Applications to probability
Master applications to probability for Year 11 Specialist Mathematics in Queensland (QCAA). When every outcome is equally likely, a probability becomes a counting problem: favourable outcomes over the total, with both counts coming from permutations and combinations.
You will learn to build probabilities from selections and arrangements, handle without-replacement draws, and turn a tricky "at least one" question into an easy one using the complement.
Theory
Applications to probability use counting — permutations and combinations — to work out exact probabilities in Year 11 Specialist Mathematics (QCAA, Queensland). When every outcome is equally likely, P(event) is the number of favourable outcomes over the total, and both counts come from \({}^{n}C_{r}\) or \(n!\). This page shows without-replacement draws, "at least one" via the complement, and arrangement probabilities.
When every outcome of an experiment is equally likely, the probability of an event is a counting problem: count how many outcomes give the event, and divide by how many outcomes there are in total.
The favourable outcomes are the outcomes that make the event happen; the total outcomes are all the equally likely possibilities. Both are found with the counting tools from combinatorics: \({}^{n}C_{r}\) for an unordered selection, and \(n!\) or \({}^{n}P_{r}\) for an ordered arrangement.
A without-replacement draw removes each item as it is taken, so the pool shrinks. Counting the selections with \({}^{n}C_{r}\) handles this automatically, because a group of \(r\) items chosen from \(n\) is counted once regardless of the order it was drawn in.
The complement of an event is "the event does not happen". Because \(P(E)+P(\text{not }E)=1\), a hard "at least one" question is usually easier as \(1-P(\text{none})\).
For an experiment whose outcomes are all equally likely:
For a without-replacement selection of \(k\) items from \(n\), where the event fixes how many come from each type (\(a\) of one type, \(b\) of another):
For an "at least one" event, use the complement:
How to find the probability
- Count the total. Work out how many equally likely outcomes there are — use \({}^{n}C_{k}\) for an unordered selection or \(n!\) for an arrangement.
- Count the favourable. Count only the outcomes that make the event happen, choosing from each type separately and multiplying (e.g. \({}^{a}C_{r}\times {}^{b}C_{s}\)).
- Consider the complement. For an "at least one" event, count \(P(\text{none})\) instead and use \(1-P(\text{none})\).
- Divide and reduce. Write favourable over total and simplify to an exact \(\dfrac{\text{a}}{\text{b}}\) fraction.
Count the favourable outcomes, then the total, then divide:
| \(\text{red (favourable)}\) | \(=\) | \(3\) |
| \(\text{total}\) | \(=\) | \(3 + 5\) |
| \(=\) | \(8\) | |
| \(P(\text{red})\) | \(=\) | \(\dfrac{3}{8}\) |
\(P(\text{red})=\dfrac{3}{8}\).
Count the total number of pairs (order does not matter):
| \(\text{total pairs}\) | \(=\) | \({}^{6}C_{2}\) |
| \(=\) | \(\dfrac{6 \times 5}{2 \times 1}\) | |
| \(=\) | \(15\) |
Count the favourable pairs — both red, chosen from the \(4\) reds:
| \(\text{red pairs}\) | \(=\) | \({}^{4}C_{2}\) |
| \(=\) | \(\dfrac{4 \times 3}{2 \times 1}\) | |
| \(=\) | \(6\) |
Divide favourable by total and reduce:
| \(P(\text{2 red})\) | \(=\) | \(\dfrac{6}{15}\) |
| \(=\) | \(\dfrac{2}{5}\) |
\(P(\text{both red})=\dfrac{2}{5}\).
Direct counting is fiddly, so use the complement — first the total:
| \(\text{total}\) | \(=\) | \({}^{10}C_{3}\) |
| \(=\) | \(120\) |
Count the “no black” outcomes (all three from the \(7\) white):
| \(\text{no black}\) | \(=\) | \({}^{7}C_{3}\) |
| \(=\) | \(35\) | |
| \(P(\text{no black})\) | \(=\) | \(\dfrac{35}{120}\) |
| \(=\) | \(\dfrac{7}{24}\) |
Subtract the complement from \(1\):
| \(P(\text{at least one black})\) | \(=\) | \(1 - \dfrac{7}{24}\) |
| \(=\) | \(\dfrac{17}{24}\) |
\(P(\text{at least one black})=\dfrac{17}{24}\).
Count the total arrangements of the \(5\) friends:
| \(\text{total}\) | \(=\) | \(5!\) |
| \(=\) | \(120\) |
Count the favourable arrangements — treat Mia and Noah as one block (\(2\) internal orders, then arrange \(4\) items):
| \(\text{favourable}\) | \(=\) | \(2 \times 4!\) |
| \(=\) | \(2 \times 24\) | |
| \(=\) | \(48\) |
Divide favourable by total and reduce:
| \(P(\text{together})\) | \(=\) | \(\dfrac{48}{120}\) |
| \(=\) | \(\dfrac{2}{5}\) |
\(P(\text{together})=\dfrac{2}{5}\).
Common pitfalls
Frequently asked questions
How do you use combinations to find a probability?
When outcomes are equally likely, count the favourable selections with \({}^{n}C_{r}\) and the total selections with \({}^{n}C_{r}\), then divide: \(P=\dfrac{\text{favourable}}{\text{total}}\).
What is the formula for probability without replacement?
Choose the required items from each type and divide by all selections, for example \(P=\dfrac{{}^{a}C_{r}\times {}^{b}C_{s}}{{}^{n}C_{k}}\). Using \({}^{n}C_{r}\) handles the shrinking pool automatically.
How do you find the probability of at least one?
Use the complement: \(P(\text{at least one})=1-P(\text{none})\). Count the single "none" outcome, divide by the total, then subtract from \(1\).
When do you multiply combinations in a probability?
Multiply \({}^{a}C_{r}\times {}^{b}C_{s}\) when the event needs a fixed number from each separate group (for example \(2\) reds and \(1\) blue); each factor counts one group.
Why keep the answer as a fraction and not a decimal?
Specialist probability answers are exact, so a \(\dfrac{a}{b}\) fraction is required. A rounded decimal loses accuracy and is treated as a distractor.
How do you find the probability that two people sit together?
Count all arrangements with \(n!\), then count the favourable ones by treating the pair as one block: \(2\times(n-1)!\). Divide to get the probability.