SL 4.7 Use discrete random-variable distributions and expected value | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Statistics and Probability

Representing chance numerically, checking distributions, and finding long-run averages

A random experiment has an uncertain outcome, but we can often describe its numerical result. For example, a game may award a score depending on a die roll. A discrete random variable assigns a number to each outcome and has a list of possible values. Its probability distribution records the chance of each value. The expected value combines these chances into a long-run average. It does not necessarily predict the result of one trial.

What you will learn

Identify a discrete random variable and describe its possible values.
Check whether a table gives a valid probability distribution.
Calculate and interpret the expected value of a discrete random variable.
Use a graphing calculator to check a calculation while showing the mathematical method.

1. From an experiment to a distribution

A random variable is a quantity whose value depends on the outcome of a random experiment. We commonly use a capital letter such as

X

for the variable and a lowercase letter such as

x

for one possible value. A variable is discrete when its possible values can be listed, as with the number of heads in three coin tosses or a prize amount.

A probability distribution can be written as a table of values and their probabilities. The notation

P(X=x)

means “the probability that

X

equals

x

.” The values must cover all possible cases, and their probabilities must add to

1

. Each probability must be between

0

and

1

, inclusive.

Connect the table to its context: what does each value mean, and what is its probability? If outcomes are equally likely, count the favourable outcomes and divide by the total number of outcomes. A distribution may also be displayed as a bar chart, with possible values on the horizontal axis and probabilities on the vertical axis. The bars represent separate possible values.

\sum P(X=x)=1

List every possible value of the random variable.
Check that each probability is between $0$ and $1$ and that the total is $1$ .
A bar chart gives a visual view of which values are more or less likely.

2. Expected value as a weighted average

The expected value is a probability-weighted average of the possible values. Multiply each value by its probability and add the products. A more likely value has greater influence on the result than a less likely value. The symbol

E(X)

means the expected value of

X

.

The expected value is a theoretical long-run average, not a promise that any single trial will produce that value. It may not be one of the possible values. If a game’s score can only be

0

or

10

, for example, an expected score of

4

means an average of

4

per play over many plays, not that a single play scores

4

.

Units matter. If

X

measures dollars, then

E(X)

is measured in dollars; if

X

counts items, its expected value is in items. In a game where the player pays to play, distinguish the prize from the net gain. The net gain is the amount received minus the cost, and its expected value describes the average net result per play.

E(X)=\sum xP(X=x)

Multiply each outcome by its probability, then add.
Use the distribution’s probabilities rather than treating all possible values as equally likely.
Interpret the result in context and include units.

3. Representations and a reliable method

The same distribution can be understood in several ways. A context explains what the values mean; a table organizes the values and probabilities; a bar chart compares their probabilities; and the expected-value calculation gives a numerical summary. These representations can help reveal errors: a table total greater than

1

cannot describe a complete probability distribution.

Use this sequence: define the random variable, list possible values, assign probabilities, check the probability conditions, and calculate the weighted sum. Keep enough digits in intermediate calculator work, then round only at the end if a question requests a particular accuracy. If the probabilities are fractions, exact arithmetic often keeps the calculation clear.

A graphing calculator can help verify the weighted sum or display the distribution. Enter values and their corresponding probabilities in paired lists, then calculate the sum of the products. The pairing matters: each probability must remain beside the value it describes. Calculator output checks arithmetic; it does not replace identifying the variable, confirming probabilities, or interpreting the result.

E(X)=x_1p_1+x_2p_2+\cdots+x_np_n

Keep values and probabilities paired throughout the calculation.
Use a calculator to check arithmetic or draw a probability bar chart.
State rounding accuracy when giving an approximate answer.

Example distribution and weighted contributions

Value $x$	Probability $P(X=x)$	Product $xP(X=x)$
$0$	$0.25$	$0$
$1$	$0.50$	$0.50$
$2$	$0.25$	$0.50$
Total	$1$	$1$

Worked example

Checking a distribution

A spinner gives a score

X

of

0

,

1

, or

2

. The listed probabilities are

0.25

,

0.50

, and

0.25

, respectively. Check that they form a valid distribution and find the expected score.

Check the probabilities
Each probability is between $0$ and $1$ . Their sum is $1$ , so the table accounts for all possible outcomes and is a valid probability distribution.
$0.25+0.50+0.25=1$
Calculate the weighted average
Multiply each score by its probability and add the products. This gives the expected score per spin.
$E(X)=0(0.25)+1(0.50)+2(0.25)=1$

Answer: The distribution is valid, and the expected score is

1

point per spin.

Check: The probability bar chart would have its tallest bar at score

1

, consistent with that score being most likely. The expected score is a long-run average, not a claim that every spin scores

1

.

