Problem 50 - Basic Statistics for Business and Economics with Student

Question

Assume the likelihood that any flight on Northwest Airlines arrives within 15 minutes of the scheduled time is .90. We select four flights from yesterday for study. a. What is the likelihood all four of the selected flights arrived within 15 minutes of the scheduled time? b. What is the likelihood that none of the selected flights arrived within 15 minutes of the scheduled time? c. What is the likelihood at least one of the selected flights did not arrive within 15 minutes of the scheduled time?

Step-by-Step Solution

Answer

a. 0.6561 b. 0.0001 c. 0.3439

Step 1: Define Probability Notations

Let the probability that a flight arrives within 15 minutes of the scheduled time be denoted by \( p = 0.9 \). The probability that a flight does not arrive within 15 minutes is \( 1 - p = 0.1 \).

Step 2: Calculate Probability of All Flights On Time

The probability that all four flights arrived within 15 minutes of the scheduled time is the product of probabilities that each individual flight arrived on time. This is given by: \( (0.9)^4 = 0.6561 \).

Step 3: Calculate Probability of No Flights On Time

The probability that none of the four flights arrived within 15 minutes of the scheduled time is the product of probabilities that each individual flight did not arrive on time. This is given by: \( (0.1)^4 = 0.0001 \).

Step 4: Calculate Probability of At Least One Late Flight

The probability that at least one flight did not arrive within 15 minutes is the complement of all flights arriving on time. This is given by: \( 1 - (0.9)^4 = 1 - 0.6561 = 0.3439 \).

Key Concepts

Probability Calculations

Probability is the measure of how likely an event is to occur. It ranges between 0 and 1, where 0 indicates an impossible event, and 1 indicates a certain event. In probability calculations, you often multiply the likelihood of independent events to find the probability of all events happening together. For example, to calculate the probability that all four flights arrive on time from the problem, each flight is an independent event with a probability of 0.9 (since each flight has a 90% chance of being on time). The combined probability of all flights being on time is obtained by multiplying:

• The probability of the first flight being on time: 0.9,
• times the probability of the second flight being on time: 0.9,
• times the probability of the third flight: 0.9,
• times the probability of the fourth flight: 0.9.

This calculation results in: \[(0.9)^4 = 0.6561\]As shown, multiplication is key to solving probability of consecutive independent events.

Complementary Probability

Complementary probability refers to the probability of an event not occurring. It's an important concept because sometimes it's easier to calculate the chance of something not happening and then subtracting it from 1 to find the likelihood of the event happening. In the exercise, we use this when calculating probabilities such as the chance that at least one flight did not arrive on time. If the probability that all flights arrive on time is 0.6561, the probability that at least one flight did not arrive on time will complement it:

• The probability of all events occurring is 0.6561.
• The complementary probability (the chance that at least one does not occur) is:\[1 - 0.6561 = 0.3439\]

Understanding complementary probability is very useful, especially in cases where calculating the direct probability can be complex or cumbersome.

Binomial Probability

Binomial probability involves situations where there are only two possible outcomes for each trial, such as success or failure. This suits our exercise because a flight can either arrive on time (success) or not (failure). The situation here involves four flights, and each one can independently be on time or late. This setup perfectly fits a binomial experiment. The formula for binomial probability is used to calculate the probability of getting a certain number of successes in a fixed number of trials. It is expressed as:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]where:

• \(n\) is the number of trials (4 flights),
• \(k\) is the number of successes we are interested in (such as all four flights on time),
• \(p\) is the probability of success (0.9),
• \(1-p\) is the probability of failure (0.1).

This formula can be adapted to answer various questions about probabilities in scenarios like this.