Compute $P(X>40\; |\; X>10)$ where $X$ has an exponential distribution Please could someone advise if I have interpreted this problem correctly
Let $X$ have an exponential distribution  with a mean of $i = 20$
(1) Compute $P(X>40 \;| \;X>10)$
I believe the correct solution here is to find $P(X>40)$ because of the inclusion / exclusion principle. That is if we define the event $A = P(X>40)$ and the event $B = P(X>10)$. In addition, define $A \cap B$ as $P(X>10)$. Then to compute (1) we find the following $P(X>40) \cup P(X>10) - P(X>10)$ which is 
$$1- (1-e^{-\frac{30}{20}})=0.2231$$
EDIT
Due to the "memorylessness" of the probability distribution 
$$P(X>40 \mid X>10)= P(X>30) $$
Thanks in advance

Yes, due to "memorylessness" of an exponential distribution, if $s>0$, $t>0$, then
$$P(X>s+t\mid X>s) = P(X>t) = e^{-\lambda t}.$$
We can find that if we simply calculate the expression, the result will show the "memorylessness":
$$
\begin{align}
P(X>s+t\mid X>s) &= \frac{P\big(X>s+t\cap X>s\big)}{P(X>s)} = \frac{P(X>s+t)}{P(X>s)} 
\\
&= \frac{\int_{s+t}^{\infty}\lambda e^{-\lambda x}dx}{\int_{s}^{\infty}\lambda e^{-\lambda x}dx} = e^{-\lambda t} = P(X>t),\quad (s>0,t>0).
\end{align}
$$