Worked example

Finding a probability from a missing entry

A discrete random variable

X

takes values

1

,

3

, and

5

with probabilities

0.2

,

q

, and

0.3

. Find

q

and then calculate

E(X)

.

Use the total probability
The probabilities must add to $1$ . Subtract the two known probabilities from $1$ to find the missing probability.
$q=1-0.2-0.3=0.5$
Calculate the expected value
Weight each possible value by its probability. The terms use the values in the same order as their probabilities.
$E(X)=1(0.2)+3(0.5)+5(0.3)=3.2$

Answer: The missing probability is

q=0.5

, and the expected value is

3.2

.

Check: The probabilities are all between

0

and

1

and sum to

1

. The expected value lies between the smallest and largest possible values,

1

and

5

.

Worked example

Expected net gain in a game

A player pays CAD 2 to play a game. A fair four-sided die numbered

1

to

4

is rolled. The player receives CAD 6 if the result is

4

and receives nothing otherwise. Find the expected net gain per play.

Define the net-gain variable
Let $X$ be the net gain in CAD. Net gain is the amount received minus the CAD 2 cost. A roll of $4$ gives a net gain of CAD 4, while any other roll gives a net gain of negative CAD 2. Since the die is fair, the chance of a $4$ is $1/4$ and the chance of another result is $3/4$ .
$P(X=4)=\frac14,\quad P(X=-2)=\frac34$
Find the expected net gain
Multiply each net gain by its probability and add. This accounts for both the prize and the cost in every outcome.
$E(X)=4\left(\frac14\right)+(-2)\left(\frac34\right)=-0.5$

Answer: The expected net gain is negative CAD 0.50 per play.

Check: The possible net gains are CAD 4 and negative CAD 2. The weighted average is closer to negative CAD 2 because that outcome occurs three times as often as the winning outcome.

Common mistakes and how to avoid them

Adding the possible values and dividing by their count.

Correction: That treats all values as equally likely. Use each value’s probability as its weight.

Accepting probabilities that do not total

1

.

Correction: Check the total before calculating an expected value; a complete distribution must have total probability

1

.

Calling the expected value the result that will occur in one trial.

Correction: Describe it as a long-run average. One trial can give any possible value in the distribution.

Using the prize as the game outcome when the question asks for net gain.

Correction: Subtract the cost from the amount received in each outcome before calculating the expected net gain.

Rounding intermediate products too early.

Correction: Retain exact fractions or adequate decimal precision until the final answer.

Lesson summary

A discrete random variable has a list of possible numerical values.
A valid probability distribution has probabilities from $0$ to $1$ whose sum is $1$ .
Expected value is found by multiplying each value by its probability and adding.
Interpret expected value as a long-run average, with appropriate units and context.

Check your understanding

Question 1

A variable takes values

0

and

4

with probabilities

0.75

and

0.25

. What is its expected value?

$1$
$2$
$3$
$4$

Show answer and explanation

1

Weight each value by its probability:

0(0.75)+4(0.25)=1

.

Question 2

A table lists probabilities

0.4

,

0.4

, and

0.3

for all possible values. What is the correct conclusion?

It is valid because every probability is less than $1$ .
It is valid if the values are equally spaced.
It is not a valid distribution because the probabilities total $1.1$ .
It is not valid because a probability must be greater than $0.5$ .

Show answer and explanation

It is not a valid distribution because the probabilities total

1.1

.

A complete distribution must have probabilities that sum to

1

; here the sum is

0.4+0.4+0.3=1.1

.

Question 3

A game has a positive expected net gain. Which statement is justified?

The player wins on every play.
The player is guaranteed to win on the next play.
The long-run average net gain per play is positive under the stated distribution.
The amount won on each play equals the expected net gain.

Show answer and explanation

The long-run average net gain per play is positive under the stated distribution.

Expected value describes a long-run average under the distribution. It does not guarantee the result of any individual play.

Key terms

Discrete random variable: A numerical quantity determined by a random experiment that can take one of a list of possible values.
Probability distribution: A list of the possible values of a random variable together with the probability of each value.
Expected value: The probability-weighted average of a random variable’s possible values, interpreted as a long-run average.
Net gain: The amount received minus the cost paid.

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SL 4.7 · Use discrete random-variable distributions and expected value

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Representing chance numerically, checking distributions, and finding long-run averages

What you will learn

1. From an experiment to a distribution

2. Expected value as a weighted average

3. Representations and a reliable method

Example distribution and weighted contributions

Checking a distribution

Finding a probability from a missing entry

Expected net gain in a game

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through IB AA SL

About this lesson and its review