Year 11 Methods (Unit 1 & 2) Counting Methods And Binomial Expansions

Applications To Probability

Q: What is the complement rule for at least one?

P of at least one equals one minus P of none. Count the none case, divide, then subtract from one.

Q: How do I find the probability that two people sit together?

Total is n factorial. Favourable glues the pair as a block, giving n minus one factorial times two factorial. Divide to get the probability.

Q: Why must the outcomes be equally likely?

The rule P equals n of E over n of S treats each outcome as having the same chance. If they are not equally likely you cannot simply divide the counts.

20 practice questions 1 video lesson Theory + worked examples

Apply counting methods to probability for Queensland Year 11 Mathematical Methods (QCAA). When every outcome is equally likely, a probability is the number of favourable outcomes divided by the total, so careful counting gives the answer.

You will learn to use combinations to count outcomes, find probabilities for committees, teams and hands of cards, and tackle "at least one" questions with the complement — a supporting application of combinations to probability.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 1), applications to probability use counting methods to find a probability as favourable over total when outcomes are equally likely. This page shows how to combine the multiplication and addition principles, permutations and combinations to count favourable and total outcomes, and how to use the complement for at-least-one problems.

When every outcome is equally likely, the probability of an event is the number of favourable outcomes divided by the total number of outcomes, \(P(E)=\dfrac{n(E)}{n(S)}\), where \(S\) is the sample space.

Both counts come from the counting methods: use the multiplication and addition principles, permutations when order matters, and combinations when it does not.

The complement rule, \(P(\text{at least one})=1-P(\text{none})\), turns a hard at-least count into a single none count.

Count favourable and total the same way. If you count the total with combinations, count the favourable with combinations too, so the ratio is valid.

\(5\) red and \(3\) blue give \(8\) equally likely draws, so \(P(\text{red})=\dfrac{5}{8}\).

Two tosses give \(4\) equally likely paths; \(P(\text{at least one H})=1-\dfrac{1}{4}=\dfrac{3}{4}\).

Equally likely outcomes — favourable over total:

\[P(E)=\dfrac{n(E)}{n(S)}\]

P (E) = \frac{n (E)}{n (S)}

Selection probability — choosing \(k\) of one kind:

\[P=\dfrac{^{w}C_{k}\times{}^{m}C_{r-k}}{^{n}C_{r}}\]

P = \frac{{}_{w}C_{k} \times {}_{m}C_{r - k}}{{}_{n}C_{r}}

Complement: \(P(\text{at least one})=1-P(\text{none})\) — far quicker than adding cases.

How to find the probability

Total: count the size of the sample space \(n(S)\) with the right counting method (order matters, use a permutation; not, use a combination).
Favourable: count \(n(E)\) the same way, applying groups and restrictions.
Divide and simplify: write \(P=\dfrac{n(E)}{n(S)}\) and reduce the fraction; for at least one, use \(1-P(\text{none})\).

Example 1 — Favourable over total

A bag holds \(5\) red and \(3\) blue counters. One is drawn at random. Find \(P(\text{red})\).

Solution

Count the sample space — all counters are equally likely:

\(n(S)\)	\(=\)	\(5+3\)
	\(=\)	\(8\)

Favourable outcomes are the \(5\) red counters:

\(P(\text{red})\)

\(=\)

\(\dfrac{5}{8}\)

\(P(\text{red})=\dfrac{5}{8}\).

P = \frac{5}{8}

Example 2 — A committee probability

A committee of \(2\) is chosen at random from \(4\) men and \(3\) women. Find the probability that both are women.

Solution

Total — choose \(2\) from \(7\), order ignored:

\(n(S)\)

\(=\)

\(^{7}C_{2}=\dfrac{7\times 6}{2}=21\)

Favourable — choose \(2\) of the \(3\) women:

\(n(E)\)

\(=\)

\(^{3}C_{2}=3\)

Divide and simplify:

\(P\)	\(=\)	\(\dfrac{3}{21}\)
	\(=\)	\(\dfrac{1}{7}\)

\(P(\text{both women})=\dfrac{1}{7}\).

P = \frac{1}{7}

Example 3 — An arrangement probability

Four people are seated at random in a row of \(4\) chairs. Find the probability that two particular people sit next to each other.

Solution

Total arrangements of \(4\) people:

\(n(S)\)	\(=\)	\(4!\)
	\(=\)	\(24\)

Favourable — glue the pair into a block (\(3!\) arrangements) and swap inside (\(2!\)):

\(n(E)\)	\(=\)	\(3!\times 2!\)
	\(=\)	\(6\times 2\)
	\(=\)	\(12\)

Divide and simplify:

\(P\)	\(=\)	\(\dfrac{12}{24}\)
	\(=\)	\(\dfrac{1}{2}\)

\(P(\text{together})=\dfrac{1}{2}\).

P = \frac{1}{2}

Example 4 — At least one (complement)

Two cards are drawn at random from \(5\) cards, of which \(3\) are winners and \(2\) are losers. Find the probability of at least one winner.

Solution

Use the complement — first find \(P(\text{no winner})\), i.e. both losers:

\(P(\text{no winner})\)	\(=\)	\(\dfrac{^{2}C_{2}}{^{5}C_{2}}\)
	\(=\)	\(\dfrac{1}{10}\)

Subtract from \(1\):

\(P(\text{at least one})\)	\(=\)	\(1-\dfrac{1}{10}\)
	\(=\)	\(\dfrac{9}{10}\)

\(P(\text{at least one winner})=\dfrac{9}{10}\).

P = \frac{9}{10}

Common pitfalls

Mixing order in the two counts. If the total uses combinations, the favourable must use combinations too; do not count one with order and the other without.

Forgetting outcomes must be equally likely. \(P=\dfrac{n(E)}{n(S)}\) only holds when every outcome in the sample space is equally likely.

Doing at-least the long way. For at least one, use \(1-P(\text{none})\) rather than adding exactly-one, exactly-two, and so on.

Frequently asked questions

How do I find a probability by counting?

When outcomes are equally likely, count the favourable outcomes and the total outcomes, then divide: \(P(E)=\dfrac{n(E)}{n(S)}\).

Do I use permutations or combinations for probability?

Use whichever fits the situation, but count the favourable and total the same way. Order matters uses permutations; order does not matter uses combinations.

What is the complement rule for at least one?

\(P(\text{at least one})=1-P(\text{none})\). Count the none case, divide, then subtract from \(1\).

How do I find the probability that two people sit together?

Total is \(n!\). Favourable glues the pair as a block: \((n-1)!\times2!\). Divide to get the probability.

Why must the outcomes be equally likely?

The rule \(P=\dfrac{n(E)}{n(S)}\) treats each outcome as having the same chance; if they are not equally likely you cannot simply divide the counts.

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