Given P(A B) = 0.12 and P(B) = 0.3, find P(A B). [2 marks]

Events have P(A) = 0.5, P(B) = 0.4, P(A B) = 0.2. Are they independent? [2 marks]

Write the general multiplication rule for P(A B). [1 mark]

CCEA CCEA 2020-style practice: In a class, P(A) = 0.6, P(B) = 0.5 and P(A B) = 0.2. Find P(A B) and determine whether A and B are independent.

The conditional probability is P(A B) = (P(A B))/(P(B)) = (0.2)/(0.5) = 0.4. For independence, check whether P(A B) = P(A) × P(B): P(A) × P(B) = 0.6 × 0.5 = 0.3, but P(A B) = 0.2 0.3. Since they are unequal, A and B are **not** independent. Markers reward the conditional formula, the value 0.4, the independence test, and the correct conclusion.

Northern IrelandMathsSyllabus dot point

How does knowing one event has happened change the probability of another?

Conditional probability, the conditional probability formula, the multiplication rule for dependent events, the test for independence, and using tree diagrams and two-way tables.

A CCEA A-Level Mathematics answer on conditional probability, the conditional probability formula, the general multiplication rule for dependent events, the formal test for independence, and using tree diagrams and two-way tables to find conditional probabilities.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

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

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

CCEA wants you to understand conditional probability, use the conditional probability formula, apply the general multiplication rule for dependent events, test formally for independence, and read conditional probabilities from tree diagrams and two-way tables. This builds on the AS probability rules and is essential for interpreting real data.

The answer

Conditional probability

The general multiplication rule

The test for independence

Tree diagrams and two-way tables

A tree diagram shows conditional probabilities directly: the second-stage branches are conditioned on the first-stage outcome, which is why they change for sampling without replacement. A two-way table cross-classifies data, and conditional probabilities are read by restricting to the relevant row or column total. The skill in a table question is identifying which total is the denominator: conditioning on a category means dividing by that category's total, not by the grand total of everyone.

Worked example: a conditional from a table

Examples in context

Example 1. Medical test interpretation. The probability of having a disease given a positive test, $P(\text{disease} \mid \text{positive})$ , differs greatly from $P(\text{positive} \mid \text{disease})$ , which is why test results must be interpreted with conditional probability and the base rate. Confusing the two directions is a famous real-world error.

Example 2. Sampling without replacement. Drawing cards without replacement makes each draw depend on the last, so the tree's later branches use conditional probabilities. The general multiplication rule, not the independent product, gives the correct chance of a sequence.

Try this

Q1. Given $P(A \cap B) = 0.12$ and $P(B) = 0.3$ , find $P(A \mid B)$ . [2 marks]

Cue. $\frac{0.12}{0.3} = 0.4$ .

Q2. Events have $P(A) = 0.5$ , $P(B) = 0.4$ , $P(A \cap B) = 0.2$ . Are they independent? [2 marks]

Cue. $P(A)P(B) = 0.2 = P(A \cap B)$ , so yes, independent.

Q3. Write the general multiplication rule for $P(A \cap B)$ . [1 mark]

Cue. $P(A \cap B) = P(B)P(A \mid B)$ .

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 20206 marksIn a class,

P(A) = 0.6

P(B) = 0.5

and

P(A \cap B) = 0.2

. Find

P(A \mid B)

and determine whether

A

and

B

are independent.

Show worked answer →

The conditional probability is $P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.5} = 0.4$ .

For independence, check whether $P(A \cap B) = P(A) \times P(B)$ :

$P(A) \times P(B) = 0.6 \times 0.5 = 0.3,$ but $P(A \cap B) = 0.2 \ne 0.3$ .

Since they are unequal, $A$ and $B$ are not independent.

Markers reward the conditional formula, the value $0.4$ , the independence test, and the correct conclusion.

CCEA 20196 marksA bag has 6 red and 4 green counters. Two are drawn without replacement. Find the probability that the second is green given that the first is red.

Show worked answer →

After drawing a red first, $9$ counters remain: $5$ red and $4$ green.

So $P(\text{2nd green} \mid \text{1st red}) = \frac{4}{9}$ .

Equivalently, using the formula with $P(\text{1st red}) = \frac{6}{10}$ and $P(\text{1st red and 2nd green}) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90}$ , the conditional probability is $\frac{24/90}{6/10} = \frac{4}{9}$ .

Markers reward recognising the reduced sample after the first draw, and the conditional probability $\frac{4}{9}$ .

Related dot points

Sources & how we know this

CCEA GCE Mathematics specification — CCEA (2018)