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Probability

Grade 12 CAPS: revise the probability rules tightly, then build the two genuinely new skills—reading a two-way contingency table, and the fundamental counting principle. Learn it as one connected route—not a list of formulas.

1. Guided Course 2. Mastery Bank 3. Quick Test
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Guided Course Slides
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Grade 12 CAPS Mathematics

Probability

From the basic rules to reading a table; from a table to counting every arrangement; from counting to probability itself. Work through the slides in order, then move to the practice bank.

Scope check:Official CAPS requirement

Your 4.5-hour learning route

A focused plan: revise the rules tightly, then build the two genuinely new Grade 12 tools.

60
min
1. Reset the rules

Complementary, sum, product, and the general addition rule.

60
min
2. Read the picture

Venn diagrams, tree diagrams, contingency tables.

75
min
3. Learn to count

The fundamental counting principle, arrangements, restrictions.

45
min
4. Combine counting with probability

Turn an arrangement count into an actual probability.

30
min
5. Recall under pressure

Complete the Mastery Bank, then the short self-test.

Study rule
Before touching any formula, decide one thing first: are the events mutually exclusive, independent, or neither? That single decision tells you exactly which rule to reach for.

What CAPS actually asks in Grade 12

Two genuinely new tools, built on Grade 11 probability.

  • 1

    Revise the rules. Dependent/independent events, the product rule, the sum rule, the general addition rule, and the complementary rule.

  • 2

    Solve problems using Venn diagrams, tree diagrams, two-way contingency tables, and other techniques for events that are not necessarily independent.

  • 3

    Apply the fundamental counting principle to solve probability problems.

The real new content
Two-way contingency tables and the fundamental counting principle are the two tools most students haven't met before Grade 12 — everything else on this page is revision of Grade 11 rules, applied in new combinations.

Revision: the four probability rules

One decision (mutually exclusive? independent? neither?) selects the right rule.

Complementary rule
\[P(\text{not }A)=1-P(A)\]

"Not A" and "A" always add to 1.

Sum rule (mutually exclusive)
\[P(A\text{ or }B)=P(A)+P(B)\]

A and B can never both happen — add.

Product rule (independent)
\[P(A\text{ and }B)=P(A)\times P(B)\]

A happening tells you nothing about B — multiply.

General addition rule
\[P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)\]

Works for ANY two events — subtracts the double-counted overlap.

Why subtract the overlap
Adding \(P(A)\) and \(P(B)\) directly counts anyone in both groups twice. Subtracting \(P(A\text{ and }B)\) once corrects for that — when there's no overlap at all, this subtracted term is simply zero.

Worked example: the general addition rule

The rule every other rule in this topic is secretly a special case of.

A class survey Level 1–2

In a class of 40 learners, 22 play soccer, 18 play cricket, and 10 play both. Calculate \(P(\text{soccer or cricket})\) and \(P(\text{neither})\) for a randomly chosen learner.

Show the complete working
  1. 1\(P(\text{soccer or cricket})=\dfrac{22+18-10}{40}=\dfrac{30}{40}=\boxed{0{,}75}\)
  2. 2\(P(\text{neither})=1-P(\text{soccer or cricket})=1-0{,}75=\boxed{0{,}25}\)

Revision: Venn diagrams for three events

The picture that makes the general addition rule visual.

Union of three sets
\[n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)\]

Add every set, subtract every pairwise overlap once, then add the triple overlap back — it was subtracted three times by mistake in the previous step.

Worked example: a three-event Venn diagram

First two events without a diagram, then three events with one — start the harder version with the triple overlap.

Two events, no diagram needed Level 1–2

In a school of 50 learners, 28 study Physical Sciences, 20 study Geography, and 8 study both. Find \(P(\text{neither Physical Sciences nor Geography})\).

Show the working
  1. 1\(n(\text{PS or Geo})=28+20-8=40\)
  2. 2Neither \(=50-40=10\), so \(P(\text{neither})=\dfrac{10}{50}=\boxed{0{,}2}\)
120 learners, three sports Level 3

\(n(N)=54\), \(n(A)=48\), \(n(S)=50\), \(n(N\cap A)=20\), \(n(N\cap S)=18\), \(n(A\cap S)=15\), \(n(N\cap A\cap S)=8\). Find the number playing none of the three, and the number playing exactly two.

