Expected value | Year 12 | Ascent Library

01 / LEARN

Understand the idea.

A probability-weighted average

The expected value is a probability-weighted average: $E (X) = \sum_{x} x P (X = x)$ . Multiply each possible value by its own probability, then add. Larger probabilities give their associated values more influence on the average.

Expectation describes the average approached over many repetitions under the same model. It need not be a possible outcome: a count may average $1.6$ even though every individual count is an integer. Its units are the units of $X$ .

Profit and a fair charge

Distinguish a gross prize from a net gain. If the entry fee is the fixed amount $c$ , the net gain is $X - c$ , so its expectation is $E (X) - c$ . Subtract the fee once after averaging prizes.

A fair fee makes expected net gain zero, giving $c = E (X)$ when other costs are ignored. This balances the long-run average only. Variability and a person's tolerance for losses still matter when comparing games.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

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).

Worked solution

1
Weight each possible value.
$E (X) = 0 (0.5) + 2 (0.3) + 5 (0.2)$
Each value contributes in proportion to how often it occurs in the model.
2
Add the contributions.
$E (X) = 0 + 0.6 + 1.0 = 1.6$
All possible values have been included once.
3
Interpret the result.
$1.6 \notin {0, 2, 5}$
A long-run average need not be an attainable individual value.

Answer

E(X) = 1.6.

EXAMPLE 02

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.

Worked solution

1
Find the expected gross prize.
$E (X) = 0 (0.6) + 4 (0.3) + 10 (0.1) = 2.20 dollars$
The prize expectation weights all three possible payments.
2
Subtract the certain entry fee.
$E (X - 2) = 2.20 - 2.00 = 0.20 dollars$
A fixed cost is incurred on every play, including a zero-prize play.
3
State the fair fee and interpretation.
$c_{fair} = E (X) = 2.20 dollars$
The actual fee gives an average gain of twenty cents per play; charging the expected prize would make that average zero.

Answer

Expected net gain = 0.20 dollars; fair fee = 2.20 dollars.

03 / YOUR TURN

Try it yourself.

Check your understanding

A prize is 12 dollars with probability 0.25 and 0 dollars otherwise. What is a fair fee?

Show answer & reasoning

3 dollars.

The expected prize is $12 (0.25) + 0 (0.75) = 3$ dollars. A fee of $3$ dollars makes the expected net gain zero.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

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 $\frac{1}{6}$ .
$E (X) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6}$ .
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 $E (X) = 0 (1 - p) + 5 p$ .
Solve $5 p = 2$ 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.

Checking saved progress…

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 ↗

In this lesson

Understand the idea.

A probability-weighted average

Profit and a fair charge

See the method in action.

An average that is not an outcome

Weight each possible value.

Add the contributions.

Interpret the result.

Prize versus profit

Find the expected gross prize.

Subtract the certain entry fee.

State the fair fee and interpretation.

Try it yourself.

Check your understanding

Keep practising