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Discrete Probability Distributions

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  Edexcel A Level Further Maths — Further Statistics 1

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.

Summary — What This Chapter Covers
  • 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.
Topic 1: E(X) of Discrete Random Variables
1What does E(X) actually mean?

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.

Formula
E(X) = Σ x·P(X = x)
In plain words: multiply each value of X by its own probability, then add all those products together.
Spot the shortcut: Symmetry
If a distribution is symmetrical — meaning the x-values are evenly spaced around a central value, and the probabilities mirror each other on either side — then the mean is automatically that central value. No calculation needed! Example: if X takes values 1, 5, 9 with probabilities 0.3, 0.4, 0.3, the shape is symmetric about 5, so E(X) = 5 by symmetry alone. This is also true that the mean equals the median here.
Practice Worked-style question
Daphne pays $15 to play a game where she wins a prize of $1, $5, $10 or $100. The random variable W represents her winnings:
w1510100
P(W = w)0.350.50.050.1
Calculate the expected value of Daphne's prize.
Topic 2: Var(X) of Discrete Random Variables
2Finding E(X²) first

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.

Formula
E(X²) = Σ x²·P(X = x)
Square each value of X, multiply by its probability, then add everything together.
The #1 mistake students make here
E(X²) is NOT the same as (E(X))². These are two completely different calculations, and mixing them up is the most common error in this whole topic.
  • 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)
Quick proof they're different: let X = 1 or X = −1, each with probability 0.5. The mean is 0, so (E(X))² = 0² = 0. But the squares of X are 1 and 1 (since (−1)² = 1 too), so E(X²) = 1. Zero ≠ one — completely different numbers!
3What is Var(X), and why do we care?

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.

Formula — the one you'll use 95% of the time
Var(X) = E(X²) − (E(X))²
The mean of the squares, minus the square of the mean. Var(X) is always positive (or zero).
Alternative formula (the formal definition — less practical)
Var(X) = E[(X − E(X))²]
The expected value of the squared distances from the mean. Mathematically identical to the formula above, but far more painful to actually compute — stick to E(X²) − (E(X))² in exams.
Rearranging when Var(X) and E(X) are known
Sometimes a question gives you Var(X) and E(X) and asks you to find E(X²) instead. Just rearrange:
E(X²) = Var(X) + (E(X))²
Practice Worked-style question
The discrete random variable X has the probability distribution:
x2357
P(X = x)0.10.30.20.4
Find (a) E(X), (b) E(X²), (c) Var(X).
Sanity check every answer
Two quick checks to catch mistakes before you move on: (1) E(X) should fall within the range of X's values — in the example above, 4.9 sits comfortably between 2 and 7. (2) Var(X) must always come out positive. If you get a negative variance, you've made an arithmetic slip somewhere — go back and check.
Topic 3: E(g(X)) — Functions of a Random Variable
4Beyond X² — any function goes

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.

Formula
E(g(X)) = Σ g(x)·P(X = x)
Apply g to each value of x, multiply by that value's probability, then sum.
Same trap, different disguise
Just like E(X²) ≠ (E(X))², in general E(g(X)) ≠ g(E(X)). You cannot find the mean of X first and then plug that single number into the function — you must apply the function to every value separately, then take the weighted average. Skipping this step is a guaranteed lost mark.
Practice Worked-style question
The random variable X has the distribution:
x1827
P(X = x)0.10.30.6
Determine which is greater: E(³√X) or E(ln X).
Topic 4: Linear Transformations of DRVs (aX + b)
5What does aX + b actually represent?

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.

The two formulae you must know cold
E(aX ± b) = a·E(X) ± b
Var(aX ± b) = a²·Var(X)

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.
Rewrite tricky expressions into the aX + b form first
Sometimes an expression doesn't obviously look like aX + b. For example, 1 − X/4 is secretly the same as −¼X + 1 (so a = −¼, b = 1). Always rearrange into this standard form before applying the formulae — it stops sign errors before they happen.
6Do these rules extend to other functions of X?

Yes — both formulae generalise beyond simple linear transformations, as long as you treat the function itself as the "variable" being transformed.

Expectation extends naturally
E(3X² + 5) = 3·E(X²) + 5
Careful: E(X²) here does NOT simplify to (E(X))² — that trap from Topic 2 still applies.
Variance extends the same way
Var(4X³) = 16·Var(X³)
where Var(X³) itself means E(X⁶) − (E(X³))² — always simplify algebra fully before taking expectations or variances.
Golden rule: simplify BEFORE you take E or Var
Example: E((X − 1)²) is much easier to handle if you expand first: E(X² − 2X + 1) = E(X²) − 2E(X) + 1. Similarly, Var((1 − 3X)²) becomes Var(1 − 6X + 9X²) = 36·Var(X) + 81·Var(X²) once expanded. Trying to apply formulae to an unexpanded bracket is where most errors creep in.
Practice Worked-style question
X is a random variable such that E(X) = 5 and Var(X) = 4. Find:
(i) E(3X + 5)   (ii) Var(3X + 5)   (iii) Var(2 − X)
Topic 5: Problem Solving with DRVs
7Turning a word problem into a distribution

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:

  1. Introduce a capital letter (like X) to represent the numerical outcome — and be crystal clear in your own mind what X actually represents.
  2. List every value X can possibly take.
  3. Calculate the probability of each value.
  4. 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.

