P(A) = 0.7, P(A B) = 0.28. Find P(B A). [2 marks]

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How do you calculate probabilities for combined events, including mutually exclusive, independent and conditional cases?

Probability of events, mutually exclusive and independent events, Venn diagrams, tree diagrams and two-way tables, the addition and multiplication laws, and conditional probability.

A focused answer to the OCR A-Level Mathematics A probability content, covering the probability of events, mutually exclusive and independent events, Venn diagrams, tree diagrams and two-way tables, the addition and multiplication laws, and conditional probability with the conditional formula.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

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

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Quick answer

Probabilities lie between $0$ and $1$ , and $P(A') = 1 - P(A)$ . The addition law is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ , reducing to a simple sum for mutually exclusive events. Independent events satisfy $P(A \cap B) = P(A)P(B)$ ; mutually exclusive (cannot co-occur) is a different idea. Conditional probability is $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ , and events are independent when $P(A \mid B) = P(A)$ . Use Venn diagrams and two-way tables for overlaps, and tree diagrams for sequences (multiply along branches, add across paths, and change the probabilities for draws without replacement).

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What this dot point is asking
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What this dot point is asking

OCR wants you to calculate probabilities of events and their complements, recognise and use mutually exclusive and independent events, work with Venn diagrams, tree diagrams and two-way tables, apply the addition and multiplication laws, and compute and interpret conditional probabilities, including testing independence and using the conditional probability formula.

The answer

Basic probability and complements

For equally likely outcomes, $P(A) = \dfrac{\text{number of favourable outcomes}}{\text{total outcomes}}$ , and every probability lies between $0$ and $1$ . The complement rule says $P(A') = 1 - P(A)$ , which often turns an awkward "at least one" into an easy "none".

The addition and multiplication laws

Mutually exclusive versus independent

These are different ideas. Mutually exclusive events cannot occur together (rolling a $2$ and a $5$ on one die). Independent events do not affect each other's probability (two separate coin tosses). Mutually exclusive events with non-zero probabilities are never independent, because if one happens the other cannot.

Conditional probability

Conditional probability is the chance of $A$ given that $B$ has already happened.

Examples in context

Venn diagrams and two-way tables

A Venn diagram makes overlaps visible: fill the intersection first, then the rest of each set, then anything outside. A two-way table organises counts by two categories, and probabilities are read as the relevant cell over the relevant total (the row or column total for a conditional probability).

Tree diagrams for sequences

A tree diagram handles a sequence of stages: multiply along the branches for a path, add across paths for an "or". For drawing without replacement, the second-stage probabilities change because the total and the counts have decreased.

A test is

95\%

accurate;

2\%

of people have a condition. Find the probability a positive test is correct

step 1 Set up the paths

$P(\text{condition}) = 0.02$ , $P(+ \mid \text{condition}) = 0.95$ ; $P(\text{no condition}) = 0.98$ , $P(+ \mid \text{no condition}) = 0.05$ .

step 2 Find the two ways to test positive

True positive $= 0.02 \times 0.95 = 0.019$ . False positive $= 0.98 \times 0.05 = 0.049$ .

step 3 Apply the conditional formula

$P(\text{condition} \mid +) = \dfrac{0.019}{0.019 + 0.049} = \dfrac{0.019}{0.068} \approx 0.28$ .

Try this

Q1. Events $A$ and $B$ are independent with $P(A) = 0.3$ and $P(B) = 0.4$ . Find $P(A \cap B)$ . [2 marks]

Cue. $P(A \cap B) = 0.3 \times 0.4 = 0.12$ .

Q2. $P(A) = 0.7$ , $P(A \cap B) = 0.28$ . Find $P(B \mid A)$ . [2 marks]

Cue. $P(B \mid A) = \dfrac{0.28}{0.7} = 0.4$ .

Exam-style practice questions

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

OCR 20196 marksFor two events

A

and

B

P(A) = 0.5

P(B) = 0.4

and

P(A \cup B) = 0.7

. Find

P(A \cap B)

, and determine whether

A

and

B

are independent. Find

P(A \mid B)

Show worked answer →

Use the addition law $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ (M1): $0.7 = 0.5 + 0.4 - P(A \cap B)$ , so $P(A \cap B) = 0.2$ (A1).

For independence check $P(A)P(B) = 0.5 \times 0.4 = 0.2$ (M1). Since this equals $P(A \cap B) = 0.2$ , the events are independent (A1).

Conditional probability (M1): $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.2}{0.4} = 0.5$ (A1).

Markers reward the addition law, the independence test by comparing $P(A)P(B)$ with $P(A \cap B)$ , and the conditional formula.

OCR 20216 marksA bag contains

5

red and

3

blue counters. Two are drawn without replacement. Draw a tree diagram and find the probability that the two counters are different colours, and the probability that the second is red given the first is blue.

Show worked answer →

First draw: $P(R) = \dfrac{5}{8}$ , $P(B) = \dfrac{3}{8}$ . Second draw probabilities depend on the first (without replacement), giving the branch values $\dfrac{4}{7}, \dfrac{3}{7}$ after red and $\dfrac{5}{7}, \dfrac{2}{7}$ after blue (M1, A1).

Different colours $= P(RB) + P(BR) = \dfrac{5}{8}\cdot\dfrac{3}{7} + \dfrac{3}{8}\cdot\dfrac{5}{7} = \dfrac{15}{56} + \dfrac{15}{56} = \dfrac{30}{56} = \dfrac{15}{28}$ (M1, A1).

Second red given first blue is read straight off the blue branch (M1): $P(R \mid B) = \dfrac{5}{7}$ (A1).

Markers reward the tree with correct without-replacement probabilities, summing the two mixed-colour paths, and reading the conditional branch.

Related dot points

Sources & how we know this

OCR A Level Mathematics A (H240) specification — OCR (2017)