How to compute $\mathbb{P}(\lambda X>4)$ directly? Given a random variable $X$ which is exponentially distributed i.e. $X\sim E(\lambda)$. Calculate $\mathbb{P}(X-\frac{1}{\lambda}>\frac{3}{\lambda})$.
My working:
$\mathbb{E}(X)=\frac{1}{\lambda}$, $Var(X)=\frac{1}{\lambda^2}$. Then $\mathbb{P}(X-\frac{1}{\lambda}>\frac{3}{\lambda})=\mathbb{P}(X>\frac{4}{\lambda})=\mathbb{P}(\lambda X>4)$. Then I am not sure how to compute this. 
The solution says its equal to $e^{-4}$, furthermore it says $\mathbb{P}(\lambda X>4)=e^{-4}$ is a direct consequence of standardised random variable. That is, it can be computed by considering the standardised random variable. 
My question is how can we relate or interpret $\lambda X>4$ to standardised X. What is standardised X in this case? is $\lambda X\sim\ E(1)$? Why and why not?
Thanks!
 A: A standard exponential, as you conjectured, is an exponential with mean $1$, and therefore parameter $\frac{1}{1}=1$. So the variance of a standard exponential is also $1$. (Referrring to a standard exponential is much less common than referring to a standard normal.) 
Let $X$ have exponential distribution with parameter $\lambda$, so with mean $\dfrac{1}{\lambda}$. Let $Y=\lambda X$. Then $Y$ has exponential distribution with mean $\lambda \cdot \dfrac{1}{\lambda}=1$.  So $Y$ has standard exponential distribution. 
Thus $Y=\lambda X$ is the standardized version of $X$. 
For your problem about $\Pr(X\gt \frac{4}{\lambda}$, where $X$ is exponentially distributed with parameter $\lambda$, there are two possible approaches.
$1.$ In general, the probability that $X\gt x$ is $e^{-\lambda x}$. Putting $x=\frac{4}{\lambda}$, we get that the probability is $e^{-4}$.
$2.$ We want the probability that $X\gt \frac{4}{\lambda}$. Now $X\gt \frac{4}{\lambda}$ if and only if $\lambda X\gt 4$. But $\lambda X$ is just the standardized exponential $Y$ discussed above. And since $Y$ has parameter $1$, we have $\Pr(Y\gt 4)=e^{-4}$.   The question really asked what is the probability of being more than $4$ "standard deviation units" from $0$. I am using that terminology to bring out the analogy with standard normal calculations. 
A: $$p(x>\frac{\lambda}{4}) =\int_{\frac{4}{\lambda}}^{\infty}\lambda\exp(-\lambda x)dx\\
= -\exp(\lambda x)|_{\frac{4}{\lambda}}^\infty\\
=\exp(-4)
$$
Not sure about the standardised part
