Use discrete random variables and associated probabilities to solve practical problems
Recognise the mean or expected value, πΈ(π) = π, of a discrete random variable π as a measure of centre, and evaluate it in simple cases
Understand that a sample mean, π₯Μ , is an estimate of the associated population mean π and that this estimate get better as the sample size increases and when we have independent observations
If we are given a probability distribution, the expected value helps us calculate the weighted average of the distribution. The expected value is commonly referred to as the expectation, mean or average.
We can find the expected value using the formula shown below:
Expected Value Formula
The expected value of a random variable with the outcomes can be expressed as:
or
An intuitive explanation of the expected value can be found below:
Intuitive Expected Value Explanation
Imagine we were given a probability distributions such as the one shown below:
The expected value of this distribution can be thought of as its balancing point or centre of mass.
An example of an expected value calculation can be found below:
Expected Value Calculation Example
Imagine we have a machine that is making a four pack of pencils. Let the random variable map each pack to the number of faulty pencils it contains. The probability distribution table of can be seen below:
We can find the expected number of faulty pencils per pack by finding .
Therefore, we would expect 0.37 out of the 4 pencils in the pack to be faulty. It is obviously impossible to have 0.37 of a pencil, however this tells us we should expect a faulty pen in ~1/3rd of the packs.
Uniform Distributions
If we are given a uniform distribution, we can find its expected value using the formula shown below:
Expected Value Of Uniform Distribution
The expected value of a uniform random variable with the outcomes can be expressed as:
or
An explanation of this formula can be found below:
Expected Value Of Uniform Distribution - Explanation
Finding The Probability Of A Single Outcome
We know that the sum of all probabilities equals to 1. Therefore we write:
Since we are dealing with a uniform distribution, all the probabilities are the same. Therefore we can add them as:
Therefore, the probability of a single event can be expressed as:
We can therefore express the distribution as:
Finding The Expected Value
We can find the expected value by adding the events multiplied by thier respective probability.
Constants can be taken out of sums, therefore we can express this as:
We can also expand the sum to express the expected value as:
An example of a uniform expected value calculation can be seen below:
Expected Value Calculation For A Uniform Distribution
Imagine we roll a dice and records the score. If we do this 100 times, what is his expected net score ? We can answer this question by finding the expected value of a single roll.
Let us define the random variable which maps a dice roll to its score. The probability distribution for can be expressed as:
1
2
3
4
5
6
1/6
1/6
1/6
1/6
1/6
1/6
Note: We are dealing with a uniform probability distribution.
We can find the expected value of this distribution as seen below:
Therefore, if we played 100 games we would expect to have a net score of .
Expected Value Properties
The expected value is a linear operator, meaning it can be split on addition and multiplication of constants.
Linearity Of Expected Values
If we have random variables and , as well as constants and
and
A proof of these properties can be seen below:
Explaining The Linearity Of Expected Values
To prove these statements we will assume that we are always working on the same event space .
Proof 1
Sums can be split on addition, we can therefore express this as:
We can also take constants out of sums as seen below:
At this stage, we should realise that the first sum is simply while the second sum is simply 1 as we are summing a probability function.
Proof 2
The sums can now be split into two seperate sums:
At this stage, we should realise that the first sum is simply while the second sum is simply .
These linear properties are very useful when calculating expected values of multiple random variables. An example of this can be seen below:
Using The Linearity Of Expected Values
A teacher is setting two tests for her students. The expected outcomes of each test are shown below:
Test 1:
Test 2:
Let us define the random variable which maps a student to their final grade. The final grade is defined as a weighted average of the first 2 exams:
We can find the expected final grade by using the linearity of expected values:
We can split expected values when they are being summed.
We can also take constants out of expected values.
We can now substitute our known expected values to get:
Lesson Loading
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SOLUTION
Difficulty
6 Mins
Time
In a standard lottery, you win the division 1 jackpot by satisfying two conditions:
From the first group of 35 balls, you need to pick the correct 7 that are drawn. There are 6,724,520 different ways this could be done.
From a group of 20 balls, only one is drawn. You have to guess it correctly.
A company advertises a $100,000,000 jackpot, and it only costs $2 to play. What is the expected return on a single lotto ticket.
SOLUTION
Defining The Random Variable
Whenever playing the Lotto there is two possible outcomes, win or loss. Let us define the random variable which maps the each outcome to the corresponding reward.
