P(A) = 0.5, P(B) = 0.4, P(A and B) = 0.1. Find P(A or B). [2 marks]

ScotlandApplications of MathematicsSyllabus dot point

How do you calculate probabilities of combined events and the expected value of an uncertain outcome?

Calculating probabilities of single and combined events using the addition and multiplication rules and tree diagrams, working with conditional probability, and finding the expected value of a situation with uncertain outcomes.

A focused answer to the SQA Higher Applications of Mathematics probability content, covering basic probability, combining events with the addition and multiplication rules, tree diagrams, conditional probability, and calculating expected value.

Generated by Claude Opus 4.89 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
Basic probability
Combining events
Tree diagrams and conditional probability
Expected value
Why this matters
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What this dot point is asking

The SQA wants you to calculate probabilities of single and combined events, use the addition and multiplication rules and tree diagrams, handle conditional probability (with and without replacement), and find the expected value of an uncertain situation. Expected value links directly to the decision-making area of the course.

Basic probability

The probability of an event is the proportion of times it is expected to happen, between $0$ (impossible) and $1$ (certain). For equally likely outcomes, $P(\text{event}) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}$ . The probability that an event does not happen is $1 - P(\text{event})$ .

Combining events

Two rules combine probabilities of events.

"And" means both happen, so you multiply; "or" means at least one happens, so you add and subtract the overlap to avoid double counting. Events are independent if one does not affect the other (for example separate dice rolls).

Tree diagrams and conditional probability

A tree diagram lays out multi-stage events. Each branch carries a probability; you multiply along a path to find the probability of that sequence and add the probabilities of all paths that give the required outcome.

A two-stage tree without replacement

A box has $4$ working and $2$ faulty bulbs. Two are taken without replacement. Find the probability that both are working.

Step 1: Identify the first-stage probabilities

There are $6$ bulbs in total. At the first draw, $P(\text{working}) = \dfrac{4}{6}$ and $P(\text{faulty}) = \dfrac{2}{6}$ . These two branches must sum to $1$ , which is a useful check.

Step 2: Condition the second stage on the first outcome

Because the bulbs are drawn without replacement, removing one changes the contents of the box. If the first draw was working, there are now $3$ working bulbs left among $5$ remaining, so the second-stage probability is conditional:

P(\text{working second} \mid \text{working first}) = \dfrac{3}{5}

This is why the denominator drops from $6$ to $5$ : one item has been removed.

Step 3: Multiply along the path to find the combined probability

Along the "working then working" path, multiply the branch probabilities together:

P(\text{both working}) = \dfrac{4}{6} \times \dfrac{3}{5} = \dfrac{12}{30} = \dfrac{2}{5}

Final answer: The probability that both bulbs drawn are working is $\dfrac{2}{5}$ .

Expected value

The expected value is the long-run average outcome of a situation with uncertain results. It weights each possible value by its probability.

Expected value is the foundation of decision making under risk: a decision with a higher expected value is, on average, the better choice over many repetitions. A negative expected profit means a long-run loss.

Find an expected value

A spinner pays $\pounds 10$ with probability $0.1$ , $\pounds 4$ with probability $0.3$ , and $\pounds 0$ with probability $0.6$ . Find the expected return per spin.

Step 1: Understand what the expected value measures

The expected value is the long-run average: if you spun the wheel many times, the average payout per spin would approach this number. It is calculated by weighting each prize by how likely it is to occur.

Step 2: Check the probabilities sum to 1

$0.1 + 0.3 + 0.6 = 1.0$ . This confirms the three outcomes are exhaustive, so nothing has been missed.

Step 3: Apply the expected value formula

Multiply each prize by its probability and add:

E = 10 \times 0.1 + 4 \times 0.3 + 0 \times 0.6 = 1 + 1.2 + 0 = \pounds 2.20

Final answer: The expected return per spin is $\pounds 2.20$ . Paying more than $\pounds 2.20$ to play would produce a long-run loss; paying less would be profitable over many plays.

Why this matters

Probability quantifies uncertainty, and expected value turns that uncertainty into a single comparable number. The course uses these tools in insurance and risk in the finance area, and in decision tables in the planning area, so a firm grasp here pays off across the course.

Try this

Q1. A fair die is rolled twice. Find the probability of two sixes. [2 marks]

Cue. Independent, so $\dfrac{1}{6} \times \dfrac{1}{6} = \dfrac{1}{36}$ .

Q2. $P(A) = 0.5$ , $P(B) = 0.4$ , $P(A \text{ and } B) = 0.1$ . Find $P(A \text{ or } B)$ . [2 marks]

Cue. $0.5 + 0.4 - 0.1 = 0.8$ .

Q3. A raffle ticket costs $\pounds 1$ . It wins $\pounds 50$ with probability $0.01$ and nothing otherwise. Find the expected profit per ticket. [2 marks]

Cue. Expected winnings $= 50 \times 0.01 = \pounds 0.50$ ; expected profit $= 0.50 - 1 = -\pounds 0.50$ .

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.

SQA Higher Apps style5 marksA bag holds

5

red and

3

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

Show worked answer →

First draw: $P(\text{red}) = \dfrac{5}{8}$ , $P(\text{blue}) = \dfrac{3}{8}$ . Without replacement the second draw depends on the first (1 mark for the tree structure).

$P(\text{both red}) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$ (2 marks).

Different colours means red then blue or blue then red: $\dfrac{5}{8} \times \dfrac{3}{7} + \dfrac{3}{8} \times \dfrac{5}{7} = \dfrac{15}{56} + \dfrac{15}{56} = \dfrac{30}{56} = \dfrac{15}{28}$ (2 marks). Markers reward correct conditional second-draw probabilities and adding both mixed orders.

SQA Higher Apps style4 marksA game costs

\pounds 2

to play. You win

\pounds 5

with probability

0.3

and nothing otherwise. Find the expected profit per game and state whether the game is worth playing in the long run.

Show worked answer →

The expected winnings are $E = 5 \times 0.3 + 0 \times 0.7 = \pounds 1.50$ per game (2 marks).

Subtracting the $\pounds 2$ cost, the expected profit is $1.50 - 2.00 = -\pounds 0.50$ per game (1 mark).

Since the expected profit is negative, on average you lose $50$ pence per game, so it is not worth playing in the long run (1 mark). Markers reward the expected winnings calculation, subtracting the cost, and a conclusion based on the sign of the expected profit.

Related dot points

Sources & how we know this

Higher Applications of Mathematics Course Specification — SQA (2023)
Higher Applications of Mathematics - Course overview — SQA (2025)