TASMath MethodsSyllabus dot point

How do we describe a discrete random variable and when does the binomial distribution apply?

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

A discrete random variable has a probability distribution from which you compute expected value and variance; the binomial distribution models the count of successes in a fixed number of independent trials.

Generated by Claude Opus 4.78 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

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.

Mean and variance

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

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

The binomial distribution

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

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

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.

Each guess is a trial with success probability $p = \tfrac{1}{4} = 0.25$ , there are $n = 10$ independent trials, so $X \sim B(10, 0.25)$ .

For exactly $3$ correct, use the binomial formula with $k = 3$ :

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

Compute the pieces: $\dbinom{10}{3} = \dfrac{10!}{3!\,7!} = 120$ ; $(0.25)^3 = 0.015625$ ; $(0.75)^7 \approx 0.133484$ .

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

The expected number correct is $E(X) = np = 10 \times 0.25 = 2.5$ questions.

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 practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 TASC5 marksThe probability of a machine in a factory producing a defective item is 1/10. X is a random variable given by 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 = 1/10, so q = 9/10. Use P(X = r) = C(n, r) p^r q^(n - r).

a) P(X = 3) = C(4, 3) (1/10)^3 (9/10)^1 = 4 times (1/1000) times (9/10) = 36/10000 = 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) = (1/10)^4 = 1/10000.
So P(X < 3) = 1 - 36/10000 - 1/10000 = 1 - 37/10000 = 9963/10000.
Markers want exact fractions, not decimals, since the question says "exact values".

2023 TASC5 marksA discrete random variable X has the distribution 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 var(X) is equal to 1.2.

Show worked answer →

a) E(X) = sum of x times P(X = x)
= (-2)(0.1) + (-1)(0.2) + (0)(0.4) + (1)(0.2) + (2)(0.1)
= -0.2 - 0.2 + 0 + 0.2 + 0.2 = 0.
The distribution is symmetric about 0, so E(X) = 0.

b) 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)
= 0.4 + 0.2 + 0 + 0.2 + 0.4 = 1.2.
Since E(X) = 0, var(X) = 1.2 - 0^2 = 1.2 as required. Show the E(X^2) sum explicitly to earn the working marks.

2024 TASC2 marksA random variable x is the earnings for a casino game with possible values x = -1, 0, 2. Calculate E(X). Is this a fair game?

Show worked answer →

For the dice game the probabilities work out (from the 24 equally likely outcomes of a 4-sided and 6-sided die rolled together) to P(X = -1) = 14/24, P(X = 0) = 4/24, P(X = 2) = 6/24.

E(X) = (-1)(14/24) + (0)(4/24) + (2)(6/24) = (-14 + 0 + 12)/24 = -2/24 = -1/12 (about -0.083).

A game is "fair" only when the expected earnings are 0. Here E(X) = -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.