Finding The Probability Of Each Event
To find the probability of the events, we must find the total possible combinations of this game. This can be found as:
Since we must pick the exact combination to win, the probability of winning can be expressed as:
And the probability of losing can be expressed as:
This information is summarised in the probability distribution table below:
Finding The Expected Value
The expected return on a single lotto ticket can be found as:
Since the expected value is negative, we should expect to lose money. However, from the perspective of the lotto company they will make $1.26 profit on each ticket, even if they pay out the prize.
This is a common issue with any form of gambling. If it wasnβt profitable for the company offering the game, they wouldnβt offer it in the first place.
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3 Mins
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Find the expected value of the probability distribution shown below:
1
2
3
5
8
9
0.1
0.15
0.2
0.05
0.2
0.3
SOLUTION
The expected value can be calculated using the following formula:
We can calculate this by finding for each outcome. This is seen below:
1
2
3
5
8
9
0.1
0.15
0.2
0.05
0.2
0.3
0.1
0.3
0.6
0.25
1.6
2.7
Therefore, the sum of is equal to:
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4 Mins
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The probability distribution for the random variable is given below:
1
2
3
4
5
The expected value of this distribution is 3.2. Find the value of .
SOLUTION
We can answer this question by finding an expression for the expected value and equating it to 3.2. The expected value can be calculated as:
We can now equate this expression to 3.2 and solve for .
Difficulty
6 Mins
Time
The image below is one of the games at the LA carnival. The game is played by spinning the wheel to determine the number of tickets you will receive.
The probability of each region is proportional to its angle.
If this wheel is spun 700 times per day, find the expected number of tickets awarded by this game.
SOLUTION
Probability Distribution Table
Let us define the random variable which represents the score of each region. The probability of each event is shown in the probability distribution table below:
1
2
4
7
10
Finding The Expected Value
The expected number of tickets received from a single game can be calculated by finding the expected value of .
Playing 700 Games
If 700 games are played the expected number of awarded tickets will be:
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6 Mins
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Christian and Mimi are going to an Italian restaurant which sells pizza, pasta, soups and risottos. The table below summarises the menu at this restaurant:
Item
Pizza
Pasta
Soup
Risotto
Number of variations
8
5
4
6
Price
$16
$21
$13
$25
Mimi cannot decide what to eat, so Christian selects a random item for her. Find the expected value of Mimiβs meal.
SOLUTION
Defining A Random Variable
Let us define the random variable which maps each menu option to its price. The probability of each price is summarised in the table below:
13
16
21
25
Note: 23 is the total number of items on the menu and each item is equally likely.
Finding The Expected Value
We can find the expected value as:
Therefore he should expect to pay around $18.91.
Difficulty
5 Mins
Time
Fred is about to play a dice game with the following rules:
If the dice rolls a 1 or 2 Fred wins
If the dice rolls a 3, 4, 5 or 6 Fred loses
It will cost Fred $5 to play one round, but if he wins he gets $20 back. Find Fredβs expected earning. Is it worth it for him to play this game 100 times ?
SOLUTION
Defining A Random Variable
Let us define the random variable which maps the possible outcomes to their earnings.
The probability distribution table for is shown below:
Note: The probabilities are found by the number of dice roll outcomes for each event.
Finding The Expected Value
The expected value calculation is shown below:
Playing The Game 100 Times
The expected earning of a single game is $1.67. Therefore, even though he may lose money on an individual round, the overall expected earning after playing 100 games is approximately .
The only reason this game is profitable is because the expected value is positive. In reality, most casino or are arcade games are designed to have a negative expected value ( you are expected to lose money ).
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5 Mins
Time
Brittanie wants to determine the average shoe size of the girls in her grade. Brittanie surveyed every girl in her grade and found the following results:
Shoe Size
5
6
7
8
9
10
11
Frequency
4
7
16
22
15
4
2
Find the average shoe size for the girls in her grade.
SOLUTION
Defining A Random Variable
Let us define the random variable , which maps a girl in Brittanieβs grade to their shoe size. The probability distribution table for will be:
5
6
7
8
9
10
11
Note: The probabilities are simply found by dividing the frequency by the total number of people surveyed.
Finding The Average Shoe Size
We can now find the average shoe size by calculating the expected value of .