Show the complete working
  1. 1Union \(=54+48+50-20-18-15+8=107\)
  2. 2None \(=120-107=\boxed{13}\)
  3. 3Exactly two \(=(20-8)+(18-8)+(15-8)=12+10+7=\boxed{29}\)

Revision: tree diagrams for dependent events

When one event changes the probability of the next.

Without replacement changes the denominator

Drawing an item and NOT putting it back means the total number of items drops by one for the next draw — every branch probability after the first must reflect this.

Reading a tree
Multiply ALONG a branch (a sequence of events happening in order). Add ACROSS different branches (different ways the same overall outcome could occur).

Worked example: drawing without replacement

The classic dependent-events picture.

Bag of 5 red, 3 blue Level 2

Two balls are drawn one at a time, without replacement. Calculate \(P(\text{one of each colour, in either order})\).

Show the working
  1. 1\(P(RB)+P(BR)=\dfrac{15}{56}+\dfrac{15}{56}=\dfrac{30}{56}\)
  2. 2\(\boxed{\dfrac{15}{28}}\) (add across the two different branches that both give "one of each")
Quick check: revision

Events A and B are mutually exclusive. Which rule finds \(P(A\text{ or }B)\)?

In a tree diagram, how do you combine probabilities ALONG one branch?

Two-way contingency tables

The first genuinely new Grade 12 tool: reading and testing a table.

Has diseaseNo diseaseTotal
Smoker304575
Non-smoker20105125
Total50150200
Marginal vs joint probability

A row or column TOTAL gives a marginal probability (e.g. \(P(\text{smoker})=\dfrac{75}{200}\)). A single inside CELL gives a joint probability (e.g. \(P(\text{smoker and disease})=\dfrac{30}{200}\)).

Testing independence
Two events from a table are independent only if \(P(A)\times P(B)\) exactly equals the joint cell probability \(P(A\text{ and }B)\). If they don't match, the events are NOT independent.

Worked example: testing independence from a table

First with bare probabilities, then reading the same test off a table.

Bare probabilities, no table Level 1–2

\(P(A)=0{,}3\), \(P(B)=0{,}5\), and \(P(A\text{ and }B)=0{,}15\). Are A and B independent?

Show the working
  1. 1Test: \(P(A)\times P(B)=0{,}3\times0{,}5=0{,}15\)
  2. 2This exactly matches the given \(P(A\text{ and }B)=0{,}15\), so \(\boxed{\text{A and B ARE independent}}\)
Are smoking and disease independent? Level 3

Using the table above, determine whether "being a smoker" and "having the disease" are independent events.

Show the complete working
  1. 1\(P(\text{smoker})=\dfrac{75}{200}=0{,}375\), \(P(\text{disease})=\dfrac{50}{200}=0{,}25\)
  2. 2Product: \(0{,}375\times0{,}25=0{,}09375\)
  3. 3Actual joint probability: \(P(\text{smoker and disease})=\dfrac{30}{200}=0{,}15\)
  4. 4Since \(0{,}09375\neq0{,}15\), \(\boxed{\text{the events are NOT independent}}\)

The fundamental counting principle

The second genuinely new Grade 12 tool — and the one most exam marks come from.

Multiply the choices at every stage
\[\text{Total outcomes}=n_1\times n_2\times n_3\times\cdots\]

If stage 1 has \(n_1\) possible choices, stage 2 has \(n_2\), and so on, multiply them all together for the total number of ways the whole sequence can happen.

First question to ask
Can an item be reused (repetition allowed), or does each choice remove one option from the next stage (no repetition)? This single decision changes every number in the calculation.

Worked example: repetition vs no repetition

Same scenario, one word changes everything.

Repetition allowed

A 4-digit PIN

Digits 0-9, repeats allowed:

\[10\times10\times10\times10=\boxed{10\,000}\]
No repetition

The same PIN, no repeats

Each digit removes one choice from the next position:

\[10\times9\times8\times7=\boxed{5\,040}\]

Arranging n distinct objects

A special, very common case of the counting principle.

Factorial notation
\[n!=n\times(n-1)\times(n-2)\times\cdots\times2\times1\]

Arranging \(n\) different objects in a row: the 1st position has \(n\) choices, the 2nd has \(n-1\) remaining, and so on down to 1.

