Finding probabilities for a discrete random variable using a CDF I have a question about notation and I want to make sure I really understand the homework.
The discrete random variable X has cdf F such that
F(x)= 0      x < 1
      1/4    1 ≤ x < 3
      3/4    3 ≤ x <4
      1      x ≥ 4

Find P(X=2), P(X=3.5), P(X=4)

Here is what I found. Can you tell me where I am making a mistake? Thanks
P(X=2) = P(X ≤ 2) - P(X < 2) = ? 

P(X=3.5) = P(X ≤ 3.5) - P( X ≤ 3) = 3/4 - 3/4 =0

P(X=4) = P(X ≤ 4) - P( X < 4) = 1 - 3/4 = 1/4

 A: You are trying to use arithmetic/algebra to solve the problem. I prefer to try to visualize the cdf, and use the visualization to tell the story. You know, of course, that $F(x)$ is the probability that $X\le x$, It is the "weight" up to and including $x$. We will have to be very careful: for these discrete distributions, there can be a big difference between $\le$ and $\lt$.
At any point $x\lt 1$, the weight of the stuff $\le x$ is $0$. Then the cdf jumps to $\frac{1}{4}$ at $1$. So there must be a weight of $\frac{1}{4}$ at $1$. In probability terms, $\Pr(X=1)=\frac{1}{4}$.
Then the cdf stays steady at $\frac{1}{4}$ for quite a while, and jumps to $\frac{3}{4}$ at $x=3$. So there must be a weight of $\frac{3}{4}-\frac{1}{4}$ at $x=3$, that is, $\Pr(X=3)=\frac{1}{2}$. Then the cdf stays at $\frac{3}{4}$ until $x=4$, and jumps to $1$. Thus $\Pr(X=4)=\frac{1}{4}$.
Our weights at $1$, $3$, and $4$ add up to $1$, so that's all there is. Adding up also provides a nice little check against minor errors.
Now we can answer any question. The probability that $X=2$ is $0$, no weight at $x=2$. Same with $3.5$. And we already knew that the weight at $4$ is $\frac{1}{4}$. 
Formal remark: Let $F(x)$ be a cdf. We give an explicit "formula" for $\Pr(X=a)$:
$$\Pr(X=a)=F(a)-\lim_{x\to a^-} F(x).$$
By $\lim_{x\to a^-} F(x)$ we mean the limit of $F(x)$ as $x$ approaches $a$ from the left. 
