Discrete Random Variables — A-Level Mathematics Stats Study Guide

For: A-Level Mathematics candidates sitting Paper 5 (Probability & Statistics 1).

Covers: Probability distribution tables, expected value $E(X)$ and variance $\text{Var}(X)$, linear transformations of random variables, binomial distribution conditions and formulas, and geometric distribution for the A-Level Mathematics Stats 1 syllabus.

You should already know: Basic probability, summation, integration (Pure 1 calculus).

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Mathematics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is a Discrete Random Variable?

A discrete random variable (DRV) is a numerical quantity that takes a countable set of distinct values, each associated with a fixed probability, as the outcome of a random experiment. "Countable" means you can list all possible values of the variable, even if the list is infinite (e.g. the number of die rolls until you get a 6).

We use uppercase letters ($X$, $Y$) to refer to the random variable itself, and lowercase letters ($x$, $y$) to refer to individual values it can take. The notation $P(X=x)$ denotes the probability that $X$ takes the specific value $x$. DRVs account for 15-20% of total marks in A-Level Mathematics Paper 5, so mastering this topic is critical for high scores.

2. Probability Distribution Table

A probability distribution (or probability mass function, PMF) for a DRV $X$ lists every possible value $x$ that $X$ can take, paired with its corresponding probability $P(X=x)$. All valid probability distributions follow two non-negotiable rules:

1. All probabilities lie between 0 and 1: $0\le P(X=x)\le 1$ for all $x$
2. The sum of all probabilities equals 1: $$\sum _{\text{all }x}P(X=x)=1$$

A probability distribution table is the standard way to present this for small sets of $x$ values. Examiners frequently ask you to fill in missing values in these tables, so always apply the sum-to-1 rule first.

Worked Example

Let $X$ be the number of heads obtained when flipping 2 fair coins. The possible values of $X$ are 0, 1, 2. The valid distribution table is:

Check validity: $\frac{1}{4}+\frac{1}{2}+\frac{1}{4}=1$, and all probabilities are positive, so the distribution is valid.

3. Expected Value $E(X)$ and Variance $\text{Var}(X)$

Expected Value

The expected value (or mean, written $\mu$) of $X$ is the long-run average value of $X$ if you repeat the random experiment thousands of times. It is calculated as a weighted average of all possible $x$ values, weighted by their probabilities: $$E(X)=\sum _{\text{all }x}x\cdot P(X=x)$$ This formula comes directly from calculating the average of repeated trials: if you flip 2 coins 4000 times, you expect ~1000 instances of 0 heads, ~2000 of 1 head, ~1000 of 2 heads, so the average is $\frac{(0\ast 1000)+(1\ast 2000)+(2\ast 1000)}{4000}=1$, which matches the formula result for the coin flip example. Note that $E(X)$ does not need to be a possible value of $X$ (e.g. the expected value of a fair 6-sided die roll is 3.5, which is not a valid die score).

Variance

Variance (written ${\sigma }^2$) measures the spread of the distribution around the mean. The definitional formula for variance is the expected value of the squared deviation from the mean: $$\text{Var}(X)=E\left[(X-\mu {)}^2\right]=\sum _{\text{all }x}(x-\mu {)}^2\cdot P(X=x)$$ For exam calculations, always use the far simpler computational formula, derived by expanding the definitional form: $$\text{Var}(X)=E(X^2)-[E(X){]}^2$$ where $E(X^2)={\sum }_{\text{all }x}{x}^2\cdot P(X=x)$. This formula reduces arithmetic errors and saves time in exams.

Worked Example

For the fair 6-sided die roll $X$:

• $E(X)=1\ast \frac{1}{6}+2\ast \frac{1}{6}+3\ast \frac{1}{6}+4\ast \frac{1}{6}+5\ast \frac{1}{6}+6\ast \frac{1}{6}=\frac{21}{6}=3.5$
• $E(X^2)=1^2\ast \frac{1}{6}+2^2\ast \frac{1}{6}+...+6^2\ast \frac{1}{6}=\frac{91}{6}\approx 15.17$
• $\text{Var}(X)=\frac{91}{6}-(3.5{)}^2=\frac{35}{12}\approx 2.92$

4. Linear Transformation — $E(aX+b),\text{Var}(aX+b)$

A linear transformation of $X$ is a new random variable $Y=aX+b$, where $a$ and $b$ are constants. This comes up frequently in real-world contexts like converting temperature units (F = 1.8C + 32) or adjusting test scores by scaling and adding bonus marks.

