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

Under the exponential distribution, the probability that Jones would get at least 20,000 additional miles out of the car is approximately 22.31%. If the car's total lifetime mileage is uniformly distributed, the probability is 25%.

Step 1: Exponential Distribution

If the car's lifetime mileage is exponentially distributed with a parameter of \(\frac{1}{20}\), its probability density function (PDF) is: \[f(x) = \frac{1}{20}e^{-\frac{1}{20}x} \text{, for x > 0}\] Since we want to calculate the probability that Jones would get at least 20,000 additional miles out of the car, which has already been driven 10,000 miles, we have to calculate the probability that the car's total lifetime mileage is greater than 30,000 miles. So, we have to find the following probability: \[P(X > 30)\] Since we're working with an exponential distribution, we can calculate this using the complementary cumulative distribution function (CCDF): \[P(X > x) = 1 - P(X \le x) = e^{-\frac{1}{20}x}\] Plug in 30 as the value of x. \[P(X > 30) = e^{-\frac{1}{20}(30)}\] Now, let's calculate the probability. \[P(X > 30) = e^{-\frac{3}{2}} \approx 0.2231\] In the exponential distribution case, the probability that Jones would get at least 20,000 additional miles out of the car is approximately 22.31%.

Step 2: Uniform Distribution

Now, let's solve the problem under the assumption that the car's lifetime mileage (in thousands of miles) is uniformly distributed over (0, 40). The probability density function (PDF) of the uniform distribution is: \[f(x) = \frac{1}{b-a} = \frac{1}{40-0} = \frac{1}{40} \text{, for 0 < x < 40}\] Now we must find the probability that the car will last for at least 20,000 more miles, given it has already been driven 10,000 miles. In other words, we need to find the probability that the car's total lifetime mileage is greater than 30,000 miles, or P(X > 30). In the uniform distribution case, the probability can be calculated as follows: \[P(X > 30) = \frac{40 - 30}{40} = \frac{10}{40} = \frac{1}{4}\] So, the probability that Jones would get at least 20,000 additional miles out of the car if the car's total lifetime mileage is uniformly distributed is 25%. In conclusion, under the exponential distribution, the probability that Jones would get at least 20,000 additional miles is approximately 22.31%, and under the uniform distribution, the probability is 25%.

Key Concepts

Exponential Distribution

The exponential distribution is a continuous probability distribution commonly used to model the time or space between events in a process that occurs at a constant rate. It is characterized by its parameter, often denoted as \( \lambda \), which is the rate of occurrence. For instance, if we have \( \lambda = \frac{1}{20} \), it means the average lifetime of an event is 20 units.

The exponential distribution has a memoryless property, meaning the probability of an event occurring in the future is independent of the past. This makes it quite useful for modeling lifetimes and waiting times.

The probability density function (PDF) of an exponential distribution is given by:

• \( f(x) = \lambda e^{-\lambda x} \) for \( x > 0 \)

To find the probability that a random variable \( X \) is greater than a certain value \( x \), we use the complementary cumulative distribution function (CCDF):

• \( P(X > x) = e^{-\lambda x} \)

In our case, the question asks for the probability of the car lasting at least an additional 20,000 miles given it has already been driven 10,000 miles. This translates to needing the probability that \( X > 30 \) when \( \lambda = \frac{1}{20} \), which computes to about a 22.31% chance.

Uniform Distribution

The uniform distribution, another continuous distribution, is straightforward and intuitive. All outcomes in its range are equally likely. If we consider a range \([a, b]\), the probability density function (PDF) of a uniform distribution is specified as:

• \( f(x) = \frac{1}{b-a} \) for \( a < x < b \)

This distribution is useful when each interval within a range has the same probability of occurrence, like the roll of a fair die yielding any number from 1 to 6.

If we want to calculate the probability of the car having a lifetime mileage exceeding 30, given a uniform distribution over the range (0, 40), we find the section of the interval greater than 30. The calculation is simple:

• \( P(X > 30) = \frac{b - 30}{b-a} = \frac{40 - 30}{40 - 0} = \frac{1}{4} \)

The result is that there is a 25% probability of the car being able to drive at least another 20,000 miles if uniformly distributed over (0, 40). Thus, under uniform distribution assumptions, things are more equally predictable, giving a slightly higher chance for this car scenario.

Lifetime Mileage

Lifetime mileage refers to the total distance a car can travel before reaching the end of its useful life. It's important because it helps potential buyers assess how much value remains in a used car. Understanding how this "remaining lifetime" is distributed can affect a buyer's decision.

A key point about lifetime mileage is that it may not always follow a specific distribution, but when it does, it often helps in making informed purchasing decisions or warranties.

For Jones, determining whether the car's lifetime mileage follows an exponential or uniform distribution impacts the perceived probability of how much more it can drive. With an exponential distribution, given a rate of \( \lambda = \frac{1}{20} \), the chance of reaching beyond 30,000 in total mileage is lower due to the memoryless nature of the distribution that naturally tails off.

Under uniform distribution, the prediction is less influenced by previous data, providing a flat expectation over its entire range. This reflects consistency in remaining usage, albeit only assuming linear wear across time and events. Understanding these models can help in making cost-effective decisions when considering a car's prospective reliability and longevity.