Year 11 Mathematical Methods — Counting & Probability

Cambridge Methods 1&2 — Counting Methods (10D)  •  Sun 14 June 2026
📚 Also for this topic: 📄 Printable worksheet ✅ Solutions (answer key)

Today's lesson

We finish Counting Methods by using everything from 10A–10C to find probabilities. When every outcome is equally likely, probability is just a counting problem: $$P(\text{event}) = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}.$$ Both the top and bottom are counted with the multiplication principle, $^nP_r$ or $^nC_r$.

Learning intentions

Part 1 — Probability with arrangements

If a row of objects is arranged at random, every ordering is equally likely. Count the total orderings (the denominator) and the favourable orderings (the numerator), then divide.

Equally-likely model. $P(\text{event}) = \dfrac{n(\text{favourable})}{n(\text{total})}$, where both counts come from the counting rules. Always simplify the fraction.

📺 Walkthrough: the digits $1,2,3,4,5$ are arranged at random — finding $P(\text{number is even})=\dfrac{2\times4!}{5!}=\dfrac{48}{120}=\dfrac{2}{5}$.

EXAMPLE 1 — arranging letters
The $6$ distinct letters of NUMBER are arranged at random in a row. Find the probability that (a) the first letter is $N$; (b) the word ends in $R$.
  1. Total arrangements $= 6! = 720$ (the denominator for both parts).
  2. (a) Fix $N$ first, arrange the other $5$: $5!=120$ favourable. $P=\dfrac{120}{720}=\boxed{\dfrac{1}{6}}$.
  3. (b) Fix $R$ last, arrange the other $5$: $5!=120$. $P=\dfrac{120}{720}=\boxed{\dfrac{1}{6}}$.

Now you try: The digits $1,2,3,4,5$ are arranged at random. Find $P(\text{the number is greater than } 30\,000)$.   Answer: first digit $3,4$ or $5$, so $\dfrac{3\times4!}{5!}=\dfrac{72}{120}=\dfrac{3}{5}$.

EXAMPLE 2 — two people together
Four people line up at random. Find the probability that two particular people, $A$ and $B$, stand next to each other.
  1. Total orderings $= 4! = 24$.
  2. Favourable (A, B together): glue them into a block → $3!$ arrangements $\times\,2!$ internal $= 6\times2 = 12$.
  3. $P = \dfrac{12}{24} = \boxed{\dfrac{1}{2}}$.

Strategy: the denominator is "all arrangements"; the numerator uses your 10B restriction skills (together / apart / end positions).

Part 2 — Probability with selections

When a committee or sample is chosen at random, every selection is equally likely. Count total selections with $^nC_r$, and favourable selections by splitting into groups ("exactly") or using the complement ("at least").

EXAMPLE 3 — committee probabilities
A committee of $3$ is chosen at random from $5$ men and $3$ women. Find the probability that the committee has (a) exactly $2$ women; (b) at least $1$ woman.
  1. Total committees $= {}^8C_3 = 56$ (the denominator).
  2. (a) Exactly $2$ women = $2$ women and $1$ man: ${}^3C_2\times{}^5C_1 = 3\times5 = 15$. So $P=\dfrac{15}{56}$.
  3. (b) At least $1$ woman — complement of "no women": all-men committees $= {}^5C_3 = 10$. So $P = 1-\dfrac{10}{56} = \dfrac{46}{56} = \boxed{\dfrac{23}{28}}$.

Now you try: $P(\text{all }3\text{ are women})$.   Answer: $\dfrac{{}^3C_3}{{}^8C_3}=\dfrac{1}{56}$.

Practice 2.1 — write each probability as a simplified fraction.

  1. The letters of MATHS (all distinct) are arranged at random. $P(\text{ends in } S)$.
  2. $4$ boys and $3$ girls stand in a row at random. $P(\text{all }4\text{ boys are together})$.
  3. $3$ students are chosen from $10$, of whom $4$ are left-handed. $P(\text{all }3\text{ left-handed})$.
  4. $2$ counters are drawn from a bag of $3$ red and $4$ blue. $P(\text{both red})$.
  5. A committee of $3$ from $5$ men, $3$ women. $P(\text{exactly }2\text{ women})$.
  6. $6$ from $40$ lottery balls. $P(\text{your single ticket wins})$.
a) $\dfrac{4!}{5!}=\dfrac{1}{5}$    b) $\dfrac{4!\,4!}{7!}=\dfrac{4}{35}$    c) $\dfrac{{}^4C_3}{{}^{10}C_3}=\dfrac{4}{120}=\dfrac{1}{30}$    d) $\dfrac{{}^3C_2}{{}^7C_2}=\dfrac{3}{21}=\dfrac{1}{7}$    e) $\dfrac{15}{56}$    f) $\dfrac{1}{{}^{40}C_6}=\dfrac{1}{3\,838\,380}$

Part 3 — Quick quiz (5 min)

Pick the correct answer for each, then click Mark.

Q1. For equally-likely outcomes, $P(\text{event})$ equals:

Q2. The digits $1,2,3,4,5$ are arranged at random. $P(\text{the number is even})$ is:

Q3. The $6$ letters of NUMBER are arranged at random. $P(\text{first letter is }N)$ is:

Q4. A committee of $3$ from $5$ men and $3$ women. $P(\text{exactly }2\text{ women})$ is:

Q5. $2$ counters are drawn from $3$ red and $4$ blue. $P(\text{both red})$ is:

Working program — Cambridge — Counting Methods Ex 10D

After the quiz, open Cambridge Methods 1&2 (Counting Methods, Chapter 10) and complete the set work:

Set workExtension
Ex 10D — all set questionsMixed chapter-review probability questions.

For each part, write the denominator (total outcomes) first, then the numerator (favourable), then simplify.

Exit ticket — write in your book

Before you pack up, write one sentence each:
  1. What goes on the top and bottom of an equally-likely probability?
  2. When counting favourable committees with "at least one", which method is usually fastest?
  3. For random arrangements of $5$ distinct letters, what is the total number of outcomes?

— Mr Wong