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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 4.88 min answerUpdated 2026-06-02

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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 $E(X^2) = \sum x^2 P(X=x)$ .

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.

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.

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.

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 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.

TCE 20245 marksCalculator-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.

TCE 20235 marksCalculator-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.

TCE 20242 marksCalculator-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.