Problem 28 - A First Course in Probability

Question

The time that it takes to service a car is an exponential random variable with rate 1. (a) If A. J. brings his car in at time 0 and M. J. brings her car in at time \(t,\) what is the probability that M. J.'s car is ready before A. J's car? (Assume that service times are independent and service begins upon arrival of the car. (b) If both cars are brought in at time \(0,\) with work starting on M. J's car only when A. J.'s car has been completely serviced, what is the probability that M. J's car is ready before time \(2 ?\)

Step-by-Step Solution

Answer

(a) The probability that M.J.'s car is ready before A.J's car when M.J. brings hers at time \(t\) is \(e^{-t}\). (b) The probability that M.J's car is ready before time \(2\) when both cars are brought in at time \(0\) with work starting on M.J.'s car only when A.J.'s car has been completely serviced is \(1 - 3e^{-2}\).

Step 1: Define exponential distribution

We're given that the servicing times are exponential random variables with rate 1. An exponential random variable is described by the probability density function (pdf): \(f(x) = \lambda e^{-\lambda x}\), where \(x \ge 0\) and \(\lambda > 0\) is the rate parameter. Here, \(\lambda = 1\), so the pdf for both A.J.'s and M.J.'s service times is \(f(x) = e^{-x}\).

Step 2: Compute the probability

Let X and Y denote the service times of A.J. and M.J.'s cars, respectively. We're to compute the probability that M.J.'s car is ready before A.J.'s car, given M.J. brings in her car at time t. Mathematically, we want to find the probability \(P(Y < X)\). To find this, we need to compute the joint pdf of X and Y, which is given by \(f(x, y) = f(x) f(y)\), since the service times are independent. Thus, \(f(x, y) = e^{-x} e^{-y} = e^{-(x+y)}.\) Next, compute the limit of integration for x and y. Since M.J. brings her car at time \(t\), we have \(Y = y + t\). Therefore, we need to integrate over the region where \(y + t < x\).

Step 3: Integrate the joint pdf

Integrate the joint pdf to compute \(P(Y < X)\): \(P(Y < X) = \int_{0}^{\infty} \int_{t}^{\infty} e^{-(x+y)} dx dy\) First, integrate with respect to x: \(\int_{t}^{\infty} e^{-(x+y)} dx = -e^{-(x+y)} |_{t}^{\infty} = e^{-(y+t)} \) Now, integrate with respect to y: \(P(Y < X) = \int_{0}^{\infty} e^{-(y+t)} dy = -e^{-(y+t)} |_{0}^{\infty} = e^{-t} \) So, the probability that M.J.'s car is ready before A.J.'s car is \(e^{-t}\). #(b) Find the probability that M. J's car is ready before time 2#

Step 4: Define the problem

We now consider the case where both cars are brought in at time \(0\), but work on M.J.'s car starts only after A.J.'s car is completely serviced. We want to find the probability that M.J.'s car is ready before time \(2\). Let \(Z\) denote the total servicing time for both cars, \(Z = X + Y\). Since X and Y are independent exponential random variables, Z is a Gamma(2, 1) random variable.

Step 5: Define Gamma distribution

The Gamma distribution with shape parameter \(k\) and rate parameter \(\lambda\) has the pdf: \(f(z) = \frac{\lambda^k}{\Gamma(k)} z^{k-1} e^{-\lambda z}\), where \(z \ge 0\), \(k > 0\), and \(\lambda > 0\). In this case, \(k = 2\) and \(\lambda = 1\), so the pdf for Z is \(f(z) = ze^{-z}\).

Step 6: Compute the probability

We want to find the probability that Z is less than 2, i.e., \(P(Z < 2)\). Integrate the pdf: \(P(Z < 2) = \int_{0}^{2} ze^{-z} dz\) To integrate, we'll use integration by parts: Let: \(u = z\) and \(dv = e^{-z}dz\) Then: \(du = dz\) and \(v = -e^{-z}\) Using integration by parts: \(P(Z < 2) = -ze^{-z} - \int_{0}^{2} -e^{-z}dz\) Now, integrate the second part: \(- \int_{0}^{2} -e^{-z}dz = e^{-z} |_{0}^{2} = e^{-2} - e^{0} = e^{-2} - 1\) Finally, we evaluate the first part in the definite integral: \(-ze^{-z} |_{0}^{2} = -(2e^{-2} - 0e^{0}) = -2e^{-2}\) Thus, the final result for \(P(Z < 2)\) is: \(P(Z < 2) = -2e^{-2} + e^{-2} - 1 = 1 - 3e^{-2}\) The probability that M.J's car is ready before time \(2\) is \(1 - 3e^{-2}\).

Key Concepts

Probability Density Function

The probability density function, often abbreviated as pdf, is a statistical term that describes the likelihood of a continuous random variable taking on a specific value. In simpler terms, it provides the probability of a random variable falling within a particular range.
For an exponential random variable, the pdf is given by the formula:\( f(x) = \lambda e^{-\lambda x} \),where:

• \( x \geq 0 \)
• \( \lambda \) is the rate parameter, which must be greater than 0.

In the case of our exercise, the rate parameter \( \lambda = 1 \), meaning the pdf simplifies to \( f(x) = e^{-x} \). This distribution is often used to model time to event scenarios, such as waiting times in queues. The pdf helps in calculating probabilities involving the service times of A.J.'s and M.J.'s cars.

Joint Probability

Joint probability involves the probability of two events occurring simultaneously. When dealing with independent random variables like the servicing times in our exercise, the joint probability distribution can be found by multiplying their individual probability density functions.
For independent exponential random variables \( X \) and \( Y \) with rate \( 1 \), the joint probability density function (pdf) is:\[ f(x, y) = f(x) f(y) = e^{-x} e^{-y} = e^{-(x+y)} \]This lets us determine combined outcomes — such as M.J.'s car being serviced before A.J.'s. By setting up the integration to evaluate the region where M.J.'s service time plus arrival time \( y + t \) is less than A.J.'s service time \( x \), we can use this joint pdf to calculate the desired probability. The result is an exponential decay in probability relative to time \( t \), summarized as \( e^{-t} \). This illustrates how integrals of joint pdfs in continuous cases give us cumulative probabilities.

Gamma Distribution

The Gamma distribution is a two-parameter family of continuous probability distributions, defined by a shape parameter \( k \) and a rate parameter \( \lambda \). It is often used to model waiting times for multiple events where the intervals of time are independent and identically distributed exponential random variables.
The pdf of a Gamma distribution is expressed as:\[f(z) = \frac{\lambda ^k}{\Gamma(k)} z^{k-1} e^{-\lambda z}\]where:

• \( z \geq 0 \)
• \( \Gamma(k) \) is the Gamma function, an extension of the factorial function to real and complex numbers.

For our scenario, \( k = 2 \) and \( \lambda = 1 \), so the pdf simplifies to \( f(z) = z e^{-z} \). In the context of servicing cars, the total service time \( Z = X + Y \) (A.J.'s service plus M.J.'s service) follows a Gamma distribution.
This model allows us to compute probabilities for scenarios such as M.J.'s car being ready before a specific time. In the exercise, we find \( P(Z < 2) \) using integration, yielding the probability expression \( 1 - 3e^{-2} \) — highlighting the versatility of the Gamma distribution in such scenarios.