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I'm struggling to see how I would manipulate the PDF so that I could find the conditional probability?

X is an exponential RV with $\lambda>0$ and the pdf=$\lambda e^{-\lambda x}$

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  • $\begingroup$ $P(A|B)=\frac{P(A\cap B)}{P(B)}$. In this case, when the event $X>x+y$ occurs, the event $X>x$ always occurs, so $A\cap B=A$. $\endgroup$ – Mike Earnest Feb 12 at 21:49
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You have $$ \mathbb{P}[X > x+y | X>x] = \frac{\mathbb{P}[X > x+y, X>x]}{\mathbb{P}[X>x]} = \frac{\mathbb{P}[X > x+y]}{\mathbb{P}[X>x]} = \frac{1-F(x+y)}{1-F(x)}. $$ where $F$ is the cdf of the r.v. in question. Can you compute $F$ from your pdf and simplify the resulting expression? It looks quite nice in the end...

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