Problem 34 - A First Course in Probability

Question

Jones figures that the total number of thousands of miles that an auto can be driven before it would need to be junked is an exponential random variable with parameter \(\frac{1}{20} .\) Smith has a used car that he claims has been driven only 10,000 miles. If Jones purchases the car, what is the probability that she would get at least 20,000 additional miles out of it? Repeat under the assumption that the lifetime mileage of the car is not exponentially distributed, but rather is (in thousands of miles) uniformly distributed over (0,40)

Step-by-Step Solution

Answer

The probability that Jones would get at least 20,000 additional miles out of the car for the exponential random variable case is \(1 - (1 - e^{-\frac{3}{2}})\), while for the uniformly distributed random variable case, the probability is \(\frac{1}{4}\).

Step 1: Define the exponential random variable

Let X be the exponential random variable representing the total number of thousands of miles a car can be driven with parameter \(\lambda = \frac{1}{20}\).

Step 2: Calculate the probability for the exponential case

We want to find the probability that Jones would get at least 20,000 additional miles out of the car. Since the car has already been driven 10,000 miles, we want to find the probability that the car has a remaining lifetime of at least 20,000 miles, or P(X - 10 >= 20). Since X has an exponential distribution, the cumulative distribution function (CDF) is given by: \(P(X \le x) = 1 - e^{-\lambda x}\). We want to find the probability that X - 10 >= 20 (a remaining lifetime of at least 20,000 miles), which can be written as P(X >= 30). We know that P(X >= 30) = 1 - P(X <= 30). Thus, we can use the CDF equation to find the probability: \(P(X \ge 30) = 1 - (1 - e^{-\lambda(30)})\) Plugging in the given parameter value (\(\lambda = \frac{1}{20}\)), we get: \(P(X \ge 30) = 1 - (1 - e^{-\frac{1}{20}(30)})\) Now calculate the probability: \(P(X \ge 30) = 1 - (1 - e^{-\frac{3}{2}})\) For the uniformly distributed case:

Step 3: Define the uniformly distributed random variable

Under the assumption that the lifetime mileage of the car is uniformly distributed over (0, 40), let Y be the uniformly distributed random variable representing the total number of thousands of miles a car can be driven.

Step 4: Calculate the probability for the uniformly distributed case

We want to find the probability that Jones would get at least 20,000 additional miles out of the car, which is P(Y - 10 >= 20). Since the car has already been driven 10,000 miles, we want to find the probability that the car has a remaining lifetime of at least 20,000 miles, or P(Y >= 30). Since Y has a uniform distribution over (0, 40), its probability density function (PDF) is given by \(\frac{1}{40 - 0} = \frac{1}{40}\) for all values of Y in the range [0, 40]. Now, we need to find the probability P(Y >= 30). This can be done by calculating the area under the PDF curve for Y in the range [30, 40]: \(P(Y \ge 30) = \int_{30}^{40} \frac{1}{40} dy\) Calculate the probability: \(P(Y \ge 30) = \left[\frac{1}{40}y\right]_{30}^{40} = \left[\frac{40}{40} - \frac{30}{40}\right] = \frac{1}{4}\) The probability that Jones would get at least 20,000 additional miles out of the car for the exponential random variable and the uniformly distributed random variable cases are \(1 - (1 - e^{-\frac{3}{2}})\) and \(\frac{1}{4}\), respectively.

Key Concepts

Exponential Random Variable

Let's start by understanding what an exponential random variable is. This type of variable is used in probability theory to model the time between events in a process that occurs continuously and independently at a constant average rate.

The mathematical representation involves the rate or the parameter \(\lambda\), which signifies how often the event occurs. The exponential random variable is characterized by its memoryless property, meaning the probability of an event occurring in the next interval is independent of how much time has already passed.

Applying this to our problem, the exponential random variable models the total mileage a car can be driven before it's junked, with the parameter \(\lambda = \frac{1}{20}\), which signifies the car's expected life span in terms of thousands of miles. To visualize this, think of \(\lambda\) as the 'thinness' or 'thickness' of the chance of the car surviving each additional mile, with a 'thicker' \(\lambda\) indicating a car more likely to break down sooner.

Cumulative Distribution Function

Next is the cumulative distribution function (CDF), a crucial concept for understanding probabilities in statistics. The CDF shows the probability that a random variable \(X\) will take a value less than or equal to \(x\). In formula form, for our exponential random variable, the CDF is expressed as \(P(X \le x) = 1 - e^{-\lambda x}\).

In the context of our car's mileage, the CDF can be used to determine the likelihood that the car will last up to a certain point. By inputting the value of \(x\), which in our scenario is the additional mileage (in thousands of miles), we can find the probability that the car will make it to at least this milestone. Notably, for extrapolating beyond a certain period (e.g., additional miles beyond the current mileage), we adjust our CDF equation to calculate the probability of the car's survival beyond that point.

Uniform Distribution

When considering uniform distribution, we're looking at a scenario where every outcome in a range is equally likely to occur. Unlike the exponential random variable, there's no decreasing probability over time or mileage.

In our automobile scenario, the assumption of uniform distribution over the interval (0, 40) implies each mileage value between 0 and 40,000 miles is equally probable. To visualize uniform distribution, imagine a flat stretch of road representing the car's lifespan, with no hills or valleys indicating more or less probable distances at which the car might fail; every point is equally likely.

When we talk about uniform distribution, we often use a probability density function (PDF), which helps us find the likelihood of the car reaching a certain mileage within this flat probability terrain.

Probability Density Function

The probability density function (PDF) is tied closely with both uniform and non-uniform distributions. It describes the relative likelihood for a random variable to take on a given value. In the case of uniform distribution, the PDF is particularly simple—it's constant across the range of possible values. For our car's uniform distribution case, the PDF equals \(\frac{1}{40}\) for values between 0 and 40,000.

To calculate the probability of the car surviving past 30,000 miles, we integrate this constant over the interval from 30 to 40. This linear nature of the PDF for uniform distribution makes it a powerful tool for calculating probabilities for ranges of values with equal likelihood, as we've seen with the problem of predicting the car's additional mileage.

Thus, by understanding the PDF, students can approach problems of probability with a solid foundation, visualizing not just the likelihood of single points but the 'weight' of entire ranges of an outcome.