The idea, step by step
A probability-weighted average
Expected value is the sum of each possible value multiplied by its probability: . Values with greater probability contribute more weight. The result has the same units as X.
An expected value need not be a possible single outcome. A count can have mean 1.6 even though no trial produces 1.6 events. The interpretation is about an average over many repetitions under an unchanged probability model.
Profit and a fair charge
Separate the gross prize from the net result. If a game pays prize X and costs c dollars to enter, net gain is X − c. Its expected value is E(X) − c. The fair entry fee is therefore E(X), ignoring other costs and risk preferences.
Expected value is one decision criterion. Two options can have the same mean but different variability. It is not a promise that a person who plays a small number of times will break even.
A common trap
The expected value is the most likely outcome.
Instead: The mode identifies a most likely outcome. Expectation is a weighted average and can lie between all listed outcomes.
See the thinking
An average that is not an outcome
X takes values 0, 2 and 5 with probabilities 0.5, 0.3 and 0.2. Find E(X).
- Multiply each value by its probability: 0 × 0.5, 2 × 0.3 and 5 × 0.2.
- Add the contributions: 0 + 0.6 + 1.0 = 1.6.
- The mean is 1.6 per trial; none of the individual outcomes needs to equal 1.6.
Prize versus profit
A game pays 0 dollars, 4 dollars or 10 dollars with probabilities 0.6, 0.3 and 0.1. It costs 2 dollars to play. Find the expected net gain.
- Expected gross prize = 0 × 0.6 + 4 × 0.3 + 10 × 0.1 = 2.20 dollars.
- Subtract the fixed entry fee once: 2.20 − 2.00 = 0.20.
- Over many repetitions, the model gives an average gain of 20 cents per play.
Check your understanding
A prize is 12 dollars with probability 0.25 and 0 dollars otherwise. What is a fair fee?
Show answer
3 dollars.
The gross expected prize is 12 × 0.25 + 0 × 0.75 = 3.
Your turn
6 questions · 13 marks · Work at your own pace.
Read the questions online
- X takes values 1, 2 and 4 with probabilities 0.2, 0.5 and 0.3. Find E(X).2 marks
- A fair die is rolled and X is its score. Find E(X).2 marks
- A ticket pays 20 dollars with probability 0.1 and 0 dollars otherwise. It costs 3 dollars. Find expected net gain.2 marks
- X is 0 or 5. If E(X) = 2, find P(X = 5).2 marks
- A machine makes 0, 1 or 2 faulty items per batch with probabilities 0.7, 0.2 and 0.1. Find the expected number over 50 batches.3 marks
- A forecast says E(X) = 2.4 calls per minute. Explain why 2.4 need not be a possible call count.2 marks
Read the worked solutions
Equivalent correct methods earn the corresponding marks. Include units where asked.
- X takes values 1, 2 and 4 with probabilities 0.2, 0.5 and 0.3. Find E(X).2 marks
E(X) = 1(0.2) + 2(0.5) + 4(0.3).
The sum is 0.2 + 1 + 1.2 = 2.4.
2.4. - A fair die is rolled and X is its score. Find E(X).2 marks
Each score has probability 1/6.
E(X) = (1 + 2 + 3 + 4 + 5 + 6)/6 = 21/6.
3.5. - A ticket pays 20 dollars with probability 0.1 and 0 dollars otherwise. It costs 3 dollars. Find expected net gain.2 marks
Expected prize is 20(0.1) = 2 dollars.
Net expected gain = 2 − 3 = −1 dollar.
−1 dollar. - X is 0 or 5. If E(X) = 2, find P(X = 5).2 marks
Let p = P(X = 5), so E(X) = 0(1 − p) + 5p.
Solve 5p = 2 to obtain p = 0.4.
0.4. - A machine makes 0, 1 or 2 faulty items per batch with probabilities 0.7, 0.2 and 0.1. Find the expected number over 50 batches.3 marks
Expected faults per batch = 0.2 + 2(0.1) = 0.4.
By linearity of expectation, the expected total is 50(0.4) = 20.
20 faulty items. - A forecast says E(X) = 2.4 calls per minute. Explain why 2.4 need not be a possible call count.2 marks
Each observed count is an integer.
A probability-weighted mean can be non-integer without contradicting the model.
It is a long-run average, not an individual count.
Curriculum reference & lesson scope
SACE Board of South Australia
Stage 2 Mathematical Methods: discrete random variables and probability distributions, expected value and variance, transformations and independent sums, Bernoulli and binomial models. Original supporting lessons and exercises follow the approved Ascent chapter structure.
Official curriculum source ↗