91Ó°ÊÓ

Airlines sometimes overbook flights. Suppose that for a plane with 50 seats, 55 passengers have tickets. Define the random variable \(Y\) as the number of ticketed passengers who actually show up for the flight. The probability mass function of \(Y\) appears in the accompanying table. \begin{tabular}{l|lllllllllll} \(y\) & 45 & 46 & 47 & 48 & 49 & 50 & 51 & 52 & 53 & 54 & 55 \\ \hline\(p(y)\) & \(.05\) & \(.10\) & \(.12\) & \(.14\) & 25 & \(.17\) & \(.06\) & \(.05\) & \(.03\) & \(.02\) & \(.01\) \end{tabular} a. What is the probability that the flight will accommodate all ticketed passengers who show up? b. What is the probability that not all ticketed passengers who show up can be accommodated? c. If you are the first person on the standby list (which means you will be the first one to get on the plane if there are any seats available after all ticketed passengers have been accommodated), what is the probability that you will be able to take the flight? What is this probability if you are the third person on the standby list?

Step by step solution

01

Define the Requirement

We need to determine the probabilities for various conditions based on the number of passengers showing up for a flight that can accommodate 50 seats. We are given probabilities for each possible value of passengers showing up, from 45 to 55.

02

Determine Probability of Accommodating All Passengers (a)

All passengers can be accommodated if at most 50 passengers show up. Thus, we need to sum the probabilities for 45 through 50. This can be expressed as:\[ P(Y \leq 50) = p(45) + p(46) + p(47) + p(48) + p(49) + p(50) = 0.05 + 0.10 + 0.12 + 0.14 + 0.25 + 0.17 = 0.83 \]

03

Determine Probability of Not Accommodating All Passengers (b)

For the scenario where not all passengers can be accommodated, more than 50 passengers must show up. Hence, we sum the probabilities for 51 through 55:\[ P(Y > 50) = p(51) + p(52) + p(53) + p(54) + p(55) = 0.06 + 0.05 + 0.03 + 0.02 + 0.01 = 0.17 \]

04

Probability for First Standby Passenger (c) - First Passenger

A standby passenger can take the flight if fewer than 50 ticketed passengers show up. Thus, we calculate the probability for fewer than 50 passengers showing up:\[ P(Y < 50) = p(45) + p(46) + p(47) + p(48) + p(49) = 0.05 + 0.10 + 0.12 + 0.14 + 0.25 = 0.66 \]

05

Probability for Third Standby Passenger (c) - Third Passenger

The third standby passenger can be accommodated if fewer than 48 passengers show up, considering they need 3 seats to be open. So, we find the probability for fewer than 48 passengers:\[ P(Y < 48) = p(45) + p(46) + p(47) = 0.05 + 0.10 + 0.12 = 0.27 \]

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Key Concepts

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

Random Variable

In probability and statistics, a random variable is a variable that can take on different values, each with an associated probability. It represents a quantity subject to variation due to chance. In our exercise, the random variable is denoted as \( Y \). It represents the number of ticketed passengers who show up for the flight.
Each potential number of passengers showing up (from 45 to 55) is a value that \( Y \) can assume, with each having a given probability.
This concept helps us understand and manage the uncertainty in various scenarios, such as predicting how many passengers will actually arrive for a flight that was overbooked.

Overbooking

Overbooking occurs when a company, like an airline, sells more tickets for a flight than there are seats available. This is a common practice intended to offset the possibility of no-shows and maximize profits. In our exercise, there are 55 ticketed passengers for a plane that only has 50 seats.
Airlines often rely on historical data to estimate the likelihood of a certain number of passengers not showing up, allowing them to sell additional tickets without necessarily requiring extra seats.

• Overbooking can create situations where not all ticketed passengers can be accommodated.
• It highlights the importance of probability mass functions (pmf) in predicting possible outcomes.

Understanding overbooking helps explain why airlines may take the risk of selling more tickets than available spots.

Standby Passengers

Standby passengers are those waiting for an available seat if all ticketed passengers do not show up. This section delves into the scenario where we involve these individuals. If some seats are unoccupied after all ticketed passengers have been accommodated, the standby passengers get a chance to board.
In the exercise, we consider the likelihood of standby passengers securing a seat:

• The first standby passenger can take a seat if fewer than 50 ticketed passengers show up.
• The third standby passenger needs fewer than 48 passengers to board, ensuring three seats are available.

Standby passengers exemplify the practical application of probability in determining their chances of boarding the plane.

Probability Calculation

Probability calculation is the process of determining the likelihood of various outcomes. It is crucial in this scenario to understand which passengers can fly. Let's break down the calculations for the exercise:

• For accommodating all passengers, we calculate \( P(Y \leq 50) \) by summing probabilities from 45 to 50, resulting in 0.83.
• For not accommodating all passengers, \( P(Y > 50) \) is calculated by summing probabilities from 51 to 55, equaling 0.17.
• For the first standby passenger, \( P(Y < 50) \) is derived by adding probabilities from 45 to 49, giving us 0.66.
• Finally, for the third standby passenger, \( P(Y < 48) \) is the sum from 45 to 47, which results in 0.27.

Each of these probability calculations is a straightforward summation of the respective probabilities derived from the probability mass function.