How I can calculate $P(10<X\leqslant 20)$ if I already have $F_X$?

Question

Let $X$: "Launch a die until you get the number 5 for the first time", a discrete random variable. I am asked to calculate $P(10<X\leqslant20) $.

Is correct to say...

If $P(10<X\leqslant20)$ then $X$ is in $(10,20]$, and that's equal to $B=(-\infty,20]$ except $A=(-\infty,10]$. But $A$ is included in $B$, so $P(10<X\leqslant20) = P(B-A)=P(B)-P(A)=F_X(20)-F_X(10)$

?

Answer 1

$P(10<X\leqslant20)=P(10<X)+P(X\leqslant20)$ is not true.

You need to consider $P(10<X\leqslant20)$ is the probabilty of $X$ to be an integer in $(10,20]$. But, $P(10<X)+P(X\leqslant20)$ is in general larger.

Answer 2

Suppose $X$ is always $15$. Then $P(10<X\le 20)=1$, but also $P(10<X)=1=P(X\le 20)$. So the answer to the question is NO.

Answer 3

You're correct. Differences are a bit tricky to deal with, but it's clear that

$$P(X \le 10) + P(10 < X \le 20) = P(X \le 20)$$

since the union of the events $\{ X \le 10 \}$ and $\{ 10 < X \le 20 \}$ is just $\{ X \le 20 \}$. Rearranging this gives you

$$P(10 < X \le 20) = P(X \le 20) < P(X \le 10)$$

and if $F_X$ is the cumulative distribution function, with $F_X(x) = P(X \le x)$, then you have

$$P(10 < X \le 20) = F_X(20) - F_X(10)$$

or more generally, when $b \ge a$,

$$P(a < X \le b) = F_X(b) - F_X(a).$$