8Combining two independent variables

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.

x24
P(X = x)0.250.75
y024
P(Y = y)0.10.20.7

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.

Two shortcut formulae for independent X and Y
E(X + Y) = E(X) + E(Y)
E(XY) = E(X)·E(Y)  (only valid when X and Y are independent!)
Repeating the same distribution
These ideas also apply when you use the same distribution more than once — like flipping a coin, then flipping it again. Just define independent random variables X₁, X₂, ... for each repetition, and treat them exactly like two separate independent variables.
9When Y depends on X (this is the tricky one!)

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−3123
P(X = x)0.250.250.250.25

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:

y058
P(Y = y)0.50.250.25
The rule you CANNOT use here
You can still safely use E(X + Y) = E(X) + E(Y) — that formula works regardless of independence. But you cannot use E(XY) = E(X)E(Y) if Y depends on X — that shortcut only works for independent variables. Instead, either (a) build a full table of every XY value and find its expectation directly, or (b) substitute Y's formula directly into XY algebraically and simplify — e.g. XY = X(9 − X²) = 9X − X³, so E(XY) = 9E(X) − E(X³), which you can find from your table of X.
10Conditional probability with DRVs

Occasionally a question asks for a probability given some restriction. The formula is the same conditional probability rule from earlier stats work:

Formula
P(A | B) = P(A ∩ B) / P(B)
Example: for integer X, P(X > 5 | X ≤ 10) = P(6 ≤ X ≤ 10) / P(X ≤ 10) — you're narrowing the "universe" down to just the values allowed by the condition, then asking what fraction of that narrowed universe satisfies the event.
Practice Worked-style question
In a game, you can earn 2, 5 or 6 points. There is a 50% chance of earning 6 points and an equal chance of earning either 2 or 5 points.

(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.
What to Memorise
Expected Value
The probability-weighted average of X. Doesn't need to be an achievable outcome.
E(X) = Σ x·P(X=x)
Expected Value of X²
Square each x FIRST, then weight by probability. NOT the same as (E(X))².
E(X²) = Σ x²·P(X=x)
Variance
Measures spread. Always ≥ 0. Use this formula, not the raw definition.
Var(X) = E(X²) − (E(X))²
Standard Deviation
The square root of variance — puts spread back into the original units of X.
SD(X) = √Var(X)
Function of X
Apply g to each x-value BEFORE weighting by probability. g(E(X)) is a trap.
E(g(X)) = Σ g(x)·P(X=x)
Linear Transform — Mean
Scaling and shifting both act directly on the mean.
E(aX ± b) = aE(X) ± b
Linear Transform — Variance
Scaling squares the effect on variance; shifting (adding b) has NO effect.
Var(aX ± b) = a²Var(X)
Independent Variables
Sums always add; products only multiply if X and Y are independent.
E(X+Y)=E(X)+E(Y)   E(XY)=E(X)E(Y)
Conditional Probability
Narrow the sample space to B, then ask what fraction also satisfies A.
P(A|B) = P(A∩B) / P(B)
Symmetry Shortcut
If x-values and probabilities are symmetric, the mean = the middle value = the median, with zero calculation needed.
E(X) = central value
Concepts Checklist
Exam Tips — Common Mistakes & Mark-Scheme Traps
Trap 1: Confusing E(X²) with (E(X))²
This is the single most common error examiners see. Always ask yourself: "am I squaring first, or averaging first?" Write out E(X²) = Σx²P(X=x) explicitly rather than trying to do it in your head.
Trap 2: Using E(XY) = E(X)E(Y) when not independent
Before using this shortcut, always check: are X and Y actually independent, or does one depend on the other? If Y is defined in terms of X, this formula is invalid — build a table or substitute algebraically instead.
Trap 3: Forgetting to square 'a' in the variance formula
Var(aX+b) = a²Var(X), NOT a·Var(X). Students often correctly square 'a' for E(X²)-type problems but then forget to do it here. Also remember: adding 'b' never changes the variance at all.
Trap 4: Not simplifying algebra before taking E or Var
Expand brackets like (X−1)² or (1−3X)² fully before applying expectation/variance rules. Trying to apply the rules to an unexpanded expression usually leads to errors or a stuck calculation.
What examiners actually reward
Always show the summation working explicitly (Σ x·P(X=x) = ...) rather than jumping straight to a final number — method marks are awarded for showing the substitution, not just the final answer. Draw a table for every random variable you define, even if it feels slow — it prevents almost every silly mistake in this topic.
Always sanity-check your final answer
E(X) should sit within the range of X's possible values. Var(X) must always be positive. If either check fails, you've made an error — go back before submitting your answer.
Discrete Probability Distributions Revision Guide · Edexcel A Level Further Maths (Further Statistics 1)
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