12. Examining Decisions
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Chapter 6
12.
Examining Decisions
This lesson delves into the role of probability and expected value in making informed decisions. It covers scenarios ranging from games to financial investments. For instance, if you're contemplating where to invest your money, understanding expected value can help you predict potential gains or losses. Similarly, in games that involve chance, knowing the probability of different outcomes can guide you on whether to take a risk or play it safe. The lesson also discusses how these mathematical concepts are used in various fields to strategize and make decisions that are not just guesses but are backed by data.
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| | 12 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Examining Decisions
Slide of 12
How are financial institutions making decisions? Are they simply predicting the future through speculation, or are they using reliable mathematical methods? In this lesson, strategizing and decision making principles will be analyzed and put into practice through using concepts of probability.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
Making Decisions Based on Probability
Kevin plans to invest his money in a cryptocurrency. He is choosing between two cryptocurrencies — Coin Q and Coin R. The table shows the probabilities of the events over the next week.
| Lose $25 | Gain $10 | Gain $50 | |
|---|---|---|---|
| Coin Q | 0.40 | 0.20 | 0.40 |
| Coin R | 0.10 | 0.75 | 0.15 |
Which cryptocurrency should Kevin choose? Justify the answer.
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Explore
Random Points on a Number Line
The applet shows the number line from 0 to 5. Each point on the number line represents a random number. Move the interval to see how the percentage of the random numbers inside the interval changes.
Create some cases where ten points are randomly distributed along the number line. Does an interval which contains half of the points also come out to be about half of the range of the number line?
Example
Using Random Points to Estimate the Value of 2
When ten points are distributed randomly on a line segment, half of the points are expected to be on each half of the segment.
In this case, on average, 5 points are expected to be between 0 and 2.5. With this in mind, take a look at the following example.
Maya thinks that she can estimate the value of 2 using probability concepts. To do so, Maya draws the graphs of y=x2 and y=2 for 0≤x≤2 on an applet that generates ten random points on y=2. 
Help Maya answer the following questions.
Now, the applet can be used to simulate ten times how the points will be distributed on the segment. The results of each trial will be recorded in a table. To do so, let NL be the number of points to the left and NR be the number of points to the right of (a,2).
Here, an example simulation is shown.
a Use the applet to simulate random points on the horizontal segment. Make a table of ten trials.
b Use the data obtained from the simulation to find an approximate value for 2.
Answer
a Example Table:
| Trial | Number of points to the left of (2,2) | Number of points to the right of (2,2) |
|---|---|---|
| 1 | 8 | 2 |
| 2 | 7 | 3 |
| 3 | 9 | 1 |
| 4 | 7 | 3 |
| 5 | 6 | 4 |
| 6 | 6 | 4 |
| 7 | 7 | 3 |
| 8 | 9 | 1 |
| 9 | 6 | 4 |
| 10 | 7 | 3 |
b About 1.44
Hint
a Mark the intersection point of the graphs. Record the number of points on the left and on the right of the point of intersection for each trial.
b On average, how many points will lie on the segment to the left of the intersection point?
Solution
a Let a represent the value of 2. Then, the intersection point of the graphs is the point (a,a2), or (a,2).
| Trial | NL | NR |
|---|---|---|
| 1 | 8 | 2 |
| 2 | 7 | 3 |
| 3 | 9 | 1 |
| 4 | 7 | 3 |
| 5 | 6 | 4 |
| 6 | 6 | 4 |
| 7 | 7 | 3 |
| 8 | 9 | 1 |
| 9 | 6 | 4 |
| 10 | 7 | 3 |
b Based on the simulation data, calculate the average of the NL values.
108+7+9+7+6+6+7+9+6+7=7.2
On average, there are 7.2 points to the left of (a,2) on the horizontal segment. In other words, the probability of a random point appearing on the left side is about the ratio of 7.2 to 10.
107.2
The probability of a point between 0 and a on the horizontal segment can also be found using the geometric probability. It is equal to the ratio of the length of the segment between 0 and a to the length of the horizontal segment.
1a
The ratio obtained using the simulation data 107.2 is an estimate of the probability obtained using geometric measures 1a. With this in mind, the value of a can be calculated.
