Loav · VCE Mathematical Methods · Units 3 and 4
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A random variable turns outcomes into numbers so that probabilities, averages and spread can be studied together. This guide develops a discrete distribution, its expected value, variance and standard deviation. It also explains the effects of defining parameters and conditional questions. You need basic probability and confident substitution into formulas. The main goal is to understand what the numbers describe, rather than to treat a probability table as an ordinary list of observations.
Formally, a random variable is a real-valued function defined on the sample space: it assigns one number to each possible outcome. For two coin tosses, a variable counting heads maps TT to 0, TH and HT to 1, and HH to 2. Different outcomes can map to the same number.
Use a capital letter, such as X, for the random variable and a lower-case letter, such as x, for a possible value. If X counts the number of damaged parcels in a delivery, X is defined before the delivery is inspected. After inspection, an observed value might be x = 2. The variable, its possible values and the observed result are related but distinct ideas.
1. Build and validate a distribution
A discrete random variable has a finite or countable collection of possible values. Counts are discrete, but a discrete variable can also take negative or fractional values, such as net earnings from several specified outcomes. The defining feature is a separated set of possibilities, not the absence of decimals.
p(x) = P(X = x)
For every possible x: p(x) ≥ 0.
The sum of all probabilities is 1.
A probability distribution lists every possible value and its probability. Outcomes of the original experiment need not be equally likely, and multiple outcomes can give the same X value. For example, several arrangements of successes and failures may lead to a count of two. Add their probabilities to build P(X = 2); do not assume every count is equally likely because it appears once in the table.
To find P(a < X ≤ b), select the table entries whose x values satisfy the inequality and add their probabilities. Because the variable is discrete, including an endpoint can change the answer. A table with support 0, 2 and 4 has no mass at 3, even though 3 lies numerically between the smallest and largest values.
2. Expected value is a probability-weighted mean
If a model is repeated many times under the same conditions, values with larger probabilities occur more frequently. The expected value weights each possible value by its probability. It is the distribution’s balance point and the long-run average predicted by the model.
μ = E(X) = Σ x p(x)
The symbol Σ means to add over every possible value. Do not divide this sum by the number of values in the table: the probabilities already supply weights that total one. An unweighted average would wrongly treat a rare outcome as equally influential as a common outcome.
An expected count can be non-integer. Expecting 1.6 faults does not claim one object has 1.6 faults; it describes an average across repetitions. Likewise, an expected net gain of $0.50 does not promise a fifty-cent gain on the next attempt. Probability models describe uncertainty, including possible losses even when the mean is positive.
3. Read the probability graph with its table
A probability-mass graph places a vertical mark or bar at each possible value, with height equal to that value’s probability. The heights, rather than geometric areas between points, are what must sum to one. The gaps have no probability unless another supported value is explicitly defined there. Label the horizontal axis with the variable and the vertical axis with P(X = x).
Two distributions can have the same mean but very different shapes. One may concentrate close to its mean while another places substantial probability far away. The expected value alone cannot distinguish those situations. A graph helps you anticipate spread before calculating it, while the variance supplies a numerical measure.
Changing a defining probability parameter moves probability between supported values. The total must remain one and each probability must remain nonnegative. If more weight moves from a low value to a high value, the mean increases. The effect on spread may be less obvious: putting nearly all the mass on one endpoint makes outcomes more predictable, even if that endpoint is far from zero. Spread is measured around the mean, not around the origin.
4. Variance measures squared distance from the mean
The average signed deviation from the mean is zero because positive and negative deviations balance. Squaring removes the cancellation and gives a measure of spread. Standard deviation returns the scale to the original units.
Var(X) = E[(X − μ)2] = Σ (x − μ)2p(x)
Var(X) = E(X2) − [E(X)]2, where E(X2) = Σ x2p(x)
sd(X) = √Var(X)
E(X2) means square each possible value before weighting. [E(X)]2 means calculate the mean and then square it. These operations are generally different. A negative calculated variance cannot be a valid answer; check the table, arithmetic and premature rounding. Variance has squared units, while mean and standard deviation have the units of X.
These formulas describe an entire probability distribution. They are not the sample-variance formula using a denominator of n − 1. A probability table gives theoretical weights, not an unlabelled sample of raw observations. Keep that distinction clear when switching between calculator statistics lists and probability calculations.
Worked examples
Worked example 1: interpret a complete distribution
An original scoring device returns X points according to the following model.
| x | −2 | 0 | 3 | 5 |
|---|---|---|---|---|
| P(X = x) | 0.20 | 0.35 | 0.30 | 0.15 |
- All probabilities are nonnegative and total 1, so the table is valid. P(X > 0) = 0.30 + 0.15 = 0.45.
- E(X) = (−2)(0.20) + 0(0.35) + 3(0.30) + 5(0.15) = 1.25 points.
- E(X2) = 4(0.20) + 0 + 9(0.30) + 25(0.15) = 7.25 points squared.
- Var(X) = 7.25 − 1.252 = 5.6875 points squared. sd(X) = √5.6875 ≈ 2.3848 points.
- P(X ≥ 0) = 0.80. Among these outcomes, positive scores have probability P(X > 0 | X ≥ 0) = 0.45/0.80 = 9/16.
The mean is 1.25 although that score is impossible in one play. Larger positive scores balance the probability at zero and below. Do not round the mean to an available score. The conditional calculation also shows why the reference group must be stated explicitly.
Worked example 2: solve an unknown probability constant
A sensor records X interruptions in a fixed interval, with possible values 0, 1, 2, 3 and probabilities k, 2k, 3k, 4k respectively. Find k, the mean, the standard deviation and P(X ≥ 2 | X > 0).
- Total probability gives 10k = 1, so k = 0.1. The four probabilities are 0.1, 0.2, 0.3 and 0.4.
