five or six individuals appear, since only four can be accommodated. The probability of exactly \( k \) people showing up is given by the binomial formula: \( P(X = k) = \binom{6}{k} (0.8)^k (0.2)^{6-k} \).Compute \( P(X = 5) + P(X = 6) \) for overbooking.

03

Solving overbooking probabilities

Compute individually:\( P(X = 5) = \binom{6}{5} (0.8)^5 (0.2)^1 = 6 \times (0.8)^5 \times 0.2 \).\( P(X = 6) = \binom{6}{6} (0.8)^6 = (0.8)^6 \).Adding them gives the total probability of overbooking.

04

Calculate Expected Available Seats

Calculate the expected number of passengers who do not show up under the assumption of a binomial distribution with parameters \( n = 6 \) and \( p = 0.2 \). The expected value is given by \( E = n \times (0.2) \). The expected number of seats available is \( 0.2 \times 6 \). Subtract from \( 4 \) to find the expected available places when the limousine departs.

05

Defining Probability Mass Function of X

Define \( X \) as the number of passengers showing up. First, determine probabilities for each reservation scenario using the given distribution table for the number of reservations (3, 4, 5, 6) and compute the probability that each number of people shows up (\( 3 \text{ to } 6 \)). Multiply by the probability of each reservation scenario.

06

Compute PMF Components

For each number of reservations (3, 4, 5, 6), compute the probability distribution:1. For 3 reservations, calculate the probability of 0 to 3 showing up.2. For 4 reservations, calculate the probability of 0 to 4 showing up.3. For 5 reservations, calculate the probability of 0 to 5 showing up.4. For 6 reservations, calculate the probability of 0 to 6 showing up.Sum the values for each passenger count to find the PMF of \( X \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Binomial Distributions

In the given problem, we are dealing with a scenario involving passengers who have made reservations for a trip. We're interested in finding out how many of those passengers actually show up. This is a classic example of a binomial distribution.

A binomial distribution helps us determine the probability of achieving a certain number of "successes" (in this case, passengers showing up) in a fixed number of trials (the total reservations made). For each reservation, there's a chance the person will show up (a success) or won't (a failure).

We know:

• Each reservation is an independent event.
• There are 6 reservations, so we have 6 trials.
• The probability of a passenger showing up (success) is 0.8.

This means we can use the binomial formula to calculate various probabilities. The formula is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
where \( n \) is the number of trials, \( k \) is the number of successes, and \( p \) is the probability of success.

Calculating Expected Value

The concept of expected value provides us with a measure of the "center" of a probability distribution. It tells us the average number that we can expect to see over many trials. In this context, we are interested in the expected number of passengers who do not show up—and by extension, the expected number of available seats.

Using the parameters of our binomial distribution, we can calculate this value using:

• The number of reservations, which is 6.
• The probability that a passenger does not show up, which is 0.2 (20%).

The formula for expected value is: \[ E = n \times p \] where \( n \) is the number of trials and \( p \) is the probability of failure (non-appearance).
Thus, in our example: \[ E = 6 \times 0.2 = 1.2 \] passengers expected not to show up. Therefore, if the limousine can take 4 passengers, we subtract 1.2 from 4 to find the expected seats available when it departs.

Understanding Probability Mass Function (PMF)

The probability mass function (PMF) is a vital concept in probability and statistics. It gives a complete summary of the probabilities associated with a discrete random variable (like the number of passengers showing up).

In the exercise, we use PMF to determine the probability of different numbers of passengers showing up based on varying numbers of reservations (3, 4, 5, or 6 reservations). The steps involved include:

• Identifying the number of reservations made.
• Calculating all possible outcomes of people showing up (0 to number reserved).
• Applying the probabilities from our binomial model to these outcomes.
• Aggregating these probabilities for each specific number of people actually showing up.

This results in the PMF for the number of passengers traveling, \( X \), helping predict the likelihood of specific passenger counts for any given trip.

Addressing Overbooking Probability

In practical scenarios, such as booking reservations for flights or limousine rides, businesses sometimes have to manage overbooking. Overbooking probability refers to the likelihood that more customers show up than there is capacity to accommodate them.

For this problem, this means finding the probability that more than four reserved passengers show up, since the limousine can only hold four passengers. Here, we're specifically interested in the probabilities of exactly five or six passengers showing up.
Using the binomial distribution, we calculate:

• \( P(X = 5) = \binom{6}{5} (0.8)^5 (0.2)^1 \)
• \( P(X = 6) = \binom{6}{6} (0.8)^6 \)

The sum of these probabilities tells us the overall probability of an overbooking situation on any given trip, which is critical for businesses to plan and potentially mitigate customer dissatisfaction.