Define a discrete random variable and its probability distribution, calculate the expected value $E(X)$ and the variance $\mathrm{Var}(X)$ and standard deviation, and recognise the Bernoulli distribution as the single-trial case

What this dot point is asking

QCAA wants you to define a discrete random variable, identify and work with its probability distribution, compute the expected value and variance, and recognise the Bernoulli distribution as the simplest case (a single yes or no trial). This dot point underpins all of Topic 3 and feeds directly into the binomial distribution.

The answer

Random variables

A random variable $X$ assigns a numerical value to each outcome of a probability experiment. A random variable is discrete if its possible values form a countable set (typically a list of integers).

The probability distribution of $X$ is the list of possible values together with their probabilities, often shown as a table.

For this to be a valid distribution, the probabilities must satisfy two conditions.

1. Non-negativity. $p_i \geq 0$ for every $i$.
2. Normalisation. $\sum_i p_i = 1$.

QCAA Paper 1 questions frequently give a partial distribution with an unknown $k$ and ask you to use the normalisation condition to solve for $k$.

Expected value

The expected value (or mean) of $X$ is the probability-weighted average of its values:

Interpretation: the long-run average of $X$ over many independent repetitions.

For any constants $a$ and $b$:

Variance and standard deviation

The variance measures spread around the mean.

The computational shortcut (the version you should use on Paper 1 and Paper 2) is

where $E(X^2) = \sum_i x_i^2 \, P(X = x_i)$.

The standard deviation is $\sigma = \sqrt{\mathrm{Var}(X)}$.

For any constants $a$ and $b$:

The $b$ disappears (a shift does not change spread) and the $a$ squares (a scale multiplies variance by $a^2$).

The Bernoulli distribution

A Bernoulli trial has exactly two outcomes, labelled success ($X = 1$) and failure ($X = 0$), with probability $p$ of success.

Write $X \sim \mathrm{Bern}(p)$. Then

The Bernoulli distribution is the building block of the binomial: a binomial $\mathrm{Bin}(n, p)$ is the sum of $n$ independent Bernoulli trials with the same $p$.

Worked examples

Validity check

A student claims the following is a probability distribution: $P(X = 1) = 0.2$, $P(X = 2) = 0.5$, $P(X = 3) = 0.4$. Is it valid?

$0.2 + 0.5 + 0.4 = 1.1 \neq 1$. Not valid; it violates the normalisation condition.

Mean and variance from a table

A discrete random variable $X$ has $P(X = 1) = 0.4$, $P(X = 2) = 0.4$, $P(X = 5) = 0.2$.

$E(X) = 1 \cdot 0.4 + 2 \cdot 0.4 + 5 \cdot 0.2 = 0.4 + 0.8 + 1.0 = 2.2$.

$E(X^2) = 1 \cdot 0.4 + 4 \cdot 0.4 + 25 \cdot 0.2 = 0.4 + 1.6 + 5.0 = 7.0$.

$\mathrm{Var}(X) = 7.0 - (2.2)^2 = 7.0 - 4.84 = 2.16$.

Standard deviation $\sigma = \sqrt{2.16} \approx 1.47$.

Linear transformation

If $E(X) = 2.2$ and $\mathrm{Var}(X) = 2.16$, find $E(3 X + 1)$ and $\mathrm{Var}(3 X + 1)$.

$E(3 X + 1) = 3 \cdot 2.2 + 1 = 7.6$.

$\mathrm{Var}(3 X + 1) = 3^2 \cdot 2.16 = 9 \cdot 2.16 = 19.44$.

Bernoulli with IMATH_60

$X \sim \mathrm{Bern}(0.4)$ has $E(X) = 0.4$ and $\mathrm{Var}(X) = 0.4 \cdot 0.6 = 0.24$.

Common traps

Forgetting the normalisation check. A discrete distribution must have probabilities that sum to exactly 1. Forgetting to set this up costs the first mark when QCAA gives an unknown $k$.

Confusing $E(X^2)$ and $[E(X)]^2$. The variance shortcut is $E(X^2) - [E(X)]^2$, where the first term squares each $x$ before weighting and the second squares after weighting. They are not equal.

Dropping the square on the scale factor. $\mathrm{Var}(a X + b) = a^2 \, \mathrm{Var}(X)$, not $a \, \mathrm{Var}(X)$.

Computing standard deviation but stopping at variance. If QCAA asks for standard deviation, take the square root.

Naming Bernoulli when you mean binomial. A Bernoulli is a single trial. A binomial is $n$ trials. Mixing the two is a frequent IA2 short response error.

Using negative probabilities. A probability cannot be negative. If your algebra produces a negative $p$, you have set up the problem wrongly.

In one sentence

A discrete random variable has a probability distribution satisfying $p_i \geq 0$ and $\sum p_i = 1$, with mean $E(X) = \sum x_i p_i$ and variance $\mathrm{Var}(X) = E(X^2) - [E(X)]^2$, and the Bernoulli distribution $\mathrm{Bern}(p)$ is the single-trial case with $E = p$ and $\mathrm{Var} = p (1 - p)$.