- E(X) = 0 + 0.2 + 0.6 + 1.2 = 2.
- E(X2) = 0 + 0.2 + 1.2 + 3.6 = 5. Thus Var(X) = 5 − 4 = 1 and sd(X) = 1.
- The conditioning event X > 0 has probability 0.9. The intersection with X ≥ 2 is simply X ≥ 2, with probability 0.7. Hence the conditional probability is 0.7/0.9 = 7/9.
The condition removes the zero-interruption case and rescales the remaining probabilities. It does not change the original distribution retrospectively; it defines a new reference group for this question.
5. Explore a parameter family
Consider a simple family where X = 0 with probability 1 − p and X = 2 with probability p, for 0 ≤ p ≤ 1. The mean is 0(1 − p) + 2p = 2p. The second moment is 4p, so the variance is 4p − 4p2 = 4p(1 − p). These expressions follow directly from the table definitions.
As p increases from zero to one, the mean increases from zero to two. The variance behaves differently: it rises to one at p = 0.5, then falls to zero at p = 1. At either endpoint, the outcome is certain and there is no spread. This is a useful reminder that increasing the mean does not necessarily increase variability.
On the graph, the probability at zero falls as the probability at two rises. The horizontal locations remain fixed, but their weights change. When a question supplies a mean or variance and asks for p, solve the resulting equation and keep only solutions in [0,1]. A variance equation can give two valid parameters because p and 1 − p produce the same spread in this family.
Worked example 3: distinguish payout from net gain
An original classroom chance game pays $9 if a fair six-sided die shows six and pays nothing otherwise. Each play costs $1. Define X as net gain, not gross payout.
- X = 8 with probability 1/6, and X = −1 with probability 5/6. The fee must be subtracted in both cases.
- E(X) = 8/6 − 5/6 = $0.50. This is the long-run mean per play under the model.
- E(X2) = 64/6 + 5/6 = 11.5. Var(X) = 11.5 − 0.52 = 11.25 dollars squared, so sd(X) ≈ $3.3541.
- If the entry fee becomes $1.50 while payout stays $9, the two net outcomes become $7.50 and −$1.50. Their mean is 7.5/6 − 1.5(5/6) = 0. Their second moment is 7.52/6 + 1.52(5/6) = 11.25, so the variance is 11.25.
A zero expected net gain makes this mathematical game fair in the expectation sense. It does not mean equal chances of winning and losing, nor does it remove the possibility of a substantial short-run loss.
Practice and reasoned answers
1. X takes values 0, 2, 4 with probabilities 0.25, 0.50, 0.25. Find mean, variance and standard deviation.
E(X) = 0 + 1 + 1 = 2. E(X2) = 0 + 2 + 4 = 6, so Var(X) = 6 − 4 = 2 and sd(X) = √2. Symmetry about 2 provides a useful mean check but does not replace the variance calculation.
2. X takes values 1, 2, 4 with probabilities 0.2, 0.5, 0.3. Find P(X > 1), mean and variance.
P(X > 1) = 0.8. E(X) = 0.2 + 1 + 1.2 = 2.4. E(X2) = 0.2 + 2 + 4.8 = 7, so Var(X) = 7 − 2.42 = 1.24. The mean is not obtained by averaging 1, 2 and 4 equally.
3. A proposed probability table contains 0.1, 0.4, k and 0.6. Can any k make it valid?
Normalization forces k = −0.1 because the known entries total 1.1. Negative probabilities are forbidden, so no valid value exists. Solving the sum equation is necessary but not sufficient: each entry must also be nonnegative.
4. X takes values 0 and 2 with probabilities 1 − p and p. Its mean is 1.6. Find p and Var(X).
E(X) = 2p = 1.6, so p = 0.8. E(X2) = 4p = 3.2 and Var(X) = 3.2 − 1.62 = 0.64. Check that the probabilities 0.2 and 0.8 are nonnegative and total one.
5. X takes 0, 1, 2 with probabilities 0.3, 0.4, 0.3. Find P(X = 2 | X ≥ 1) and Var(X).
The condition has probability 0.7, so the conditional answer is 0.3/0.7 = 3/7. For the original distribution, E(X) = 1 and E(X2) = 0.4 + 1.2 = 1.6, giving Var(X) = 0.6. Do not silently replace the original distribution by the conditional one for the variance.
6. X takes 0 and 2 with probabilities 1 − p and p. Its variance is 0.75. Find all possible p values.
The variance equation is 4p(1 − p) = 0.75, giving p2 − p + 0.1875 = 0. Its roots are p = 0.25 and p = 0.75, both valid. Their means are 0.5 and 1.5 respectively. Equal variance does not imply equal mean or an identical distribution.
7. X equals 0 or 10 with equal probabilities. Y equals 4 or 6 with equal probabilities. Compare their means and variances.
Both means are 5. For X, E(X2) = 50, so Var(X) = 50 − 25 = 25. For Y, E(Y2) = (16 + 36)/2 = 26, so Var(Y) = 1. Their standard deviations are 5 and 1. The second distribution is more concentrated around the shared mean.
Learning check
Build a table, check its total, and explain each weight. Calculate a mean and a second moment separately before subtracting. Describe why an expected value need not be possible in one trial. Finally, predict how changing a defining probability parameter moves the graph’s probability weights and affects its mean and spread. The next guide studies the particularly useful discrete distribution obtained by counting independent successes.
Curriculum and study route
Loav Mathematical Methods Units 3 and 4. These module divisions are a teaching sequence. Use the current VCAA Mathematics study design, Mathematical Methods Units 3 and 4, Area of Study 4, and the VCAA Mathematical Methods FAQ for course requirements. Scope checked 15 October 2026. All worked scenarios, questions and explanations in this guide are original.
