Discrete random variables and the binomial distribution - TCE Mathematics Methods (Tasmania)

Syllabus dot point

“Construct probability distributions for discrete random variables, find their mean and variance, and apply the binomial distribution.”

TCEMath MethodsUnit 38 min readUpdated 2 Jun 2026

Quick answer

A discrete random variable takes separate values each with a probability that sums to 1; its expected value $E(X)=\sum x\,P(X=x)$ and variance measure centre and spread. The binomial distribution applies when you count successes in $n$ fixed, independent trials with constant success probability $p$ .

What this dot point is asking

A random variable assigns a number to each outcome of a random process. A discrete random variable takes countable, separate values (often $0, 1, 2, \dots$ ). Its probability distribution lists each value with its probability.

Two requirements of a distribution

For any probability distribution: every probability satisfies $0 \le P(X=x) \le 1$ , and all the probabilities sum to exactly $1$ , that is $\sum P(X=x) = 1$ .

Mean and variance

The expected value (mean) is the long-run average:

Expected value and variance

$E(X) = \mu = \displaystyle\sum x\,P(X=x)$

$\operatorname{Var}(X) = \sigma^2 = \displaystyle\sum (x-\mu)^2 P(X=x) = E(X^2) - \mu^2$

Standard deviation: $\sigma = \sqrt{\operatorname{Var}(X)}$

The form $E(X^2) - \mu^2$ is usually faster to compute, where $E(X^2) = \sum x^2 P(X=x)$ .

Mean and variance from a distribution table

A discrete random variable $X$ has $P(X = x) = 0.2, 0.5, 0.3$ for $x = 1, 2, 3$ respectively. Find $E(X)$ , $\operatorname{Var}(X)$ and the standard deviation.

Step 1: Compute the mean $E(X)$

The expected value is the weighted average of the possible values, where each value is weighted by its probability. Multiply each $x$ by its probability and sum the results:

E(X) = \sum x\,P(X = x) = (1)(0.2) + (2)(0.5) + (3)(0.3) = 0.2 + 1.0 + 0.9 = 2.1.

Step 2: Compute $E(X^2)$

To use the shortcut formula $\operatorname{Var}(X) = E(X^2) - \mu^2$ , we need the average of the squared values. Replace each $x$ with $x^2$ and repeat the weighted sum:

E(X^{2}) = (1^2)(0.2) + (2^2)(0.5) + (3^2)(0.3) = 0.2 + 2.0 + 2.7 = 4.9.

Step 3: Find the variance and standard deviation

Subtract the square of the mean from $E(X^2)$ . This shortcut avoids computing each $(x - \mu)^2$ term separately and is the form examiners expect:

\operatorname{Var}(X) = E(X^{2}) - \mu^{2} = 4.9 - 2.1^{2} = 4.9 - 4.41 = 0.49.

Final answer: $E(X) = 2.1$ , $\operatorname{Var}(X) = 0.49$ , and $\sigma = \sqrt{0.49} = 0.7$ .

The binomial distribution

The binomial distribution counts the number of successes in a fixed number of trials.

Conditions for a binomial distribution

There are a fixed number $n$ of trials; each trial has only two outcomes (success or failure); the probability of success $p$ is constant across trials; and the trials are independent.

If $X$ is binomial with parameters $n$ and $p$ , written $X \sim B(n,p)$ , then

Binomial probability, mean and variance

$P(X = k) = \dbinom{n}{k} p^k (1-p)^{n-k}$ , for $k = 0, 1, \dots, n$ .

Mean: $E(X) = np$ .

Variance: $\operatorname{Var}(X) = np(1-p)$ .

Here $\dbinom{n}{k} = \dfrac{n!}{k!(n-k)!}$ counts the number of ways to choose which trials are the successes.

Worked example: a binomial calculation

A multiple-choice quiz has $10$ questions, each with $4$ options. A student guesses every answer. Find the probability of getting exactly $3$ correct, and the expected number correct.

Step 1: Identify the distribution

Each guess is an independent trial with only two outcomes (correct or incorrect), and the probability of a correct guess is constant at $p = \tfrac{1}{4} = 0.25$ . With $n = 10$ trials, the count of correct answers follows a binomial distribution: $X \sim B(10, 0.25)$ .

Step 2: Set up the binomial probability for exactly 3 correct

The binomial formula is $P(X = k) = \dbinom{n}{k}p^k(1-p)^{n-k}$ . Here $k = 3$ , $p = 0.25$ and $1 - p = 0.75$ :

P(X = 3) = \binom{10}{3}(0.25)^3 (0.75)^7.

Step 3: Evaluate each factor and multiply

Compute each piece separately to avoid errors. The binomial coefficient counts the number of ways to choose which 3 of the 10 questions are correct:

\dbinom{10}{3} = \frac{10!}{3!\,7!} = 120, \quad (0.25)^3 = 0.015625, \quad (0.75)^7 \approx 0.133484.

P(X=3) = 120 \times 0.015625 \times 0.133484 \approx 0.2503.

Step 4: Find the expected number correct

The mean of a binomial distribution is $E(X) = np$ , which gives the long-run average number of successes without any lengthy calculation:

E(X) = np = 10 \times 0.25 = 2.5 \text{ questions}.

Final answer: $P(X = 3) \approx 0.2503$ and the expected number correct is $2.5$ questions.

Confusing "exactly", "at least" and "at most"

"Exactly 3" is a single term $P(X=3)$ . "At least 3" means $P(X \ge 3) = 1 - P(X \le 2)$ , and "at most 3" means $P(X \le 3) = P(0)+P(1)+P(2)+P(3)$ . Read the wording carefully and use the complement $1 - P(\text{other side})$ when it saves adding many terms.