Therefore the average shoe size is approximately 7.81
Difficulty
8 Mins
Time
One of the games in the Red Dragon Casino is a simple slot machine that can show whole numbers from 1 to 8 .
The probability of showing an 8 is 7%
The probability of showing a 7 is 10%
The probabilities of all the other numbers are equal to each other
Each spin of this game costs $5. A score of 8 wins the player $25 , a score of 6 or 7 wins the player $10 , otherwise the player wins no money.
In a typical day, a gambling addict has 100 spins on this spinner. Find the gamblers expected loss.
SOLUTION
Finding The Probability Of Each Event
There are 8 total events in our sample space:
We know that , and the remaining six probabilities are equal. Based on this we can express the sum of probabilities as:
Where represents the probability of the remaining numbers. We can solve for as seen below:
Therefore, we know that:
Defining A Random Variable
Let the random variable map the outcomes { max win, mid win, loss } to their respective earnings.
Max win: Make $20 ( Win 25 but spent $5 to play )
Mid win: Make $5 ( Win 10 but spent $5 to play )
Loss: Lose $5 ( The $5 it costs to play )
We can therefore find the probability of each outcome. This is shown in the probability distribution table below:
20
5
-5
Since the probabilities are mutually exclusive, we can simply add the probability of the unions to get:
20
5
-5
0.07
0.2383
0.6917
Finding Expected Value
We can find the expected value as:
Therefore we expect to lose ~87c per spin.
Finding The Total Loss
If a gambler spins this game 100 times, they should expect to lose around per day.
Difficulty
6 Mins
Time
Fairfield city council is conducting a survey to find the average household temperature during the month of April.
Let the random variable represent the temperature of a given household in Β°C. The probability distribution table for is shown below:
16Β°
17Β°
18Β°
19Β°
20Β°
21Β°
23Β°
24Β°
4%
6%
15%
20%
25%
13%
8%
9%
Note: Outliers have been removed from the data
Find the expected value of . What does this value represent ?
Let the random variable represent the temperature of a given household in Β°F. The relationship between and is given as:
Find the expected value of .
SOLUTION
Part 1
The expected value of can be found using the steps shown below:
Therefore, if we randomly selected a house the most likely temperature of that house would be .
Part 2
We can find the expected value of by utilising the following property of expected values:
Where and are constants ( A proof of this property can be found in the notes ). This property is utilised below:
We can substitute our value for from part 1 to get:
Difficulty
5 Mins
Time
The probability distribution for the random variable is shown below:
1
3
4
6
9
12
0.1
0.3
0.15
0.05
0.25
0.15
The random variable is defined by the following relationship:
Create a probability distribution table for
Find the expected value of .
SOLUTION
Part 1
We can map the values of the random variable to the outputs of the random variable using the function below:
1
3
4
6
9
12
It is important to realise that the probability of each outcome remains the same. Therefore, the probability distribution table can be expressed as:
0.1
0.3
0.15
0.05
0.25
0.15
Part 2
We can now find the expected value of , as shown below:
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Expected Value
Cambridge Advanced Year 11 - 11B
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Calculating Random Variables
Review Questions
If we play a game with a negative expected value, we cannot make money. Is this statement true or false ?
False. It is possible to make money, however it is more likely to lose money.
True. If the expected value is negative you will always lose money.
Γ
Calculating Random Variables
Review Questions
If we play a game with a negative expected value, we cannot make money. Is this statement true or false ?
True. If the expected value is negative you will always lose money.
False. It is possible to make money, however it is more likely to lose money.
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Introducing Expected Values
Review Questions
The expected value gives us
The βaverageβ value in a probability distribution
The spread of a probability distribution
The outcome with the highest probability in a distribution
The expected value can be calculated by:
Adding the outcomes and dividing by the total number of outcomes
Multiplying each outcome by its respective probability and summing the results.
Adding all the probabilities in a distribution
The formula for calculating an expected value can be written as:
Γ
Introducing Expected Values
Review Questions
The expected value gives us
The outcome with the highest probability in a distribution
The βaverageβ value in a probability distribution
The spread of a probability distribution
The expected value can be calculated by:
Adding all the probabilities in a distribution
Multiplying each outcome by its respective probability and summing the results.
Adding the outcomes and dividing by the total number of outcomes
The formula for calculating an expected value can be written as:
