Discrete Random Variable vs Continuous Random Variable

A discrete random variable takes values you could write down in a list, with a gap between each possible value and the next. Every value carries its own probability $P(X = x)$, which is a genuine positive number rather than zero, and those probabilities total 1 across the list. Counting is the usual source: how many, how often, how many out of $n$.

Let $X$ be the number of heads in 3 flips of a fair coin. There are 8 equally likely sequences. TTT gives $X = 0$; HTT, THT and TTH give $X = 1$; HHT, HTH and THH give $X = 2$; and HHH gives $X = 3$. So the probabilities are $\frac{1}{8}$, $\frac{3}{8}$, $\frac{3}{8}$ and $\frac{1}{8}$, which add to $\frac{8}{8} = 1$, and $P(X = 2) = 0.375$.

"Discrete means whole numbers" is the misreading, and it is wrong. A scratch card paying 0, 2.50 or 10 dollars is discrete. So is a shoe size that steps 7, 7.5, 8. What makes a variable discrete is that its values sit apart from each other with nothing in between, not that they happen to be integers.

Discrete does not mean finite either. Count how many cars pass before the first red one and the possible values run 0, 1, 2 and onward without end, yet they are still listable and their probabilities still total 1. The real dividing line is against continuous variables. For those, $P(X = x) = 0$ at every single value and probability exists only over intervals, which is why a weight rounded to the nearest tenth of a kilogram is a discrete record of a quantity that is not.

That difference decides the machinery you use. A discrete variable gets a table of probabilities, a probability histogram, and sums; a continuous one gets a density curve and areas. Discrete random variables enter at topic 2.8 of Unit 2, and their mean and standard deviation at topic 2.9, Parameters of Random Variables.

A random variable is continuous when its possible values fill an interval instead of sitting in a list, so between any two of them there is always another. Probability is not attached to the values at all. It is attached to a density curve, and $P(a < X < b)$ is the area under that curve between $a$ and $b$, with the area under the whole curve equal to 1.

Suppose your wait for a bus is equally likely anywhere from 0 to 10 minutes. The density is flat at height $1/10 = 0.1$, so $P(X < 3) = 3(0.1) = 0.30$, and $P(4 < X < 7) = 3(0.1) = 0.30$ as well, since both windows are three minutes wide.

Now the fact this entry exists for: $P(X = 3) = 0$, and the same holds at every single value. The window from 2.99 to 3.01 carries $0.02(0.1) = 0.002$; shrink its width toward zero and the area goes with it. "So the wait can never be exactly 3 minutes" is the conclusion students draw, and it is false. The bus arrives at some exact instant every time, and whatever instant that turns out to be had probability 0 beforehand. Across values that cannot be written as a list, probability 0 and impossible stop meaning the same thing. Only the values outside 0 to 10 are genuinely ruled out here.

The companion trap is reading the height of the curve as a probability. Density is probability per unit of $x$, so it can exceed 1: a normal curve with standard deviation 0.2 peaks at $\frac{1}{0.2\sqrt{2\pi}} \approx 1.99$. Nothing is broken. The curve is tall because it is narrow, and its total area is still 1.

Continuous random variables reach the course through the normal distribution, topic 2.11 of Unit 2, where every question is an area between two boundaries rather than a value read off a table of probabilities.

Every one of those between-two-boundaries areas is a definite integral of the density function, which is why a single point has probability 0: an integral across an interval of zero width is zero. CalcLearn covers the same object from the other side in area under a curve.