Exponential distribution wait time probability I would like to check my answer, I have been asked to work out the probability of value greater then 10 given an exponential distribution with a mean of 10. 
My intuition would be that this is equal to 0.5 but when I go to do a definite integral between 10 and infinity I get a decimal answer so small I would round it to zero.
What is the correct answer and what are the steps. Sorry if this is too simple.
 A: I hate parameters and always try to get rid of them.
Let $Y$ have exponential distribution with parameter $\lambda=1$. Then $1$ is its mean and:
$$P(Y>1)=\int_1^\infty e^{-y}dy=\left[-e^{-y}\right]_1^\infty=e^{-1}$$
Now realize that $X:=10Y$ has exponential distribution with mean $\lambda=10$ and: $$P(X>10)=P(Y>1)=e^{-1}$$
A: If $X \sim Exp(\mu = 10),$ then $P(X > 10) = e^{-1} =  0.3678794.$
Compare with the cumulative distribution function (CDF) of this distribution for details.
This should also be $\int_{10}^\infty f(x)\,dx,$ where $f(x)$ is the
density function of $X$. Are you sure you are integrating the
correct density function? Some books write the density function
in terms of the rate $\lambda = 1/\mu.$
If you wanted $P(X > \eta) = 0.5,$ then take logs to find $\eta = 6.931472.$
In this case $\eta$ is the median of $X.$
A: Let the pdf be $p(x)=\lambda e^{-\lambda x}$.  
Mean, $\mu=1/\lambda =10$.
$$P(X>10)=\int_\mu^\infty p(x)dx
=\int_{1/\lambda}^\infty \lambda e^{-\lambda x}dx
=\left[\frac{\lambda e^{-\lambda x}}{-\lambda}\right]_{1/\lambda}^\infty=\frac 1e\qquad\blacksquare$$
Note that the answer is independent of $\mu$. 