Expected Value of Linear Transformation

Expectation is a linear operator, so scaling and shifting apply directly to the mean: $$E(aX+b)=aE(X)+b$$ Intuitively, if you double every value of $X$ and add 3, the long-run average will also double and increase by 3.

Variance of Linear Transformation

Variance measures spread, so adding a constant $b$ does not change the spread at all (it shifts all values equally, so gaps between them stay the same). Scaling by $a$ increases spread by a factor of $a^2$, since variance is a squared measure: $$\text{Var}(aX+b)=a^2\text{Var}(X)$$ The standard deviation of $aX+b$ is $|a|\sqrt{\text{Var}(X)}$, since standard deviation is the positive square root of variance.

Worked Example

For the die roll $X$ with $E(X)=3.5$, $\text{Var}(X)=\frac{35}{12}$, let $Y=2X+3$ (payout for a die roll game):

• $E(Y)=2\ast 3.5+3=10$
• $\text{Var}(Y)=2^2\ast \frac{35}{12}=\frac{35}{3}\approx 11.67$
• Standard deviation of $Y=\sqrt{\frac{35}{3}}\approx 3.42$

5. Binomial Distribution — Conditions and Formula

The binomial distribution is the most frequently tested DRV in Paper 5, used to model the number of successes in a fixed number of independent identical trials. A random variable $X$ follows a binomial distribution, written $X\sim B(n,p)$, if all 4 of these conditions are met:

1. There is a fixed number of trials $n$
2. Each trial has only two outcomes: success or failure
3. The probability of success $p$ is constant for all trials
4. All trials are independent of each other

Examiners will often ask you to justify using a binomial distribution, so you must link each condition to the question context, not just list them.

The PMF, expected value, and variance for the binomial distribution are: $$P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}\text{ for }k=0,1,2,...,n$$ $$E(X)=np,\text{Var}(X)=np(1-p)$$ where $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}$ is the number of ways to choose $k$ successes out of $n$ trials.

Worked Example

A fair coin is flipped 10 times, $X$ is the number of heads, so $X\sim B(10,0.5)$:

• $P(X=3)=\left(\genfrac{}{}{0ex}{}{10}{3}\right)(0.5{)}^3(0.5{)}^7=120\ast \frac{1}{1024}\approx 0.117$
• $E(X)=10\ast 0.5=5$
• $\text{Var}(X)=10\ast 0.5\ast 0.5=2.5$

6. Geometric Distribution (where applicable)

The geometric distribution models the number of independent trials required to get the first success. For A-Level Mathematics, the standard parameterization defines $X$ as the number of trials up to and including the first success, written $X\sim \text{Geo}(p)$, where $p$ is the probability of success per trial. The conditions for a geometric distribution are:

1. Trials are independent
2. Probability of success $p$ is constant for all trials
3. Each trial has two outcomes: success or failure
4. Trials continue until the first success is observed

The PMF, expected value, and variance for the geometric distribution are: $$P(X=k)=(1-p{)}^{k-1}p\text{ for }k=1,2,3,...$$ $$E(X)=\frac{1}{p},\text{Var}(X)=\frac{1-p}{p^2}$$ Note that $k$ starts at 1 for the A-Level syllabus: if a question asks for the number of failures before the first success, use the variable $Y=X-1$ to adjust the distribution.

Worked Example

You roll a fair die repeatedly until you get a 6, so $X\sim \text{Geo}(\frac{1}{6})$:

• $P(X=4)=(\frac{5}{6}{)}^3\ast \frac{1}{6}=\frac{125}{1296}\approx 0.0965$
• $E(X)=\frac{1}{1/6}=6$
• $\text{Var}(X)=\frac{5/6}{(1/6{)}^2}=30$

