SQA AH style: conditional-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 state whether A and B are independent.

Conditional probability: P(A B)=P(A B)P(B)=0.20.4=0.5 (2 marks). For independence, check whether P(A B)=P(A)P(B): here P(A)P(B)=0.5× 0.4=0.2, which equals P(A B) (1 mark). Since they are equal, A and B are independent; equivalently P(A B)=P(A)=0.5. Markers reward the conditional formula, the correct value and the independence check.

ScotlandStatisticsSyllabus dot point

How do you calculate probabilities for combined, conditional and independent events?

Apply the addition and multiplication laws of probability, calculate conditional probabilities and use tree diagrams, the total probability rule and Bayes' theorem, and test events for independence and mutual exclusivity.

A focused answer to the SQA Advanced Higher Statistics probability content: the addition and multiplication laws, conditional probability, independence and mutual exclusivity, tree diagrams, the total probability rule and Bayes' theorem for reversing a conditional probability.

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

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

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What this dot point is asking
The addition and multiplication laws
Conditional probability and independence
Tree diagrams and the total probability rule
Bayes' theorem
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What this dot point is asking

Probability is the language the rest of Advanced Higher Statistics is written in: every distribution, confidence interval and hypothesis test rests on it. The SQA wants you to combine probabilities with the addition and multiplication laws, to work with conditional probability, to use tree diagrams and the total probability rule, and to reverse a conditional probability with Bayes' theorem.

The addition and multiplication laws

Two laws build every compound probability.

The subtraction in the addition law removes the overlap $P(A\cap B)$ that would otherwise be counted twice. If the events are mutually exclusive then $P(A\cap B)=0$ and the law simplifies.

Mutual exclusivity and independence are different ideas; two events with non-zero probabilities cannot be both mutually exclusive and independent, because exclusivity forces $P(A\cap B)=0$ while independence needs $P(A\cap B)=P(A)P(B)>0$ .

Conditional probability and independence

Conditioning restricts attention to the outcomes in which an event has occurred.

Find a conditional probability and test independence

Conditional probability restricts attention to outcomes in which event $C$ has already occurred. Once we compute $P(D\mid C)$ , we test for independence by checking whether knowing $C$ happened changes the probability of $D$ .

Step 1: Write down the conditional probability formula and identify the values

The formula is $P(D\mid C)=\dfrac{P(C\cap D)}{P(C)}$ . From the given information: $P(C)=0.6$ , $P(D)=0.3$ , $P(C\cap D)=0.18$ .

Step 2: Compute the conditional probability

Divide the joint probability by the probability of the conditioning event.

P(D\mid C)=\dfrac{P(C\cap D)}{P(C)}=\dfrac{0.18}{0.6}=0.3

Step 3: Test for independence by comparing with the unconditional probability

Events $C$ and $D$ are independent if and only if $P(D\mid C)=P(D)$ , equivalently $P(C\cap D)=P(C)P(D)$ . Here, $P(D\mid C)=0.3=P(D)$ , and indeed $P(C)P(D)=0.6\times 0.3=0.18=P(C\cap D)$ .

Final answer: $P(D\mid C)=0.3$ ; the events $C$ and $D$ are independent because conditioning on $C$ does not change the probability of $D$ .

Tree diagrams and the total probability rule

A tree diagram lays out a sequence of stages, with the branches at each stage labelled by conditional probabilities; multiply along a path and add across paths.

Bayes' theorem

Bayes' theorem reverses a conditional probability: it turns $P(B\mid A)$ into $P(A\mid B)$ using the total probability of $B$ as the denominator.

The striking lesson is that when an event is rare, even a positive test result can leave the probability of the cause low, because the many false positives from the large healthy group swamp the few true positives. This is why the disease example above gives only about 16%.

Try this

Q1. $A$ and $B$ are mutually exclusive with $P(A)=0.35$ and $P(B)=0.25$ . Find $P(A\cup B)$ . [1 mark]

Cue. Mutually exclusive means $P(A\cap B)=0$ , so $P(A\cup B)=0.35+0.25=0.6$ .

Q2. A fair coin is tossed three times. Find the probability of at least one head. [2 marks]

Cue. Use the complement: $P(\text{at least one head})=1-P(\text{no heads})=1-\left(\tfrac{1}{2}\right)^3=1-\tfrac{1}{8}=\tfrac{7}{8}$ .

Exam-style practice questions

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

AH style: conditional3 marksEvents

A

and

B

satisfy

P(A)=0.5

P(B)=0.4

and

P(A\cap B)=0.2

. Find

P(A\mid B)

and state whether

A

and

B

are independent.

Show worked answer →

Conditional probability: $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.2}{0.4}=0.5$ (2 marks).

For independence, check whether $P(A\cap B)=P(A)P(B)$ : here $P(A)P(B)=0.5\times 0.4=0.2$ , which equals $P(A\cap B)$ (1 mark). Since they are equal, $A$ and $B$ are independent; equivalently $P(A\mid B)=P(A)=0.5$ . Markers reward the conditional formula, the correct value and the independence check.

AH style: Bayes4 marksA test for a disease is 95% sensitive and 90% specific. The disease affects 2% of a population. A randomly chosen person tests positive. Find the probability they actually have the disease.

Show worked answer →

Let $D$ be "has the disease" and $+$ be "tests positive". Given $P(D)=0.02$ , $P(+\mid D)=0.95$ , $P(+\mid D')=1-0.90=0.10$ (1 mark).

Total probability of a positive test: $P(+)=P(+\mid D)P(D)+P(+\mid D')P(D')=0.95\times 0.02+0.10\times 0.98=0.019+0.098=0.117$ (2 marks).

Bayes' theorem: $P(D\mid +)=\dfrac{P(+\mid D)P(D)}{P(+)}=\dfrac{0.019}{0.117}\approx 0.162$ (1 mark). Only about 16% of positives are true cases because the disease is rare. Markers reward the correct conditional probabilities, the total-probability denominator and the final Bayes calculation.

Related dot points

Sources & how we know this

SQA Advanced Higher Statistics Course Specification (C803 77) — SQA (2023)
SQA Advanced Higher Statistics Data Booklet — SQA (2019)