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Counting and probability: QCE Math Methods Unit 1 Year 11

Syllabus dot point

“Counting techniques (multiplication principle, permutations and combinations), simple probability, conditional probability and the addition and multiplication rules”

9 min read

Quick answer

Counting uses the multiplication principle, permutations P(n,k)=n!(n−k)!P(n,k) = \dfrac{n!}{(n-k)!} (order matters) and combinations (nk)=n!k!(n−k)!\binom{n}{k} = \dfrac{n!}{k!(n-k)!} (order does not). Probability uses P(A)=∣A∣∣S∣P(A) = \dfrac{|A|}{|S|}, the addition rule P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B), conditional P(A∣B)=P(A∩B)P(B)P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}, and independence P(A∩B)=P(A)P(B)P(A \cap B) = P(A)P(B).

Jump to a section
  1. What this dot point is asking
  2. Counting principles
  3. Set notation
  4. Simple probability
  5. The addition rule
  6. Conditional probability and the multiplication rule
  7. In one sentence
  8. Exam-style questions

What this dot point is asking

QCAA wants Year 11 students to apply counting techniques (the multiplication principle, permutations and combinations) to compute probabilities, to use set notation fluently, and to apply conditional probability, the addition and multiplication rules, and independence. These ideas underpin the discrete-probability and distribution work later in the course, so secure technique here pays off across the subject.

Counting principles

The multiplication principle says that if a first task can be done in mm ways and a second independently in nn ways, the two together can be done in m×nm \times n ways. This generalises to any number of stages.

A permutation is an arrangement where order matters. The number of ordered selections of kk objects from nn distinct objects is

P(n,k)=n!(n−k)!.P(n, k) = \frac{n!}{(n - k)!}.

A combination is a selection where order does not matter. The number of unordered selections of kk from nn is

(nk)=n!k! (n−k)!.\binom{n}{k} = \frac{n!}{k!\,(n - k)!}.

The factor k!k! in the denominator removes the k!k! orderings that a permutation would count separately.

Set notation

Probability is built on sets. The sample space SS is the set of all outcomes; an event AA is a subset of SS. The union A∪BA \cup B contains outcomes in AA or BB (or both); the intersection A∩BA \cap B contains outcomes in both; and the complement A′A' contains outcomes not in AA. A Venn diagram is often the fastest way to organise overlapping events before applying a rule.

Simple probability

For equally likely outcomes, the probability of an event is the proportion of the sample space it occupies:

P(A)=∣A∣∣S∣.P(A) = \frac{|A|}{|S|}.

Every probability satisfies 0≤P(A)≤10 \leq P(A) \leq 1, and the complement rule P(A′)=1−P(A)P(A') = 1 - P(A) is often the quickest route, especially for "at least one" questions.

The addition rule

For any two events,

P(A∪B)=P(A)+P(B)−P(A∩B).P(A \cup B) = P(A) + P(B) - P(A \cap B).

The overlap is subtracted so it is not counted twice. When the events are mutually exclusive they cannot both occur, so P(A∩B)=0P(A \cap B) = 0 and the rule simplifies to P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B).

Conditional probability and the multiplication rule

The conditional probability of AA given that BB has occurred is

P(A∣B)=P(A∩B)P(B),P(B)>0.P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \qquad P(B) > 0.

Rearranging gives the multiplication rule P(A∩B)=P(A∣B) P(B)P(A \cap B) = P(A \mid B)\,P(B). Two events are independent when the occurrence of one does not change the probability of the other, that is P(A∣B)=P(A)P(A \mid B) = P(A), in which case P(A∩B)=P(A) P(B)P(A \cap B) = P(A)\,P(B). Sampling without replacement is dependent, because removing an item changes the composition of what remains.

A selection probability

A bag holds 55 red and 77 blue marbles. Three are drawn at random without replacement. Find the probability that exactly two are red.

