compute $P(X\le 2 \textrm{ or } 3Suppose X is exponentially distributed with $\lambda=2$.
I want to compute $P(X\le 2 \textrm{ or } 3<X<8)$.
The probability density function of $x$ is $f(x)=2e^{-2x}$.
Then I will find its distribution function as $F(x)=1-e^{-2x}$.
I don’t know exactly, but I think that $P(X\le 2 \textrm{ or } 3<X<8)$ can be calculated as follows:
$$P(X\le 2 \textrm{ or } 3<X<8) = P(X\le 2)+ P(3<X<8)-P(X\le 2, 3<X<8)$$
$$P(X\le 2)=F(2)= 1-e^{-4}$$
$$P(3<X<8)=F(8)-F(3)= e^{-16}-e^{-6}$$
But what is $P(X\le 2, 3<X<8)$?
Or what is the correct way to solve this question?
 A: As $X$ is random variable i.e. in particular usual function, then $\{\omega\colon X(\omega)\leqslant 2\} \cap \{\omega\colon 3<X(\omega)< 8\}=\emptyset$, so $P(X\le 2, 3<X<8) = 0$.
A: $X<2$ is an event.  So too is $3 <X <8$.  Since the intervals do not overlap, they are mutually exclusive.
So $P(X <2 \text{ or }3 <X<8) = P(X<2)+P(3<X<8)$ which is $P(X<2) +P(X<8)-P(X\le 3)$.  Since the exponential distribution is continuous, this is also equal to   $P(X\le 2) +P(X\le 8)-P(X\le 3)$ and you can calculate each of these terms from the cumulative distribution function and gives $1-e^{-
4}+e^{-6}-e^{-16}$
Meanwhile $P(X <2,3 <X<8)$ is read as  $P(X <2 \text{ and  }3 <X<8)$ and since the events are mutually exclusive, this is $0$
A: You're almost there.  Note that the events $\{X \le 2\}$ and $\{3 < X < 8 \}$ are mutually exclusive.
$$P(X \le 2 \text{ or } 3 < X < 8) \\
= P(\{X \le 2\} \cup \{3 < X < 8\})
\\ = P(X \le 2) + P(3 < X < 8) \\= (1 - e^{-4}) + (e^{-6} - e^{-16})
\\ = 1 - e^{-4} + e^{-6} - e^{-16}$$
Note that you've inverted the sign when calculating $P(3 < X < 8)$.