Therefore, a, the value of 2, is about 1.44. Notice that this is close to the value of 2.
2=1.414213…
Explore
Random Points in a Square
Consider a square whose sides have a length of 5 units. The applet shows 100 random points inside and a unit inside the square. Move the unit square to see how the percentage of the random numbers inside the unit square changes.
Example
Estimating the Area Under the Graph of y=x2
Using a similar thought process, Maya now attempts to approximate the area under the graph of y=x2 between 0 and 1. The following applet generates 100 random points inside a unit square.
The results can be recorded in a table. Here is an example table.
On average, there are 33.2 points under the curve. In other words, the probability of a random point appearing under the curve is about the ratio of 33.2 to 100.
Help Maya estimate the area under the curve.
a Use the applet to make ten simulations of the distribution of the points in the unit square. In a table, record the number of points under the curve for each trial.
b Using the data obtained from simulation, find an approximate value for the area under the curve between 0 and 1.
Answer
a Example Table:
| Trial | Number of Points Under the Graph | Number of Points Above the Graph |
|---|---|---|
| 1 | 35 | 65 |
| 2 | 39 | 61 |
| 3 | 34 | 66 |
| 4 | 33 | 67 |
| 5 | 31 | 69 |
| 6 | 28 | 72 |
| 7 | 27 | 73 |
| 8 | 37 | 63 |
| 9 | 33 | 67 |
| 10 | 35 | 65 |
b About 0.332 square units
Hint
a Count the number of points under the curve.
b Calculate the average of the numbers found.
Solution
a Start by simulating random points inside the unit square. Then, count the number NU of random points under the graph of y=x2 after each trial. The number of random points above the graph NA will also be recorded. Recall that NA is the total number of random points subtracted by NU.
| Trial | NU | NA |
|---|---|---|
| 1 | 35 | 65 |
| 2 | 39 | 61 |
| 3 | 34 | 66 |
| 4 | 33 | 67 |
| 5 | 31 | 69 |
| 6 | 28 | 72 |
| 7 | 27 | 73 |
| 8 | 37 | 63 |
| 9 | 33 | 67 |
| 10 | 35 | 65 |
b Based on the simulation data, calculate the average of the NU values.
Number of NU ValuesSum of NU Values
SubstituteValues
Substitute values
1035+39+34+33+31+28+27+37+33+35
33.2
10033.2
The probability of a point being under the curve can also be found using the geometric probability. It is equal to the ratio of the area under the curve AU to the area of the unit square 1.
1AU
The ratio obtained using the simulation data 10033.2 is an estimate of the probability obtained using geometric measures 1AU. With this in mind, the value of AU can be calculated.
Therefore, the area under the graph of y=x2 from 0 to 1 is about 0.332 square units. This result is close to the exact value of the area.
Example
Rolling a Die 60 Times
Vincenzo and his friends decide to play a game. They roll a die up to 60 times. They have to pay 2 coins if 1, 2, 3, or 4 appears. They get 5 coins if 5 or 6 appears.
a Find the number of coins that Vincenzo and his friends obtain at the end of the first trial.
c How many coins will they expect to have if they simulate the game infinitely many times?
Answer
a Number of Coins: 27
b Example Table:
| Trial | Number of Coins |
|---|---|
| 1 | 27 |
| 2 | -1 |
| 3 | 41 |
| 4 | 34 |
| 5 | 6 |
| 6 | 69 |
| 7 | -8 |
| 8 | -8 |
| 9 | 34 |
| 10 | 27 |
| Average | 22.1 |
c Number of Coins: 20
Hint
a Multiply the frequencies of 1, 2, 3, and 4 by -2 and the frequencies of 5 and 6 by 5. Then, find the sum of the products.
b Repeat the procedure followed in the previous part.
c What are the frequencies expected to be when the game is simulated infinitely many times?
Solution
a Consider the values in the given table.
∙∙ If 1,2,3, or 4 appear, pay 2 coins If 5 or 6 appear, get 5 coins
For the given table, the number of coins Vincenzo and his friends can get is found as follows. Since paying 2 coins is a loss, its effect will be negative.
(-2)11+(-2)9+(-2)11+(-2)8+(5)9+(5)12
Multiply
Multiply
-22−18−22−16+45+60
AddSubTerms
Add and subtract terms
27
b By repeating the procedure in Part A ten times, the number of coins Vincenzo can get or give after each trial can be recorded.