7. Common Pitfalls (and how to avoid them)

• Wrong move: Using a linear factor of $a$ instead of $a^2$ for variance of linear transformations, or including the $b$ term in variance calculations. Why it happens: Students mix up expectation and variance rules. Correct move: Write the formula $\text{Var}(aX+b)=a^2\text{Var}(X)$ on your exam formula sheet before starting questions, and remember that variance only measures spread, so shifts do not affect it.
• Wrong move: Using the geometric distribution parameterization where $k$ starts at 0 for A-Level questions. Why it happens: Some textbooks use the "number of failures" parameterization. Correct move: Check question wording: if it asks for number of trials to first success, $k\ge 1$, use $P(X=k)=(1-p{)}^{k-1}p$. If it asks for failures before first success, subtract 1 from $X$.
• Wrong move: Calculating variance with the $(x-\mu {)}^2$ definitional formula instead of the $E(X^2)-[E(X){]}^2$ computational formula. Why it happens: Students memorize the definitional form first. Correct move: Always use the computational formula for variance: it has fewer steps and eliminates sign errors from squared differences.
• Wrong move: Failing to link binomial conditions to question context when justifying use of the distribution. Why it happens: Students only list the 4 conditions without applying them. Correct move: For a question about defective items in a sample of 20, state: "Fixed $n=20$ trials, each item is defective or not, defect probability $p=0.05$ is constant, items are independent, so binomial is appropriate."
• Wrong move: Assuming $E(X)$ must be a possible value of $X$. Why it happens: Students confuse expected value with the mode (most common value). Correct move: Expected value is a long-run average, it can be a non-integer even if $X$ only takes integer values (e.g. 3.5 for a die roll is completely valid).

8. Practice Questions (A-Level Mathematics Paper 5 Style)

Question 1

The discrete random variable $X$ has the following probability distribution:

Solution

(a) Sum of probabilities = 1: $0.1+0.2+k+0.25+0.15=1$ → $0.7+k=1$ → $k=0.3$ (b) $E(X)=(1\ast 0.1)+(2\ast 0.2)+(3\ast 0.3)+(4\ast 0.25)+(5\ast 0.15)=3.15$ $E(X^2)=(1^2\ast 0.1)+(2^2\ast 0.2)+(3^2\ast 0.3)+(4^2\ast 0.25)+(5^2\ast 0.15)=11.35$ $\text{Var}(X)=11.35-(3.15{)}^2=1.4275$ (c) $E(Y)=3\ast 3.15-2=7.45$ $\text{Var}(Y)=9\ast 1.4275=12.8475$ Standard deviation of $Y=\sqrt{12.8475}\approx 3.58$ (3 significant figures)

Question 2

A multiple-choice quiz has 12 questions, each with 4 options, only one correct. A student guesses all answers randomly. (a) Explain why the number of correct answers $X$ follows a binomial distribution. (3 marks) (b) Find the probability the student gets exactly 3 correct answers. (3 marks) (c) Find the mean and variance of $X$. (2 marks)

Solution

(a) 1) Fixed $n=12$ trials (questions). 2) Each trial has two outcomes: correct (success) or incorrect (failure). 3) Probability of success $p=0.25$ is constant for all guesses. 4) Guesses are independent, so outcome of one question does not affect others. All conditions met, so $X\sim B(12,0.25)$. (b) $P(X=3)=\left(\genfrac{}{}{0ex}{}{12}{3}\right)(0.25{)}^3(0.75{)}^9=220\ast 0.015625\ast 0.07508\approx 0.258$ (3 sf) (c) $E(X)=12\ast 0.25=3$, $\text{Var}(X)=12\ast 0.25\ast 0.75=2.25$

Question 3

A basketball player has a 0.3 probability of making a free throw per attempt, independent of previous attempts. (a) Find the probability the player makes their first successful free throw on their 4th attempt. (3 marks) (b) Find the expected number of attempts needed to make their first successful free throw, and the variance. (2 marks)

Solution

(a) Let $X$ = number of attempts to first success, so $X\sim \text{Geo}(0.3)$ $P(X=4)=(0.7{)}^3\ast 0.3=0.343\ast 0.3=0.1029\approx 0.103$ (3 sf) (b) $E(X)=\frac{1}{0.3}\approx 3.33$ (3 sf), $\text{Var}(X)=\frac{0.7}{0.3^2}\approx 7.78$ (3 sf)

9. Quick Reference Cheatsheet

10. What's Next

This topic is the foundation for all other probability content in A-Level Mathematics Paper 5. You will apply the expectation and variance rules you learned here to continuous random variables (including the normal distribution) later in the syllabus, and solve combined probability problems that use multiple distributions in the same question. DRV concepts are also used for high-mark real-world application questions, including game theory, quality control, and risk calculation, which make up the final sections of most Paper 5 exams.

If you are struggling with any of the concepts, formulas, or practice questions in this guide, you can ask Ollie for step-by-step explanations, additional practice problems, or clarification of exam marking conventions at any time. Head to [the homepage](/ to access Ollie and other A-Level Mathematics Paper 5 study resources, including past paper walkthroughs and topic quizzes tailored to your weak areas.

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