Can someone help me with this probability density function? I'm wondering how to find the probability that it will take longer than 10 minutes to finish a test when the given function is:
$$ f(t) = \frac{1}{9} e^{-\frac{1}{9}t} $$
I'm not sure how to do set notation on latex, but for f(t):
$ \frac{1}{9}e^{-\frac{1}{9}}$  when $ t >= 0 $
$ 0 $ otherwise
I know how to do it for a set number, ie I know how to determine if it is less than 8 minutes. I would simply set the bounds on the integral from 0 to 7.
But how do I do it for an improper integral like 10 to infinity?
 A: When you take the anti derivative, you get the same e-power back, meaning with the same exponent. But due to the chain rule, your coefficient is to be multiplied by a factor -1/9. Now when you plug in -infinity in the anti derivative, then that value becomes zero (why?) and thus plugging in t=10 should give you the answer (Both negatives cancel, your answer is positive). I have given you the recipe, you try to work it out.
A: Let $X$ be the amount of time taken. We want the probability that $X\gt 10$. Let us first calculate something you know how to do, $\Pr(X\le 10)$.   We have
$$\Pr(X\le 10)=\int_0^{10} \frac{1}{9}e^{-t/9}\,dt.$$
To find the probability that the time taken is greater than $10$, find $1-\Pr(X\le 10)$.
Remark: Another way of putting it is that the probability that the time is greater than $10$ is 
$$\int_{10}^\infty \frac{1}{9}e^{-t/9}\,dt.$$
For the exponentially distributed random variable $X$ of the problem, we have 
$$\Pr(X\le x)=\int_0^{x} \frac{1}{9}e^{-t/9}\,dt=1-e^{-x/9}.$$
The expression for $\Pr(X\gt x)$ is a little simpler, it is $e^{-x/9}$. 