Step 1: Count the favourable selections

We need exactly 22 red and 11 blue. Because the order of drawing does not matter, use combinations. The number of ways to choose 22 red marbles from the 55 available, and independently 11 blue from the 77 available, are:

(52)=10(ways to choose 2 red from 5)\binom{5}{2} = 10 \qquad \text{(ways to choose 2 red from 5)}

(71)=7(ways to choose 1 blue from 7)\binom{7}{1} = 7 \qquad \text{(ways to choose 1 blue from 7)}

By the multiplication principle, the number of favourable selections is the product of these independent choices:

10×7=7010 \times 7 = 70

Step 2: Count all possible selections

Any selection of 33 marbles from the 1212 in the bag is equally likely. The total number of unordered draws is:

(123)=12!3! 9!=12×11×103×2×1=220\binom{12}{3} = \frac{12!}{3!\,9!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220

Step 3: Calculate the probability

Divide the number of favourable outcomes by the total number of equally likely outcomes:

P(exactly two red)=70220=722≈0.318P(\text{exactly two red}) = \frac{70}{220} = \frac{7}{22} \approx 0.318

Final answer: The probability of drawing exactly two red marbles is 722≈0.318\dfrac{7}{22} \approx 0.318, or about 31.8%31.8\%.

Conditional probability from a table

Read the joint probabilities

Of all commuters, 40%40\% take the train and arrive on time, 10%10\% take the train and are late. Find the probability a commuter is on time given they took the train.

The train total is P(train)=0.40+0.10=0.50P(\text{train}) = 0.40 + 0.10 = 0.50.

Apply the conditional formula

P(on time∣train)=P(on time∩train)P(train)=0.400.50=0.8.P(\text{on time} \mid \text{train}) = \frac{P(\text{on time} \cap \text{train})}{P(\text{train})} = \frac{0.40}{0.50} = 0.8.

So 80%80\% of train commuters arrive on time.

Common errors
Permutation versus combination
If order matters (arrangements, rankings, podium finishes), use a permutation; if it does not (committees, hands of cards), use a combination.
Forgetting the overlap
The addition rule subtracts P(A∩B)P(A \cap B); omitting it double-counts the shared outcomes.
Inverting a conditional
P(A∣B)P(A \mid B) is not the same as P(B∣A)P(B \mid A) in general; keep the conditioning event in the denominator.
Treating dependent events as independent
Without-replacement sampling changes the pool after each draw, so multiply conditional probabilities rather than unconditional ones.

In one sentence

Counting combines the multiplication principle with permutations (P(n,k)=n!(n−k)!P(n,k) = \tfrac{n!}{(n-k)!}, order matters) and combinations ((nk)=n!k!(n−k)!\binom{n}{k} = \tfrac{n!}{k!(n-k)!}, order does not), and probability is governed by P(A)=∣A∣∣S∣P(A) = \tfrac{|A|}{|S|}, the addition rule, conditional probability P(A∣B)=P(A∩B)P(B)P(A \mid B) = \tfrac{P(A \cap B)}{P(B)}, and independence.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks
Paper 2 (complex familiar). From a class of 3030 students (1818 boys, 1212 girls), a committee of 55 is chosen at random. Determine the probability that exactly 33 girls are chosen.
Show worked answer →

Choose 33 girls from 1212: (123)=220\binom{12}{3} = 220. Choose 22 boys from 1818: (182)=153\binom{18}{2} = 153.

Favourable selections: 220×153=33 660220 \times 153 = 33\,660. Total selections: (305)=142 506\binom{30}{5} = 142\,506.

Probability =33 660142 506≈0.236.= \dfrac{33\,660}{142\,506} \approx 0.236.

Markers reward the two combinations, the multiplication principle to combine them, and dividing by the total number of committees.

Practice4 marks
Paper 2 (complex familiar). At a school, 60%60\% of students study a language and 35%35\% study music; 20%20\% study both. A student is chosen at random. (a) Determine the probability the student studies a language or music. (b) Given the student studies music, determine the probability they also study a language.
Show worked answer →

Let LL be language and MM be music: P(L)=0.60P(L) = 0.60, P(M)=0.35P(M) = 0.35, P(L∩M)=0.20P(L \cap M) = 0.20.

(a) Addition rule: P(L∪M)=0.60+0.35−0.20=0.75.P(L \cup M) = 0.60 + 0.35 - 0.20 = 0.75.

(b) Conditional probability: P(L∣M)=P(L∩M)P(M)=0.200.35≈0.571.P(L \mid M) = \dfrac{P(L \cap M)}{P(M)} = \dfrac{0.20}{0.35} \approx 0.571.

Markers reward the addition rule with the overlap subtracted and the conditional formula with the correct conditioning event in the denominator.

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