A discrete random variable takes values you can list, each with a probability. Two descriptions carry the same information: the probability mass function lists $P(X = x)$, and the cumulative distribution function adds them up. Moving between them, with care at the endpoints, is the skill here.
Key idea
The pmf is $p_X(x) = P(X = x)$: never negative, adding to 1 over all values. The CDF is $F_X(x) = P(X \le x)$: it climbs in steps from 0 to 1, jumping by $P(X = a)$ at each value $a$. For $a < b$: $$P(a < X \le b) = F(b) - F(a).$$
Why it works
The events $\{X \le a\}$ and $\{a < X \le b\}$ are disjoint, and together they make $\{X \le b\}$. By the addition axiom, $F(b) = F(a) + P(a < X \le b)$. The size of a jump is a special case: $P(X = a) = F(a) - P(X < a)$, the CDF at $a$ minus its value just before $a$. Between two values of $X$ nothing happens, so $F$ is flat there: if $X$ is a whole number, $F(2.5) = F(2)$.
On the CDF a filled dot sits at the top of each jump and an open dot at the bottom: at $x = a$ the CDF already includes $P(X = a)$.
Worked examples
Worked example: Reading the CDF from the pmf
$X$ is the number of heads in three fair flips: $p(0) = \frac{1}{8}$, $p(1) = \frac{3}{8}$, $p(2) = \frac{3}{8}$, $p(3) = \frac{1}{8}$. Find $F(1)$, $F(1.5)$ and $P(1 \le X \le 2)$.
- $F(1) = p(0) + p(1) = \frac{1}{2}$, and $F(1.5) = \frac{1}{2}$ too: no value lies between 1 and 1.5.
- $P(1 \le X \le 2) = p(1) + p(2) = \frac{3}{4}$.
- With the CDF: $F(2) - F(0) = \frac{7}{8} - \frac{1}{8} = \frac{3}{4}$. Subtract $F(0)$, not $F(1)$, so that $X = 1$ stays in.
Answer: $\frac{1}{2}$, $\frac{1}{2}$ and $\frac{3}{4}$
Worked example: From a CDF table
A whole-number $X$ has $F(0) = 0.1$, $F(1) = 0.3$, $F(2) = 0.6$, $F(3) = 0.85$, $F(4) = 1$. Find $P(X = 2)$, $P(1 \le X \le 3)$ and $P(X \ge 2)$.
- $P(X = 2)$ is the jump at 2: $0.6 - 0.3 = 0.3$.
- $$P(1 \le X \le 3) = F(3) - F(0) = 0.85 - 0.1 = 0.75$$
- $$P(X \ge 2) = 1 - P(X \le 1) = 1 - 0.3 = 0.7$$
Answer: $0.3$, $0.75$ and $0.7$
Common mistake
Using $F(b) - F(a)$ for $P(a \le X \le b)$ when $X$ is discrete. That difference is $P(a < X \le b)$ and leaves out $P(X = a)$, which can be large. Check by listing the values that should count and adding their probabilities directly. (For continuous variables single points have probability 0, so there the endpoints do not matter.)