Let $X$ be exponential distributed with parameter $\lambda>0$, i.e. it holds that $P[X\leq x]=1-e^{-\lambda x}$.
Calculate $P[X\cdot E[X]\geq 1]$.
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It holds that $E[X]=\frac{1}{\lambda}$, right?
So we get \begin{align*}P[X\cdot E[X]\geq 1]&=P\left [X\cdot \frac{1}{\lambda}\geq 1\right ]=P\left [X\geq \lambda \right ]\\ & =1-P[X<\lambda ]=1-P[X\leq\lambda ]+P[X=\lambda ] \\ & =1-\left (1-e^{-\lambda \cdot \frac{1}{\lambda}}\right )+\frac{\lambda^{\lambda}}{e^{-\lambda}}\cdot \lambda!\\ & =e^{-1}+\frac{\lambda^{\lambda}}{e^{-\lambda}}\cdot \lambda!\end{align*} Is that correct?
It is correct that $E[X]= \frac{1}{\lambda}$. But the equation $$P[X\geq \lambda] = 1-P[X \leq\lambda] + P[X=\lambda]$$ is not correct since for a continuos distribution the term $P[X=\lambda]$ is not really defined (as A rural reader pointed out). Because of the definition of the distribution function with the help of an integral, namely $$P[X \leq x] := \int_{0}^x \lambda e^{-\lambda t} dt$$ for $x \geq 0$ and $P[X \leq x] = 0$ for $x \leq 0$ one can see that $$P[X \geq \lambda] = 1- \int_0^{\lambda} \lambda e^{-\lambda t} dt = 1- (1-e^{-\lambda^2}) = e^{-\lambda^2}$$