# Solved – Exponential random variable

The time it takes a printer to print a job is an exponential random variable with mean of 10 seconds. You send a job to the printer at 9:00 am, and it appears to be the fourth in line. What is the probability that your job will be ready before 9:01 (i.e. before 60 seconds)

My solution:

\$E[X] = 1/λ = 10\$

\$λ=1/10\$

\$P(X≥a) = e^{-λa}\$

\$P(X<60) = 1 – P(X≥60) = 1 – e^{-60/10} = 1 – e^{-6} = 1 -1/e^6 = 1 – 0.0024 = 0.9976\$

Would the printer being "the fourth in line" would make a difference?

Contents

To understand what is required, consider a slightly simpler problem: how much time does it take two jobs to finish? Call these jobs \$X\$ and \$Y\$, which are independent \$Exp(1/10)\$ random variables. What is the distribution of \$T = X+Y\$?

The pdf of an \$Exp(1/10)\$ r.v. is \$f(x) = frac{1}{10}e^{frac{-x}{10}}\$, which implies that if the value of \$X\$ was known, say \$X=x\$, then the pdf for \$T = X+Y\$ given \$X=x\$ would be \$g(t) = frac{1}{10}e^{frac{-(t-x)}{10}}\$, where \$t\$ must now be greater than \$x\$. That is the same as the original distribution only shifted \$x\$ seconds to the right.

Since we know the pdf for \$X\$ and the pdf for \$T|X=x\$, then we find the marginal distribution of \$T\$ by integrating/averaging over our uncertainty about \$X\$. In what follows, think of fixing \$T\$ at \$t\$ and taking a weighted average of the possible values of the conditional pdf \$T|X=x\$ where the weights are given by the pdf for \$X\$ over \$0<x< t\$:

\$\$int_0^{t} frac{1}{10}e^{frac{-(t-x)}{10}} f(x) ;dx = int_0^{t} frac{1}{10^2}e^{frac{-(t-x)}{10}}e^{frac{-x}{10}};dx\$\$ \$\$=frac{1}{10^2}e^{-t}int_0^{t};dx\$\$ \$\$=frac{1}{10^2}t;e^{-frac{t}{10}}\$\$

This pdf has the form of a gamma, namely \$Gamma(2,1/10)\$. It is not hard to see that if we iterate the procedure above to extend to the required four jobs, the power in the constant term and in the \$t\$ term will increment by 1 each time, so that the pdf will be \$frac{1}{10^4}t^3;e^{-frac{t}{10}}\$, a \$Gamma(4,1/10)\$. Now you can integrate to find \$P(tle 60)\$;

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