Discrete random variables
In Year 12 Mathematical Methods (Queensland, QCAA), a discrete random variable \(X\) takes separate values, each with a probability \(P(X=x)\). The values and probabilities form its probability distribution, valid when every \(P(X=x)\ge 0\) and the total is \(1\). You solve for an unknown probability, read off cumulative and interval probabilities, use the discrete uniform distribution, and show a distribution as a table or column graph.
A discrete random variable \(X\) takes separate, countable values. Its probability function \(P(X=x)\) gives the probability of each value, and the full list of values with their probabilities is the probability distribution. It is displayed as a table or a column graph (one column per value, its height the probability).
A distribution is valid exactly when every \(P(X=x)\ge 0\) and the probabilities sum to \(1\). Both conditions matter: a total of \(1\) with a negative entry is not a valid distribution. Setting the total equal to \(1\) lets you solve for an unknown constant \(k\) or \(c\).
When the outcomes are equally likely the distribution is discrete uniform, with \(P(X=x)=\dfrac{1}{n}\) for each of the \(n\) values (a fair die, a fair spinner). A non-uniform distribution is often built from a context using relative frequencies — a count divided by the total. Once you have the distribution, add entries to get cumulative and interval probabilities.
| \(x\) | 1 | 2 | 3 | 4 |
| \(P(X=x)\) | 0.1 | 0.3 | 0.4 | 0.2 |
A valid probability distribution satisfies both conditions:
Cumulative and interval probabilities are sums of entries:
Discrete uniform distribution on \(n\) equally likely values:
How to work with a discrete probability distribution
- Check or enforce validity. Confirm every \(P(X=x)\ge 0\) and that the probabilities sum to \(1\); if a constant is unknown, set the total equal to \(1\) and solve for it.
- Reject impossible values. For a quadratic in \(k\) or \(c\), discard any solution that gives a negative probability or one above \(1\).
- Read cumulative and interval probabilities. Add the relevant entries, watching the boundary: \(P(X\le k)\) includes \(x=k\); \(P(X>k)\) excludes it; \(P(a\le X\le b)\) adds every value from \(a\) to \(b\).
- Recognise uniform or build from data. Equally likely outcomes give \(P(X=x)=\dfrac{1}{n}\); otherwise use relative frequencies from a context to construct the table, then represent it as a column graph if asked.
| \(0.1+0.3+k+0.2\) | \(=\) | \(1\) |
| \(0.6+k\) | \(=\) | \(1\) |
| \(k\) | \(=\) | \(0.4\) |
So \(k=\mathbf{0.4}\), which is non-negative, so the distribution is valid.
| \(P(X\le 3)\) | \(=\) | \(0.15+0.25+0.3=0.7\) |
| \(P(X>3)\) | \(=\) | \(0.2+0.1=0.3\) |
Note \(P(X>3)=1-P(X\le 3)\), so \(\mathbf{0.7}\) and \(\mathbf{0.3}\).
| \(P(X=x)\) | \(=\) | \(\dfrac{1}{5}=0.2\) |
| \(P(X\le 3)\) | \(=\) | \(3\times 0.2=0.6\) |
Each value has probability \(\mathbf{0.2}\); \(P(X\le 3)=\mathbf{0.6}\).
| \(P(1),P(2),P(3)\) | \(=\) | \(0.5,\ 0.3,\ 0.2\) |
| \(P(X\ge 2)\) | \(=\) | \(0.3+0.2=0.5\) |
The probabilities total \(1\), and \(P(X\ge 2)=\mathbf{0.5}\).
Common pitfalls
Frequently asked questions
What is a discrete random variable?
A variable \(X\) that takes separate, countable values, each with a probability \(P(X=x)\); the values and probabilities form its distribution.
When is a distribution valid?
When every \(P(X=x)\ge 0\) and \(\sum P(X=x)=1\). Both conditions are needed.
How do you solve for an unknown probability?
Set the sum of all probabilities equal to \(1\) and solve for the constant, rejecting any value that makes a probability negative.
How do you find \(P(X\le k)\) or \(P(a\le X\le b)\)?
Add the relevant entries. \(P(X\le k)\) includes \(x=k\); \(P(a\le X\le b)\) adds every value from \(a\) to \(b\) inclusive.
What is a discrete uniform distribution?
One where the \(n\) outcomes are equally likely, so \(P(X=x)=\dfrac{1}{n}\) for each value, as for a fair die or spinner.
How do you build a distribution from data?
Use relative frequencies: each probability is the count of that value divided by the total number of outcomes.