What is not reducing for without replacement?

After removing an item, both the favourable count and the total fall, so update the second fraction.

CCEA CCEA Unit 3 (style)-style practice: Events A and B have P(A) = 0.4, P(B) = 0.5 and P(A B) = 0.2. Find P(A B) and state whether A and B are independent.

Use the addition law: P(A B) = P(A) + P(B) - P(A B) = 0.4 + 0.5 - 0.2 = 0.7. Test independence: A and B are independent if P(A B) = P(A) × P(B). Here P(A) × P(B) = 0.4 × 0.5 = 0.2, which equals P(A B). So A and B are independent. Marks are for the addition law, the value, the independence test, and the conclusion.

Northern IrelandFurther MathsSyllabus dot point

How do you use the addition and multiplication laws and conditional probability to solve probability problems?

Calculate probabilities: use the addition and multiplication laws, mutually exclusive and independent events, tree diagrams, and conditional probability.

A CCEA GCSE Further Mathematics answer on probability, covering the addition and multiplication laws, mutually exclusive and independent events, tree diagrams, and conditional probability in the Statistics unit.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-14

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

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What this dot point is asking
The addition law
The multiplication law and independence
Tree diagrams
Conditional probability
Why this matters

What this dot point is asking

Probability measures how likely an event is, and CCEA GCSE Further Mathematics develops it beyond ordinary GCSE to conditional probability. You must use the addition law for combined events, the multiplication law for events in sequence, recognise mutually exclusive and independent events, use tree diagrams for multi-stage experiments, and calculate conditional probabilities where one event depends on another. These ideas underpin the distribution work that follows in the unit.

The addition law

The addition law finds the probability that one event or another (or both) happens. You add the separate probabilities but subtract the overlap, because the intersection would otherwise be counted twice.

The multiplication law and independence

The multiplication law finds the probability that two events both happen, one after the other. In general the second probability is conditional on the first, but for independent events the first has no effect on the second.

So for independent events $P(A \cap B) = P(A) \times P(B)$ , while for dependent events you must use the reduced probability for the second event.

Tree diagrams

A tree diagram lays out a multi-stage experiment branch by branch, with each branch carrying its probability. You multiply along a path to find the probability of that complete outcome, and add the probabilities of all the paths that satisfy the question.

Tree diagram without replacement

A box has $4$ green and $2$ yellow sweets. Two are eaten one after the other. Find the probability that exactly one is green.

step 1 Set up the first-stage probabilities

First sweet green: $\dfrac{4}{6}$ ; first sweet yellow: $\dfrac{2}{6}$ .

step 2 Set up the second-stage probabilities

If the first was green, $3$ green and $2$ yellow remain; if yellow, $4$ green and $1$ yellow remain. The denominators drop to $5$ .

step 3 Identify the routes for exactly one green

Green then yellow: $\dfrac{4}{6} \times \dfrac{2}{5} = \dfrac{8}{30}$ . Yellow then green: $\dfrac{2}{6} \times \dfrac{4}{5} = \dfrac{8}{30}$ .

step 4 Add the routes

$\dfrac{8}{30} + \dfrac{8}{30} = \dfrac{16}{30} = \dfrac{8}{15}$ . "Exactly one" covers both orders, so both routes count.

Conditional probability

Conditional probability is the probability of one event given that another has already happened, which narrows the sample space to the cases where the condition holds. It is found by dividing the probability of both events by the probability of the condition. For example, if $P(A \cap B) = 0.2$ and $P(B) = 0.5$ , then $P(A \mid B) = \dfrac{0.2}{0.5} = 0.4$ . Conditional probability is exactly what the second branches of a without-replacement tree diagram represent, so the two ideas are closely linked.

Why this matters

Probability is the foundation of the whole Statistics unit. The multiplication law for independent trials leads directly to the binomial distribution, the addition and conditional laws appear in interpreting distributions, and the careful reasoning about with or without replacement trains the precision that statistics demands. Tree diagrams and conditional probability also reward clear, structured working that secures method marks.

Exam-style practice questions

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

CCEA Unit 3 (style)4 marksA bag has

5

red and

3

blue counters. Two are taken without replacement. Find the probability that both are red.

Show worked answer →

The first counter is red with probability $\dfrac{5}{8}$ .

After removing a red, $4$ red and $3$ blue remain, so the second is red with probability $\dfrac{4}{7}$ .

Multiply along the branch (events in sequence): $\dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$ .

Marks are for both fractions with the reduced denominator after removal, and for multiplying. The common error is using $\dfrac{5}{8}$ twice, which would be sampling with replacement.

CCEA Unit 3 (style)4 marksEvents

A

and

B

have

P(A) = 0.4

P(B) = 0.5

and

P(A \cap B) = 0.2

. Find

P(A \cup B)

and state whether

A

and

B

are independent.

Show worked answer →

Use the addition law: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.5 - 0.2 = 0.7$ .

Test independence: $A$ and $B$ are independent if $P(A \cap B) = P(A) \times P(B)$ . Here $P(A) \times P(B) = 0.4 \times 0.5 = 0.2$ , which equals $P(A \cap B)$ .

So $A$ and $B$ are independent. Marks are for the addition law, the value, the independence test, and the conclusion.

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Sources & how we know this

CCEA GCSE Further Mathematics specification (2330) — CCEA (2017)