Discrete probability distributions: QCE Math Methods Unit 2 Year 11

What this dot point is asking

QCAA wants Year 11 students to be introduced to discrete probability distributions, to compute expected value and variance, and to recognise the Bernoulli and uniform discrete distributions. This is the foundation for the binomial distribution studied in Unit 3.

Discrete random variable

A discrete random variable $X$ takes values in a finite or countable set, such as the number rolled on a die or the number of heads in three coin tosses. Its probability mass function (pmf) assigns a probability $P(X = x)$ to each possible value $x$. A valid pmf satisfies two conditions: every probability lies in $[0, 1]$, and the probabilities sum to exactly $1$. The total-probability condition is frequently used to find a missing probability, and checking it is a quick sanity test on any table.

Expected value

The expected value is the long-run average of $X$ over many repetitions, computed as the probability-weighted sum of the values:

It need not be a value the variable can actually take (a fair die has $E(X) = 3.5$). Expectation is linear, $E(aX + b) = aE(X) + b$, which lets you find the mean of a rescaled or shifted variable without recomputing the whole sum. This is the discrete analogue of the average, and it is the centre about which the distribution's spread is measured.

Variance and standard deviation

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

Standard deviation $\sigma = \sqrt{\text{Var}(X)}$.

Property: $\text{Var}(aX + b) = a^2 \text{Var}(X)$.

The Bernoulli distribution

The Bernoulli distribution models a single trial with two outcomes: success with probability $p$ and failure with probability $1 - p$, coded as $X = 1$ and $X = 0$. Its mean is $E(X) = p$ and its variance is $\text{Var}(X) = p(1 - p)$, the latter largest when $p = 0.5$ (maximum uncertainty) and zero at $p = 0$ or $p = 1$ (a certain outcome). A sum of independent Bernoulli trials is the binomial distribution studied in Unit 3, so the Bernoulli is the single building block of that theory.

The uniform discrete distribution

In a uniform discrete distribution each of $n$ outcomes is equally likely, with probability $\dfrac{1}{n}$. For the values $1, 2, \ldots, n$ the mean is $E(X) = \dfrac{n + 1}{2}$ (the midpoint) and the variance is $\text{Var}(X) = \dfrac{n^2 - 1}{12}$. A fair die is the standard example ($n = 6$, mean $3.5$). Recognising a uniform distribution lets you write down the mean and variance immediately rather than summing term by term.

Reading a distribution table

Most questions present the distribution as a table of values and probabilities. The workflow is always the same: first check (or enforce) that the probabilities sum to $1$, using this to solve for any unknown probability; then compute $E(X)$ as the weighted sum of values; then compute $E(X^2)$ as the weighted sum of squared values; and finally find the variance from the working formula. Laying the calculation out in columns for $x$, $P(X = x)$, $xP$ and $x^2 P$ keeps it organised and is exactly how markers expect the working to appear.

Why the working formula for variance

Variance measures the average squared distance of $X$ from its mean. The defining form $\text{Var}(X) = E[(X - \mu)^2]$ is correct but awkward, so the algebraically equivalent working formula $\text{Var}(X) = E(X^2) - [E(X)]^2$ is used in practice: compute $E(X^2)$ by squaring each value before weighting, then subtract the square of the mean. The standard deviation $\sigma = \sqrt{\text{Var}(X)}$ returns the spread to the original units, which is why it, rather than the variance, is usually quoted in context.