Expected Outcomes in Probability Experiments Revision Notes | GCSE AQA Higher Maths

1. Why Expected Outcomes Matter

You roll a fair die 600 times. How many sixes should you expect? Probability lets you answer that question before you roll a single time — and that predictive power is exactly what this lesson is about.

Expected outcomes sit at the heart of real decisions: insurance companies use them to set premiums, game designers use them to balance difficulty, and scientists use them to spot when experimental results look suspicious. At GCSE, you need to calculate them confidently and interpret what they tell you.

2. The Core Idea: Probability Meets Repetition

Think back to what probability actually measures: the long-run relative frequency of an event. If the probability of landing heads on a fair coin is $\frac{1}{2}$, that means in a very large number of flips, roughly half will be heads. The expected outcome formula turns that idea into a concrete number.

Notice that the formula is a multiplication — nothing more. The challenge in exam questions is identifying $P(\text{event})$ correctly, especially when the event is described in words rather than given directly as a number.

3. Worked Examples: Building the Method

Theory is straightforward to state but straightforward to misapply. Walk through these examples step by step, covering each line and trying the calculation yourself before revealing the answer.

Worked example: Expected Outcomes: Fair Die

A fair six-sided die is rolled 300 times. Calculate the expected number of times a 4 is rolled. [2 marks]

1. $P(4) = \frac{1}{6}$ — A fair die has 6 equally likely outcomes. Only one face shows a 4. [M1 — correct probability identified]
2. $\text{Expected outcomes} = P(\text{event}) \times n$ — State the formula before substituting — this earns the method mark even if arithmetic slips.
3. $\text{Expected outcomes} = \frac{1}{6} \times 300$ — n = 300 trials, P = 1/6.
4. $\frac{1}{6} \times 300 = \frac{300}{6} = 50$ — 300 ÷ 6 = 50. [A1 — correct answer]

Answer: 50 fours are expected in 300 rolls.

That example used a simple unit fraction. Exam questions often use probabilities that are not unit fractions — the method is identical, but the arithmetic needs more care.

Worked example: Expected Outcomes: Biased Spinner

A spinner has four sections: Red, Blue, Green, and Yellow. The probability of landing on Red is 0.35. The spinner is spun 200 times. Calculate the expected number of times it lands on Red. [2 marks]

1. $P(\text{Red}) = 0.35$ — The probability is given directly in the question. [M1 — correct probability used]
2. $\text{Expected outcomes} = 0.35 \times 200$ — Substitute P = 0.35 and n = 200.
3. $0.35 \times 200 = 70$ — 0.35 × 200: think of it as 35 × 2 = 70. [A1 — correct answer]

Answer: 70 times landing on Red is expected.

Notice the check built into that diagram: the expected counts for all outcomes must sum to the total number of trials. If they do not, you have made an error somewhere — either in the probabilities or in the arithmetic.

4. Randomness, Fairness, and Equally Likely Events

The formula works perfectly when events are equally likely — but you need to check that assumption before using it. A fair coin, a fair die, a well-shuffled deck: these all have equally likely outcomes by design. A biased spinner or a loaded die does not, and the probability must be given or estimated from experimental data instead.

Fairness also matters when comparing expected outcomes for different players in a game. A game is fair if every player has the same expected number of wins over many rounds. If one player's expected wins are higher, the game favours them — and you can prove this using the formula.

Worked example: Evaluating Fairness Using Expected Outcomes

Aisha and Ben play a game. They take turns rolling a fair six-sided die. Aisha wins a point if the die shows 1 or 2. Ben wins a point if the die shows 5. They play 60 rounds each. Is the game fair? Show your working. [3 marks]

1. $P(\text{Aisha}) = \frac{2}{6} = \frac{1}{3}$ — Two favourable outcomes (1 or 2) out of 6 equally likely outcomes. [M1]
2. $P(\text{Ben}) = \frac{1}{6}$ — One favourable outcome (5) out of 6 equally likely outcomes.
3. $\frac{1}{3} \times 60 = 20$ — 60 ÷ 3 = 20. [M1 — correct application of formula]
4. $\frac{1}{6} \times 60 = 10$ — 60 ÷ 6 = 10.
5. Aisha expects 20 points; Ben expects 10 points. — Aisha's expected score is double Ben's. The game is not fair — it favours Aisha. [A1 — correct conclusion with justification]

Answer: The game is not fair. Aisha expects 20 points and Ben expects only 10 points over 60 rounds.

Exam questions on fairness always require a comparison and a conclusion. Writing only the expected values without stating whether the game is fair will cost you the final mark.

5. Expected Outcomes vs Actual Results

Here is something that trips up many students: the expected outcome is a prediction, not a guarantee. If you expect 50 sixes in 300 rolls, you might actually get 47 or 53 — that is perfectly normal randomness at work. The expected value tells you the average result over many, many repetitions of the whole experiment.

This distinction matters in exam questions that ask you to comment on results. If an actual result is close to the expected value, that supports the model (e.g., the die appears fair). If the actual result is very different, it might suggest the die is biased — but you cannot be certain from a small number of trials.

As the number of trials increases, the actual result gets closer and closer to the expected value as a proportion. This is the law of large numbers in action — you do not need to name it at GCSE, but understanding the pattern helps you interpret experimental data questions.

6. Multi-Event Expected Outcomes

Sometimes a question asks for the expected number of outcomes across several different events — for example, how many times you expect to roll either a 3 or a 4. The approach is the same: find the combined probability first, then multiply by the number of trials.

Worked example: Expected Outcomes for a Combined Event

A fair eight-sided die (faces numbered 1 to 8) is rolled 240 times. Calculate the expected number of times the result is a prime number. [3 marks]

1. Primes from 1 to 8: 2, 3, 5, 7 — 1 is not prime. 4, 6, 8 are not prime. There are 4 prime faces. [M1 — correct identification]
2. $P(\text{prime}) = \frac{4}{8} = \frac{1}{2}$ — 4 favourable outcomes out of 8 equally likely outcomes. Simplify the fraction.
3. $\text{Expected outcomes} = \frac{1}{2} \times 240$ — [M1 — correct formula applied with their probability]
4. $\frac{1}{2} \times 240 = 120$ — 240 ÷ 2 = 120. [A1 — correct answer]

Answer: 120 prime results are expected in 240 rolls.

7. Exam Technique and Common Mistakes

With the core method secure, the final step is learning how examiners test it — and where marks are most commonly dropped. The mistakes below are drawn from the patterns that appear repeatedly in mark schemes.