Q.In a pack of $52$ playing cards, two cards are drawn at random simultaneously. If the number of black cards drawn is a random variable, find the values of the random variable and the number of points in its inverse images.

A random variable $X$ is a function defined on a sample space $S$ into the real numbers $\mathbb R$, such that the inverse image of every point, subset or interval of $\mathbb R$ is an event in $S$ (Definition 11.1). It turns the outcomes of a random experiment — which need not themselves be numbers, e.g. a coin's head/tail — into numbers we can compute with. Capital letters $X,Y,Z$ denote the random variable itself; small letters $x,y,z$ denote its possible values. If $x$ is a possible value, its inverse image $X^{-1}(x)=\{\omega\in S:X(\omega)=x\}$ is always an event of $S$, so it has a probability — this is what makes "$P(X=x)$" meaningful.

Two flavours are studied:

• Discrete random variable (Definition 11.2): its range is countable — finite, or a sequence $x_1,x_2,x_3,\dots$ — with every value carrying positive probability and the whole range's probabilities summing to $1$. Used for counting a quantity: number of heads, number of defectives, a sum of dice faces, a winning amount taking finitely many values. A discrete random variable can even live on a continuous sample space (e.g. a step function of $\omega\in[0,20]$ taking only two values is still discrete, because its range, not its domain, is what is tested for countability).
• Continuous random variable (Definition 11.5): $X:S\to\mathbb R$ takes any value in a set $I\subseteq\mathbb R$, and $P(X=x)=0$ for every $x\in I$. Used for measuring a quantity: a lifetime, a waiting time, a distance from a centre — probability only accumulates over an interval, never at a single point.

Counting inverse images. For a discrete random variable built from a finite sample space, the standard technique is: (1) list/describe the full sample space $S$ (or its size, via combinatorics); (2) work out every value $X$ can take; (3) for each value, count how many sample points map to it — this count is the size of that value's inverse image, and dividing by $|S|$ (when outcomes are equally likely) gives its probability. A quick check is that the inverse-image counts across all values of $X$ must add up to $|S|$.