Random variable $X$ has exponential distribution with parameter $1$. What's the probability $P(X\leq2)$
How can I calculate $P(X=1)$ etc? Do I have to get aid from Poisson distribution like this:
$P(X=k)=\frac{\lambda^k}{k!}e^{-k}$
$P(X=0)=\frac{1^0}{0!}e^{-1}=\frac{1}{e}$
$P(X=1)=\frac{1^1}{1!}e^{-1}=\frac{1}{e}$
$P(X=2)=\frac{1^2}{2!}e^{-1}=\frac{1}{2e}$
$P(X=0)+P(X=1)+P(X=2)=\frac{5}{2e}$
Does it make sense?
An exponential variable is absolutely continuous and thus have a density which is a function $f$ that can be used to define the probability that the variable lies in an interval:
$$ \mathbb P(a \leq X \leq b ) = \int_a^b f(x)dx $$
For absolutely continuous variable, it is more relevant to compute the probability of the variable falling in a given interval rather than computing the probability of the variable being equal to a number because $\mathbb P(X=x)=0$ for any real number $x$.
In the case of an exponential variable with a parameter $\lambda$ we have $f(x) = \lambda e^{-\lambda x} I_{x \geq 0}$.
Thus,
\begin{align*} \mathbb P(X \leq 2 ) &= \lim_{x \to -\infty} \mathbb P(x \leq X \leq 2) \\ &= \lim_{x \to -\infty} \int_x^2 \lambda e^{-\lambda x} I_{x \geq 0} dx\\ &= \int_{- \infty}^2 \lambda e^{-\lambda x} I_{x \geq 0} dx \\ &= \int_0^2 \lambda e^{-\lambda x}dx \\ &= \lambda \left [ \frac{-1}{\lambda} e^{-\lambda x} \right]_0^2 \\ &= 1-e^{-2 \lambda} \approx 0.865 \ \text{if} \ \lambda = 1 \end{align*}
Note that since $\mathbb P(X=2) = 0$ we have $\mathbb P(X \leq 2 ) = \mathbb P(X<2)$.
