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Let $X$ be an exponential random variable with parameter $\lambda > 0$. Find the probabilities $$P(X>2/\lambda)$$ and $$P(|X-1/\lambda|<2/\lambda)$$

I am not sure where to start with this. I know that $P(X>x)=1-F(x)=e^{-\lambda x}$. Does this help with the first one in some way? Do I plug $2/\lambda$ in for $x$ and that gets rid of the $\lambda$'s so the answer would be ${e^-}^2$? Lost on the second.

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Yes, the fact that you know that $$\Pr(X\gt x)=e^{-\lambda x}$$ (if $x\ge 0$) is very useful. Equivalently, $$F(x)=1-e^{-\lambda x}$$ when $x\ge 0$, and $F(x)=0$ elsewhere. Being aware of the "elsewhere" part can help avoid error.

For the first problem, just substitute $2/\lambda$ for $x$. Note that $\lambda(2/\lambda)=2$.

For the second, let's "unwrap" what $|X-1/\lambda|\lt 2/\lambda$ says. It says that $X$ is not too far from $1/\lambda$, indeed within $2/\lambda$ of $1/\lambda$. More precisely, it says that $$\cfrac{1}{\lambda}-\cfrac{2}{\lambda} \lt X \lt \cfrac{1}{\lambda}+\cfrac{2}{\lambda}.$$ Some simplification reduces this to $$-\cfrac{1}{\lambda}\lt X\lt \cfrac{3}{\lambda}.$$ But note that an exponentially distributed random variable can never be negative. So all we want is $$\Pr\left(X\lt \cfrac{3}{\lambda}\right).$$ Another way of arriving at the answer is to note that our probability is $F(3/\lambda)-F(-1/\lambda)$. But now we must be careful. Since $-1/\lambda$ is negative, $F(-1/\lambda)=0$, so our answer is just $F(3/\lambda)$. We get this by substituting $3/\lambda$ for $x$ in the formula for $F(x)$.

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