Worked example: arranging people in a row

The most direct application of \(n!\).

Seven friends, one bench Level 2

In how many different ways can 7 friends be seated in a row?

Show the working
  1. 17 distinct people: \(7!=7\times6\times5\times4\times3\times2\times1\)
  2. 2\(\boxed{5\,040}\) different seating arrangements
Quick check: counting basics

A 3-digit code uses digits 0-9 with repetition allowed. How many codes are possible?

In how many ways can 4 distinct books be arranged on a shelf?

Restrictions: "together" and "not together"

The most commonly tested counting-principle pattern.

Must sit together

Glue the group into ONE block. Arrange the blocks, then multiply by the ways to arrange people WITHIN the block.

Must NOT sit together

Find the "together" count first, then subtract it from the total number of arrangements.

Worked example: two people must sit together

A small case first, then the full-size version. Treat the pair as a single unit either way.

4 people, 2 must be together Level 1–2

In how many ways can 4 people be arranged in a row if two particular people must sit next to each other?

Show the working
  1. 1Glue the pair into one block: now \(4-2+1=3\) items to arrange.
  2. 2Arrange the 3 items: \(3!=6\). Arrange the 2 people inside the block: \(2!=2\).
  3. 3Total: \(6\times2=\boxed{12}\)
6 people, 2 must be together Level 3

In how many ways can 6 people be arranged in a row if two particular people must sit next to each other?

Show the complete working
  1. 1Glue the pair into one block: now there are \(6-2+1=5\) items to arrange.
  2. 2Arrange the 5 items: \(5!=120\)
  3. 3Arrange the 2 people inside the block: \(2!=2\)
  4. 4Total: \(120\times2=\boxed{240}\)

Worked example: two people must NOT sit together

Total minus "together" — never try to count "not together" directly.

Same 6 people, now kept apart Level 3

In how many ways can the same 6 people be arranged in a row if those same two particular people must NOT sit next to each other?

Show the complete working
  1. 1Total arrangements of 6 people: \(6!=720\)
  2. 2"Together" arrangements (from the previous slide): \(240\)
  3. 3Not together \(=720-240=\boxed{480}\)
Common mistake
Never try to place the two people "apart" directly by counting gaps — it's easy to double-count or miss cases. Total minus together is always safe and always correct.

Arranging a word with repeated letters

When identical letters make some arrangements look the same.

Divide out the repeats
\[\text{Arrangements}=\dfrac{n!}{r_1!\times r_2!\times\cdots}\]

\(n\) is the total number of letters; each \(r_i!\) divides out the ways one repeated letter's copies could swap with each other without changing what the word looks like.

Worked example: repeated letters combined with probability

Just the counting skill first, then counting combined with a probability.

The word LETTER, counting only Level 1–2

How many different arrangements are there of the letters in the word LETTER?

Show the working
  1. 1LETTER has 6 letters: L, E, T, T, E, R (E repeated twice, T repeated twice).
  2. 2Arrangements: \(\dfrac{6!}{2!\,2!}=\dfrac{720}{4}=\boxed{180}\)
The word ADDRESS Level 4

The letters of the word ADDRESS are arranged in a random order. Calculate the probability that the two S's are next to each other.

Show the complete working
  1. 1ADDRESS has 7 letters: A, D, D, R, E, S, S (D repeated twice, S repeated twice).
  2. 2Total arrangements: \(\dfrac{7!}{2!\,2!}=\dfrac{5\,040}{4}=1\,260\)
  3. 3Glue the two S's into one block: now 6 items (A, D, D, R, E, [SS]), with D still repeated twice: \(\dfrac{6!}{2!}=360\)
  4. 4Probability \(=\dfrac{360}{1\,260}=\boxed{\dfrac{2}{7}}\)
Exam habit
Probability from counting is always \(\dfrac{\text{arrangements satisfying the condition}}{\text{total arrangements}}\) — get both counts exactly right, including every repeated-letter correction, before dividing.
Quick check: restrictions and repeats

To count arrangements where two specific people must sit together, the first step is to:

The word "LEVEL" has 5 letters with E repeated twice and L repeated twice. How many different arrangements are there?

Mistake clinic: repair the exact error

These are small slips, but each one can break an otherwise correct probability question.

