Probability for Matric Maths

Venn diagrams, tree diagrams, dependent/independent events and counting principles for NSC Paper 1. CAPS-aligned with AI explanations.

Probability in Paper 1 covers Venn diagrams, tree diagrams, mutually exclusive and independent events, conditional probability, and the fundamental counting principle including arrangements. Questions are often set in a familiar context, such as a survey of learners, a sports team or number plates, and the first step is to organise the information. A Venn diagram suits overlapping events, a two-way contingency table suits survey data, and a tree diagram suits events that happen one after another. Counting-principle questions ask how many arrangements are possible, often with a condition such as letters that must stay together or digits that may not repeat, and then use that count to find a probability.

Key concepts

  • P(A or B) = P(A) + P(B) − P(A and B)
  • Independent events: P(A and B) = P(A) · P(B)
  • Tree diagrams for sequential events
  • Fundamental counting principle and arrangements
  • Mutually exclusive events: P(A and B) = 0
  • Complementary events: P(not A) = 1 − P(A)
  • Two-way contingency tables to test for independence
  • Arrangements with n! and with conditions (items together or in fixed positions)

Worked examples

P(A) = 0.6, P(B) = 0.5, P(A and B) = 0.3. Are A and B independent?

Independent ⇔ P(A and B) = P(A) · P(B) = 0.6 · 0.5 = 0.3. Yes, independent.

P(A) = 0.4, P(B) = 0.35 and A and B are mutually exclusive. Find P(A or B) and P(not A).

Mutually exclusive means P(A and B) = 0, so P(A or B) = 0.4 + 0.35 = 0.75. P(not A) = 1 − 0.4 = 0.6.

A bag holds 3 red and 2 blue balls. Two balls are drawn without replacement. Find P(both red).

P(first red) = 3/5. One red ball is gone, so P(second red) = 2/4. P(both red) = 3/5 × 2/4 = 6/20 = 3/10.

In how many ways can the letters of NUMBER be arranged if U and E must be next to each other?

Treat UE as one block, giving 5 items to arrange: 5! = 120. The block can be UE or EU, so multiply by 2: 240 arrangements.

Six friends sit in a row. What is the probability that Thabo and Lerato sit next to each other?

Total arrangements: 6! = 720. Keep the pair together as one block: 5! × 2 = 240. P = 240/720 = 1/3.

Common mistakes

  • Confusing 'mutually exclusive' with 'independent'.
  • Adding probabilities of intersecting events without subtracting the overlap.
  • Treating draws without replacement as independent events.
  • Forgetting to multiply by the internal arrangements of a block of items that must stay together.
  • Using n! when only some of the items are arranged.

Exam tips

  • Draw a Venn diagram even if the question doesn't ask for one — it prevents arithmetic errors.
  • For 'at least one' questions, calculate 1 − P(none).
  • Work out the total number of arrangements first: it is the denominator of most counting-principle probabilities.

In past papers

  • NSC Nov 2024 Paper 1 Q11 — counting principle

Frequently asked questions

How many marks is probability worth?

Roughly 15 marks in NSC Paper 1.

What is the difference between mutually exclusive and independent events?

Mutually exclusive events cannot happen together, so P(A and B) = 0. Independent events do not affect each other, so P(A and B) = P(A) × P(B). Two events with non-zero probabilities cannot be both.

When should I use a tree diagram?

When events happen in stages, such as two draws from a bag or two matches in a row. Multiply along each branch, then add the branches that satisfy the question.

How do I handle items that must stay together in an arrangement?

Treat the group as a single item, arrange all the items, then multiply by the number of ways the group can be arranged internally.

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