Cumulative probabilities and technology

In the calculator-assumed section you will use the cumulative binomial function to evaluate ranges quickly. The cumulative probability $P(X \le k)$ adds up all terms from $0$ to $k$ , and most graphics or CAS calculators provide it directly as a "binomial CDF". Then $P(X \ge k) = 1 - P(X \le k - 1)$ and $P(a \le X \le b) = P(X \le b) - P(X \le a - 1)$ . Translating the English wording into these cumulative expressions is half the skill: write the inequality first, then read off which cumulative values you need.

Where the binomial is used

The binomial underpins quality-control sampling (the number of defective items in a batch), opinion polling at small scale (the number of supporters in a fixed sample), and any " $n$ independent yes/no trials" scenario. In Unit 4 it returns as the foundation for the sampling distribution of a proportion: dividing a binomial count by $n$ gives $\hat{p}$ , whose mean $p$ and standard deviation $\sqrt{p(1-p)/n}$ come straight from $E(X) = np$ and $\operatorname{Var}(X) = np(1-p)$ .

Summary

For any discrete random variable, build the distribution table, confirm the probabilities sum to $1$ , then compute $E(X)$ and $\operatorname{Var}(X)$ . Before using the binomial formula, check all four conditions hold. Decide whether the question wants an exact value, a cumulative range, or a complement, and use $E(X)=np$ and $\operatorname{Var}(X)=np(1-p)$ as quick shortcuts for binomial summaries.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice5 marks

Calculator-free. The probability of a machine producing a defective item is

\dfrac{1}{10}

X

is the number of defective items produced when

4

items are made. Calculate exact values for: (a)

P(X = 3)

; (b)

P(X < 3)

Show worked answer →

$X$ is binomial with $n = 4$ trials and success probability $p = \dfrac{1}{10}$ , so $q = \dfrac{9}{10}$ . Use $P(X = r) = \binom{n}{r}p^{r}q^{n-r}$ .

(a) $P(X = 3) = \binom{4}{3}\left(\dfrac{1}{10}\right)^{3}\left(\dfrac{9}{10}\right)^{1} = 4\times\dfrac{1}{1000}\times\dfrac{9}{10} = \dfrac{36}{10000} = \dfrac{9}{2500}$ . One mark for the correct binomial coefficient and powers, one for the exact value.

(b) $P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$ . It is fastest to use $P(X < 3) = 1 - P(X = 3) - P(X = 4)$ .
$P(X = 4) = \left(\dfrac{1}{10}\right)^{4} = \dfrac{1}{10000}$ .
So $P(X < 3) = 1 - \dfrac{36}{10000} - \dfrac{1}{10000} = 1 - \dfrac{37}{10000} = \dfrac{9963}{10000}$ .
Markers want exact fractions, not decimals, since the question asks for exact values.

Practice5 marks

Calculator-free. A discrete random variable

X

has

P(X = x) = 0.1, 0.2, 0.4, 0.2, 0.1

for

x = -2, -1, 0, 1, 2

respectively. (a) Find

E(X)

. (b) Show that

\operatorname{Var}(X) = 1.2

Show worked answer →

(a) $E(X) = \sum x\,P(X = x)$

= (-2)(0.1) + (-1)(0.2) + (0)(0.4) + (1)(0.2) + (2)(0.1) = 0.

The distribution is symmetric about

0

, so

E(X) = 0

(b) $\operatorname{Var}(X) = E(X^{2}) - [E(X)]^{2}$ .

E(X^{2}) = (4)(0.1) + (1)(0.2) + (0)(0.4) + (1)(0.2) + (4)(0.1) = 1.2.

Since

E(X) = 0

\operatorname{Var}(X) = 1.2 - 0^{2} = 1.2

as required. Show the

E(X^{2})

sum explicitly to earn the working marks.

Practice2 marks

Calculator-assumed. A casino game has net earnings

X

dollars with

P(X = -1) = \dfrac{14}{24}

P(X = 0) = \dfrac{4}{24}

and

P(X = 2) = \dfrac{6}{24}

. Calculate

E(X)

and state whether this is a fair game.

Show worked answer →

E(X) = (-1)\frac{14}{24} + (0)\frac{4}{24} + (2)\frac{6}{24} = \frac{-14 + 0 + 12}{24} = \frac{-2}{24} = -\frac{1}{12} \approx -0.083.

A game is fair only when the expected earnings are $0$ . Here $E(X) = -\dfrac{1}{12}$ dollars is negative, so on average the player loses about $8.3$ cents per play. It is not a fair game; it favours the casino. One mark for the correct $E(X)$ , one for the justified conclusion.

Practise this

Sources & how we know this

Mathematics Methods Level 4 (MTM415117) course document (2017 onwards) — TASC

What this dot point is asking

Mean and variance

Step 1: Compute the mean E(X)E(X)E(X)

Step 2: Compute E(X2)E(X^2)E(X2)

Step 3: Find the variance and standard deviation

The binomial distribution

Step 1: Identify the distribution

Step 2: Set up the binomial probability for exactly 3 correct

Step 3: Evaluate each factor and multiply

Step 4: Find the expected number correct

Cumulative probabilities and technology

Where the binomial is used

Summary

Exam-style questions

Practise this

Sources & how we know this

Step 1: Compute the mean $E(X)$

Step 2: Compute $E(X^2)$