Rule choice

Check the relationship first

Never add when you should multiply, or multiply when you should add — decide mutually exclusive / independent / neither before writing anything.

Without replacement

The denominator shrinks

After removing one item, both the numerator AND the total for the next draw must change.

Independence

Never assume it

Always test \(P(A)\times P(B)=P(A\text{ and }B)\) explicitly — don't assume independence just because a table has clean categories.

Restrictions

"Not together" is never direct

Always calculate total minus "together" — don't try to count "apart" arrangements directly.

Repeated letters

Divide out every repeat

Every letter that repeats needs its own factorial in the denominator — missing even one repeated letter changes the whole answer.

Your Paper 1 checklist

Use this before submitting an exam answer.

Avoid this

Adding when events are independent.

Assuming independence without testing it.

Counting "not together" directly.

Do this

Decide the relationship between events first.

Test independence with \(P(A)\times P(B)\) vs \(P(A\text{ and }B)\).

Use total minus "together" for every "not together" question.

Apply the complementary, sum, product and general addition rules correctly.
Solve problems using Venn diagrams (up to three events) and tree diagrams for dependent events.
Read a two-way contingency table and test statistical independence.
Apply the fundamental counting principle, with and without repetition.
Handle "together" and "not together" restrictions and repeated-letter arrangements.
Combine a counting-principle result with a probability calculation.
More explanation and exercises:Siyavula Grade 12 Fundamental Counting Principle
Summary complete

You now have the route.

Use the Mastery Bank to build fluency. Then take the test without notes and use the result to choose the exact slide to revisit.

Next step 1

Mastery Bank

17 original questions from a direct rule application through to combined counting-and-probability problems. Answers reveal only after your attempt.

Next step 2

Quick Test

A short exam-style self-check. Mark it, then revisit the exact slide that matches a missed skill.

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Learn It in Short Videos

Two free, independent videos — not made by Equation Station SA.

The Fundamental Counting Principle

A clear introduction to the flagship new Grade 12 skill.

Mindset · Probability: Fundamental Counting Principal

Counting Principle Involving Probability

Shows the counting principle combined directly with a probability question.

JB Education · The Counting Principle involving Probability

Practise in the right order

The core teaching is above. These are the next steps, not a replacement for it.

01
Built-in practice
Probability Mastery Bank

17 original questions by skill, with concise reveal answers and methods.

Start after the slides
Open Mastery Bank
02
Built-in check
Test Your Knowledge

Use the short exam-style self-check when you want a fast confidence check.

Then target one weak skill
Take the Quick Test
CAPS
Free textbook chapter
Siyavula: Grade 12 Probability

Use its full counting-principle and probability sequence for extra explanation and exercises.

Free • CAPS aligned
Open Siyavula
DBE
Official free books
DBE Grade 12 Textbooks

Official state-owned learner books and teacher support for Grade 12 Mathematics.

Official • free access
Open DBE Books
NSC
Official past papers
DBE NSC Examination Archive

Use official papers after the course and original practice are secure.

Past papers • memos
Open DBE Archive
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Related Topics

These skills strengthen the reasoning underneath every probability question.

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Algebra

Rearranging equations and fractions

Study this →
Σ

Statistics

Data, distributions and interpretation

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Σ

Patterns & Sequences

Systematic counting and series

Study this →
Student Questions

Frequently Asked Questions

Short answers for the checks learners make while preparing for the Grade 12 CAPS exam.

What is genuinely new in Grade 12 Probability?

Two-way contingency tables as a technique for testing independence, and the fundamental counting principle (systematic counting of arrangements, including restrictions), which then combines with probability itself.

How do I know whether to add or multiply two probabilities?

Add for "or" when events are mutually exclusive (or use the general addition rule for any two events). Multiply for "and" when events are independent, or follow a tree diagram's branches for dependent events.

How do I test if two events from a table are independent?

Calculate \(P(A)\times P(B)\) and compare it to the actual joint probability \(P(A\text{ and }B)\) read from the table. They must be exactly equal for the events to be independent.

How do I handle a "must not sit together" question?

Never count it directly. Calculate the total number of arrangements, calculate the "together" arrangements, then subtract.

Where should I practise next?

Finish the interactive slides, open the Mastery Bank, then take the short self-test without notes.