Here is an example table.
To find the average of these values, add them and divide the sum by 10.
It looks like they will get 22.1 coins on average.
| Trial | Number of Coins |
|---|---|
| 1 | 27 |
| 2 | -1 |
| 3 | 41 |
| 4 | 34 |
| 5 | 6 |
| 6 | 69 |
| 7 | -8 |
| 8 | -8 |
| 9 | 34 |
| 10 | 27 |
1027+(-1)+41+34+6+69+(-8)+(-8)+34+27
AddSubTerms
Add and subtract terms
10221
CalcQuot
Calculate quotient
22.1
c When an experiment is repeated many times, the result will tend to the event's theoretical probability. In this case, the theoretical probability of each possible outcome is 61. Therefore, in theory, the frequency of the possible outcomes should be equal to 60⋅61=10. 
Under these conditions, the number of coins Vincenzo can get will be equal to 20.
(-2)10+(-2)10+(-2)10+(-2)10+(5)10+(5)10
Multiply
Multiply
-20−20−20−20+50+50
AddSubTerms
Add and subtract terms
20
Discussion
Random Variable and Expected Value
Consider an experiment involving the rolling of a die 60 times in which the outcome of each roll determines the number of coins that will be received or lost. If 5 or 6 appears, 5 coins are received. Otherwise, 2 coins are lost. This situation can be represented by a variable, it can be called X, for example.
X={-2,5,if 1,2,3, or 4 appear,if 5 or 6 appear.
The variable X has a special name in probability. Concept
Random Variable
A random variable is a variable whose values depend on the outcomes of a probability experiment. Random variables are usually denoted by capital letters such as X. For an experiment, numerous random variables can be defined. Consider the following random variables and their respective experiments.
As can be seen, depending on the outcomes of the experiment, the value of a random variable can vary. Additionally, the random variables X and Y can be classified as discrete because both can take a finite number of distinct values. Contrarily, a continuous random variable takes any value within a continuous interval such as when measuring height, weight, and time.
Concept
Expected Value of a Random Variable
The expected value of a random variable X, often denoted E(X), can be thought of as the average value of the random variable.
In this formula, n is the number of all possible values of X. The right-hand side of the formula can also be written using sigma notation.
Expected Value of X: E(X)
The expected value of a discrete random variable is the weighted mean of the values of the variable. It is evaluated by finding the sum of the products of every possible value x of the random variable X and its associated probability P(x). The expected value uses theoretical probability to give an idea of what is expected in the long run.
E(X)=i=1∑nxi⋅P(xi)
The expected value of a random variable does not necessarily have to be equal to a possible value of the random variable. The alternative notations for the expected value are shown.
EX, E(X), E(X), E[X], E(X), E[X]
Pop Quiz
Practice Finding Expected Values
The table shows the random variable X assigned to outcomes of a probability experiment and the corresponding probabilities. Calculate the expected value of the random variable. If necessary, round the answer to two decimal places.
Example
Finding the Expected Value When Guessing on a Test
Jordan will take a multiple-choice test in which each question has five choices. She receives 1 point for each correct answer and loses 0.5 points for an incorrect answer. Help Jordan decide whether guessing is advantageous or not for the questions she does not know the correct answer.
Answer
No, it is not advantageous to guess. See solution.
Hint
Define a random variable for the situation. Then, find the expected value of it.
Solution
When Jordan does not know the correct answer and guesses, she either receives 1 point or loses 0.5 points. Let X be the random variable that represents the point value assigned to answers. Then, X can be written as follows.
From here, the expected value of the random variable X can be calculated.
Because the expected value for each guess is -0.2, Jordan would lose 0.2 points for each guess. Therefore, she should not guess.
X={ 1if the answer is correct-0.5if the answer is incorrect
Since there are 5 choices for each question, the probability of selecting the right answer is 51 and the probability of selecting an incorrect one is 54. The following table shows probabilities. | X | 1 | -0.5 |
|---|---|---|
| P(X) | 51 | 54 |
E(X)=x1⋅P(x1)+x2⋅P(x2)
SubstituteValues
Substitute values
E(X)=1⋅51+(-0.5)⋅54
▼
Evaluate right-hand side
E(X)=-0.2
Example
Analyzing Decisions
Tearrik and six more people apply for a job. The company announces that four of the applicants will be hired using random selection. It is known that four out of the seven applicants are females.
