Discrete Probability Distributions
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Discrete Probability Distributions
The big idea: a discrete random variable is just a list of numbers with probabilities attached, and once you can find its mean (E(X)) and spread (Var(X)), you can predict, transform, and combine it with total confidence.
- E(X) — the expected value (mean) of a discrete random variable: multiply each value by its probability and add them up.
- E(X²) and E(g(X)) — the expected value of a function of X, found by applying the function to each x-value first, then weighting by probability.
- Var(X) — the variance (spread) of X, calculated as E(X²) − (E(X))².
- Linear transformations — how E(X) and Var(X) change when you do aX + b to every value of X.
- Problem solving — building your own distribution tables from word problems, combining two independent variables, and handling a variable Y that depends on X.
E(X) stands for the expected value — but that name is a little misleading, because it doesn't mean "the value you should expect to see." It's really the long-run average if you repeated the random experiment thousands of times and averaged your results.
Here's the classic example that makes this click: if you flip a fair coin 5 times, the expected number of tails is 2.5. You will never actually get 2.5 tails on a single go — you'll get 0, 1, 2, 3, 4, or 5. But if you did this experiment thousands of times and averaged all your results, you'd land right around 2.5. E(X) doesn't have to be an achievable outcome — it's a theoretical balancing point.
Think of it like a physical seesaw. Imagine each value of X sitting on a number line, and its probability is how "heavy" that value is. E(X) is exactly where you'd need to place the pivot so the whole seesaw balances perfectly. Values with bigger probabilities pull the balance point more strongly toward them.
| w | 1 | 5 | 10 | 100 |
|---|---|---|---|---|
| P(W = w) | 0.35 | 0.5 | 0.05 | 0.1 |
Before you can find the variance, you need E(X²) — the mean of the squares of X. The process is almost identical to finding E(X), except you square each x-value before multiplying by its probability.
- E(X²) = the mean of the squared values of X (square first, then average)
- (E(X))² = the square of the mean of X (average first, then square)
Var(X) tells you how spread out the values of X are around the mean. A high variance means the outcomes are wildly unpredictable and scattered far from the average — low consistency. A low variance means outcomes tend to cluster tightly around the mean — high consistency, low surprise.
Think about two games: Game A always pays out exactly $10. Game B pays $0 half the time and $20 half the time. Both have the same mean ($10), but Game B has a much higher variance — its outcomes are far more spread out, even though the "average" result is identical. Variance is what tells the two games apart.
| x | 2 | 3 | 5 | 7 |
|---|---|---|---|---|
| P(X = x) | 0.1 | 0.3 | 0.2 | 0.4 |
E(X²) is really just a special case of a bigger idea. Let g(X) be any function of X — it could be ³√X, ln X, 1/X, X³, whatever the question throws at you. The exact same logic applies: apply the function to each x-value first, then weight by probability and add.
| x | 1 | 8 | 27 |
|---|---|---|---|
| P(X = x) | 0.1 | 0.3 | 0.6 |
This is one of the most useful ideas in the whole chapter, because it means you almost never have to rebuild a whole distribution table from scratch when a variable gets scaled or shifted.
Picture this: let X be the number of hours you work each day. Your company pays you by tripling your hours and adding a flat £60 bonus, so your daily pay is Y = 3X + 60. If you already know your expected hours per day, E(X), you don't need to rebuild an entirely new table for Y just to find your expected pay — there's a direct shortcut.
Here's the intuition behind why they're different:
- Multiplying by a stretches or shrinks every value of X, so naturally the mean scales by a too — and the spread scales by a as well. But since variance is a "squared" measure of spread (it's built from squared distances), it scales by a², not just a.
- Adding b shifts every value of X up or down by the same fixed amount, like sliding the whole distribution along the number line. This shifts the mean by b too — but it doesn't change how spread out the values are relative to each other at all, so the variance is completely unchanged.
Yes — both formulae generalise beyond simple linear transformations, as long as you treat the function itself as the "variable" being transformed.
(i) E(3X + 5) (ii) Var(3X + 5) (iii) Var(2 − X)
This is the skill that ties everything together, and exam questions love to disguise it in a story. The method is always the same four steps:
- Introduce a capital letter (like X) to represent the numerical outcome — and be crystal clear in your own mind what X actually represents.
- List every value X can possibly take.
- Calculate the probability of each value.
- Put it all into a table — this single step prevents almost every careless mistake in this topic.
If a question then asks you to "comment" on the results, use E(X) to talk about the typical/average outcome, and Var(X) to talk about how spread out or unpredictable the outcomes are.
Sometimes a question gives you two independent random variables — say X from flipping a coin and Y from spinning a 3-sided spinner. "Independent" means the outcome of one has zero influence on the other.
| x | 2 | 4 |
|---|---|---|
| P(X = x) | 0.25 | 0.75 |
| y | 0 | 2 | 4 |
|---|---|---|---|
| P(Y = y) | 0.1 | 0.2 | 0.7 |
To find something like P(X + Y = 4), list every combination of x and y that adds to 4 — here that's (X=2, Y=2) or (X=4, Y=0) — then multiply the probabilities within each pair, and add the pairs together: 0.25 × 0.2 + 0.75 × 0.1.
Sometimes a new variable Y is built directly from X — for example Y = 9 − X². Because Y's value is completely determined by X, they are not independent, and this changes what you're allowed to do.
| x | −3 | 1 | 2 | 3 |
|---|---|---|---|---|
| P(X = x) | 0.25 | 0.25 | 0.25 | 0.25 |
Substituting Y = 9 − X² into each x-value gives y = 0, 8, 5, 0 — note two different x-values (−3 and 3) both map to y = 0, so once you group and simplify, that probability of 0.25 doubles to 0.5:
| y | 0 | 5 | 8 |
|---|---|---|---|
| P(Y = y) | 0.5 | 0.25 | 0.25 |
Occasionally a question asks for a probability given some restriction. The formula is the same conditional probability rule from earlier stats work:
(a) Find the probability of earning more than 9 points when playing the game twice.
(b) The amount of money (£) won is found by multiplying the points X by a variable Y, where Y = 3 if X < 5, or Y = X − 3 if X ≥ 5. Find the expected amount of money won per game.
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