Q. 6.28 - 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 of M.J's car being ready before A.J's car is $\frac{e^{-t}}{2}$

(b) the probability for M.J's car is ready before time 2 is $1-3e^{-2}$

Step 1: Given information (part a)

Time for servicing is an exponential random variable with a rate 1.

A.J brings her car in at time 0.

M.J brings her car in at time t.

Step 2: Explanation (part a)

Let, X be the time for A.J's car and Y be the time for M.J's car.

Then,

$f(x,y)=\left\{\begin{array}{l}{e}^{-(x+y)} \text{if} 0<y+t<x,0<x<\infty \\ 0 \text{otherwise}\end{array}\right.$

Thus,

$$\begin{gathered} P(X+Y<2)={\int }_0^{\infty }{\int }_{1+y}^{\infty }{e}^{-(x+y)}dxdy \\ =-{\int }_0^{\infty }{\left[e^{-(x+y)}\right]}_{t+y}^{\infty }dy \\ =-{\int }_0^{\infty }\left[e^{-(x+y)}-e^{-(-(t+y)+y)}\right]dy \\ =-{\left[e^{-(\infty +y)}+\frac{e^{-\left(\right(t+y)+y)}}{2}\right]}_0^{\infty } \\ =-{\left[e^{-(\infty +\infty )}+\frac{e^{-\left(\right(t+\infty )+\infty )}}{2}\right]}_{}^{} \\ =\frac{e^{-t}}{2} \end{gathered}$$

Step 3: Given information (part b)

Time for servicing is an exponential random variable with rate 1.

Both cars are brought in at time 0.

M.J's car is serviced only after complete service of A.J's car.

Step 4: Explanation (part b)

Let, x be the time for A.J's car and y be the time for M.J's car.

Now, we are required to find $P(X+Y<2)$

$$\begin{gathered} P(X+Y<2)={\int }_0^2{\int }_0^{2-y}{e}^{-(x+y)}dxdy \\ =-{\int }_0^2{\left[e^{-(x+y)}\right]}_0^{2-y}dy \\ =-{\int }_{}^{}\left[e^{-(2-y+y)}-e^{-(0+y)}\right]dy \\ =-{\left[e^{-2}-e^{-y}\right]}_0^2 \\ =-{\left[e^{-2}-e^{-y}\right]}_0^2 \\ =\left(-e^{-2}-e^{-2}\right)-\left(e^{-2}-e^0\right) \\ =1-3e^{-2} \end{gathered}$$