Tearrik hears that the company filled the four opening with all four female applicants. Tearrik wonders if this is an unlikely outcome. By answering the following questions, help Tearrik understand if the outcome, indeed, was unlikely.
a What is the probability of selecting 4 females out of 7 people?
b How many females are expected to be hired? If necessary, round the answer to the nearest integer.
Answer
a P(selecting 4 females)=351
b E(X)≈2
Hint
b Start by defining a random variable for the possible number of females.
Solution
a Recall that the probability of an event A is the ratio of the number of favorable outcomes to the number of possible outcomes.
Number of favorable outcomes=1
Since the order of selecting people does not matter, the number of possible outcomes can be found using combinations. The number of combinations when selecting r items out of n is given by the Combination Formula.
nCr=r!(n−r)!n!
By substituting 7 for n and 4 for r, the number of combinations can be calculated. nCr=r!(n−r)!n!
SubstituteII
n=7, r=4
7C4=4!(7−4)!7!
▼
Evaluate right-hand side
SubTerm
Subtract term
7C4=4!⋅3!7!
Write as a product
7C4=4!⋅3!7⋅6⋅5⋅4!
CancelCommonFac
Cancel out common factors
7C4=4!⋅3!7⋅6⋅5⋅4!
SimpQuot
Simplify quotient
7C4=3!7⋅6⋅5
Write as a product
7C4=3⋅2⋅17⋅6⋅5
Multiply
Multiply
7C4=6210
CalcQuot
Calculate quotient
7C4=35
P(selecting 4 females)=351
b Since there are 3 male applicants out of 7 applicants, there will be at least one female hired when 4 people are selected. With this in mind, a random variable X can be defined.
X=⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧1 if one female is selected2 if two females are selected3 if three females are selected4 if four females are selected
To find the expected value of X, the probabilities of each case needs to be determined. To do so, combinations and the Fundamental Counting Principle will be used. Now, consider the case in which 1 female and 3 males are selected. 1 Female ⇓4C1and⇓⋅3 Males⇓3C3
Here, 4C1 represents the number of possible ways 1 female is selected out of 4 females and 3C3 represents the number of possible ways 3 males are selected out of 3 males. Compute the value of the product.
4C1⋅3C3
▼
Evaluate
Rewrite
Rewrite 4C1 as 1!(4−1)!4!
1!(4−1)!4!⋅3C3
Rewrite
Rewrite 3C3 as 3!(3−3)!3!
1!(4−1)!4!⋅3!(3−3)!3!
SubTerm
Subtract term
1!⋅3!4!⋅3!⋅0!3!
MultFrac
Multiply fractions
1!⋅3!⋅3!⋅0!4!⋅3!
Write as a product
1!⋅3!⋅3!⋅0!4⋅3!⋅3!
CancelCommonFac
Cancel out common factors
1!⋅3!⋅3!⋅0!4⋅3!⋅3!
SimpQuot
Simplify quotient
1!⋅0!4
1!=1
1⋅0!4
0!=1
1⋅14
MultByOne
a⋅1=a
14
DivByOne
1a=a
4
| Number of Possible Outcomes, 35 | |||
|---|---|---|---|
| 1 Female and 3 Males |
2 Females and 2 Males |
3 Females and 1 Male |
4 Females and 0 Males |
| 4C1⋅3C3 | 4C2⋅3C2 | 4C3⋅3C1 | 4C4⋅3C0 |
| 4⋅1 | 6⋅3 | 4⋅3 | 1⋅1 |
| 4 | 18 | 12 | 1 |
Considering the above table, the probabilities can be determined.
