What is not reducing counts without replacement?

After each draw without replacement, both the favourable and total counts decrease.

If P(A) = 0.3, find P(A'). [1 mark]

Events A and B are mutually exclusive with P(A) = 0.2, P(B) = 0.5. Find P(A B). [2 marks]

SEAB Original-style practice: Events A and B satisfy P(A) = 0.5, P(B) = 0.4 and P(A B) = 0.2. Find P(A B) and P(A' B').

Addition rule: P(A B) = P(A) + P(B) - P(A B) = 0.5 + 0.4 - 0.2 = 0.7. A' B' is the complement of A B (by De Morgan), so P(A' B') = 1 - P(A B) = 1 - 0.7 = 0.3. Markers reward the addition rule, recognising A' B' = (A B)', and both values.

SingaporeMathsSyllabus dot point

How do we calculate probabilities of combined events using the basic rules?

Use the probability rules for the complement, union and intersection of events, and apply Venn diagrams and tree diagrams to combined events

A focused answer to the H2 Mathematics outcome on probability rules. The complement, addition and multiplication rules, mutually exclusive events, and using Venn and tree diagrams for combined events.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

SEAB wants you to use the fundamental probability rules - the complement, the addition rule for unions, and the multiplication rule for intersections - and to model combined events with Venn diagrams and tree diagrams.

The answer

Basic definitions

Probability measures likelihood on a scale from $0$ (impossible) to $1$ (certain). For equally likely outcomes, $\mathrm{P}(A) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}$ .

The complement rule

The complement $A'$ is " $A$ does not happen":

\mathrm{P}(A') = 1 - \mathrm{P}(A).

This is the key to "at least one" problems, where the complement "none" is easier.

The addition rule

For the union (either event):

\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B).

Subtracting the intersection avoids double counting the overlap. If $A$ and $B$ are mutually exclusive ( $A \cap B = \varnothing$ ), the rule reduces to $\mathrm{P}(A) + \mathrm{P}(B)$ .

The multiplication rule

For the intersection (both events):

\mathrm{P}(A \cap B) = \mathrm{P}(A)\,\mathrm{P}(B \mid A),

the probability of $A$ times the probability of $B$ given $A$ . This is the rule you follow along a tree-diagram branch.

Diagrams

A Venn diagram shows overlapping regions, ideal for union, intersection and complement problems.
A tree diagram shows sequential events; multiply along branches and add across the branches that satisfy the event.

Worked example

A box has $5$ defective and $15$ good items. Three are drawn without replacement. Find the probability that at least one is defective.

Step 1: Use the complement

"At least one defective" is the complement of "none defective" (all three good).

Step 2: Probability all three are good

\mathrm{P}(\text{all good}) = \frac{15}{20} \times \frac{14}{19} \times \frac{13}{18}

multiplying along the tree branch with reducing counts.

Step 3: Compute

= \frac{15 \times 14 \times 13}{20 \times 19 \times 18} = \frac{2730}{6840} = \frac{91}{228}

Step 4: Apply the complement

\mathrm{P}(\text{at least one defective}) = 1 - \frac{91}{228} = \frac{137}{228} \approx 0.601

Examples in context

Example 1. Quality control. A factory uses the complement rule to find the chance that a batch has at least one faulty unit, the practically important figure, by computing one minus the chance every unit passes.

Example 2. Medical screening. Sequential test results are modelled on a tree diagram, multiplying along branches for the joint probability of a result pattern, the foundation of interpreting screening outcomes.

Try this

Q1. If $\mathrm{P}(A) = 0.3$ , find $\mathrm{P}(A')$ . [1 mark]

Cue. $1 - 0.3 = 0.7$ .

Q2. Events $A$ and $B$ are mutually exclusive with $\mathrm{P}(A) = 0.2$ , $\mathrm{P}(B) = 0.5$ . Find $\mathrm{P}(A \cup B)$ . [2 marks]

Cue. $0.2 + 0.5 = 0.7$ (no overlap to subtract).

Q3. Explain why "at least one" problems are often solved by the complement. [2 marks]

Cue. The complement "none" is a single product, simpler than summing the many ways of getting one or more.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksEvents

A

and

B

satisfy

\mathrm{P}(A) = 0.5

\mathrm{P}(B) = 0.4

and

\mathrm{P}(A \cap B) = 0.2

. Find

\mathrm{P}(A \cup B)

and

\mathrm{P}(A' \cap B')

Show worked answer →

Addition rule: $\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B) = 0.5 + 0.4 - 0.2 = 0.7$ .

$A' \cap B'$ is the complement of $A \cup B$ (by De Morgan), so $\mathrm{P}(A' \cap B') = 1 - \mathrm{P}(A \cup B) = 1 - 0.7 = 0.3$ .

Markers reward the addition rule, recognising $A' \cap B' = (A \cup B)'$ , and both values.

Original4 marksA bag has

4

red and

6

blue balls. Two are drawn without replacement. Find the probability that both are red.