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?

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#### Best Answer

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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