| X | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| P(X) | 354 | 3518 | 3512 | 351 |
E(X)=x1⋅P(x1)+x2⋅P(x2)+x3⋅P(x3)+x4⋅P(x4)
SubstituteValues
Substitute values
E(X)=1⋅354+2⋅3518+3⋅3512+4⋅351
▼
Evaluate right-hand side
MoveLeftFacToNum
a⋅cb=ca⋅b
E(X)=354+3536+3536+354
AddFrac
Add fractions
E(X)=3580
CalcQuot
Calculate quotient
E(X)=2.285714…
RoundInt
Round to nearest integer
E(X)≈2
Closure
Making Decisions Based on Probability
The expected values of situations can be compared when making decisions. Now, returning to the initial challenge, the most profitable cryptocurrency for Kevin to invest in can be determined. Recall the table showing the probabilities of the events over the next week.
| Lose $25 | Gain $10 | Gain $50 | |
|---|---|---|---|
| Coin Q | 0.40 | 0.20 | 0.40 |
| Coin R | 0.10 | 0.75 | 0.15 |
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Hint
Start by calculating the expected value of the gain or loss for each cryptocurrency.
Solution
The expected value of gain or loss for each cryptocurrency will be calculated one at a time.
The Expected Value of Gain or Loss for Coin Q
Let X be the random variable that represents the gains and losses for Coins Q.
| X | -25 | 10 | 50 |
|---|---|---|---|
| P(X) | 0.4 | 0.2 | 0.4 |
E(X)=x1⋅P(x1)+x2⋅P(x2)+x3⋅P(x3)
SubstituteValues
Substitute values
E(X)=-25⋅0.4+10⋅0.2+50⋅0.4
▼
Evaluate right-hand side
E(X)=12
The Expected Value of Gain or Loss for Coin R
Let Y be the random variable that represents the gains and losses for Coin R.
| Y | -25 | 10 | 50 |
|---|---|---|---|
| P(X) | 0.1 | 0.75 | 0.15 |
E(Y)=y1⋅P(y1)+y2⋅P(y2)+y3⋅P(y3)
SubstituteValues
Substitute values
E(Y)=-25⋅0.1+10⋅0.75+50⋅0.15
▼
Evaluate right-hand side
E(Y)=12.5
Conclusion
Since the expected gains from Coin R are greater than the expected gains from Coin Q, the favorable decision is to invest in Coin R.E(Y)12.5 >E(X)> 12
Tadeo has challenged Kriz to a game of marbles.
▶ ▶ ▶ ▶ Tadeo sets up a pyramid of four marbles.Kriz throws one of his marbles to the pyramid.Kriz loses the marble he throws.If Kriz hits the pyramid, he wins all themarbles that are in the pyramid.
The probability of hitting the target decrease linearly by about 10 percentage units for every four feet away Kriz stands from the pyramid. How far away should Kriz stand from the pyramid when throwing his marbles for the game to be fair? Round the answer to the nearest foot.
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First, we will find the probability of Kriz hitting the pyramid that makes the game fair. Then, using the relationship between the probability p of hitting the pyramid and the distance d to the pyramid, we will find d.
Finding the Probability of Kriz Hitting the Pyramid
We know that Kriz loses the marble he throws. We can think of that marble as the cost of playing the game. Let's define a random variable for the possible number of marbles Kriz can get. X = 4 & if Kriz hits the pyramid 0 & if Kriz misses the pyramid For the game to be fair, the expected value of X minus 1, the number of marble that Kriz gives to play the game, should be 0. E(X)-1=0 Next, we need to calculate the expected value of X. Let p be the probability that Kriz hits the pyramid. Then, the probability of Kriz missing the pyramid is 1-p by the Complement Rule. Therefore, the expected value of X can be written as follows. E(X) = 4 * p + 0(1-p) ⇕ E(X) = 4p Since the expected value minus 1 must be 0, we can find p that makes the game fair.
Finding the Distance
Now, we need to use the following information.
- If Kriz stands 0 feet away, he has a 100 % chance of hitting the target.
- The probability of hitting the target decreases linearly by 10 percentage units for every 4 feet away he stands.
With this information, we can express the relationship between the probability of Kriz hitting the pyramid p in percent and the distance to the pyramid d as follows. p = 100 - 10/4d We have found that p should be 14, or 25 % in order for the game to be fair. By substituting p=25 into the above equation, we can find the distance.
Therefore, when Kriz stands 30 feet away, the probability of hitting the target is 25 %. This would make the game fair. We can also check our result by analyzing the graphs of each equation used.
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Examining Decisions
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