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.
| Student Learning Objectives: |
|---|
|
| | 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.
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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.
{"type":"choice","form":{"alts":["Coin Q","Coin R"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":1}
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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.
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Example
Using Random Points to Estimate the Value of sqrt(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 sqrt(2) using probability concepts. To do so, Maya draws the graphs of y=x^2 and y=2 for 0 ≤ x ≤ 2 on an applet that generates ten random points on y=2.
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 sqrt(2).
Answer
a
Example Table:
| Trial | Number of points to the left of (sqrt(2),2) | Number of points to the right of (sqrt(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 sqrt(2). Then, the intersection point of the graphs is the point (a,a^2), or (a,2).
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 N_L be the number of points to the left and N_R be the number of points to the right of (a,2).
Here, an example simulation is shown.
| Trial | N_L | N_R |
|---|---|---|
| 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 N_L values.
8+7+9+7+6+6+7+9+6+7/10 = 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. 7.2/10 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. a/1 The ratio obtained using the simulation data 7.210 is an estimate of the probability obtained using geometric measures a1. With this in mind, the value of a can be calculated.
Therefore, a, the value of sqrt(2), is about 1.44. Notice that this is close to the value of sqrt(2). sqrt(2) = 1.414213 ...
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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.
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Example
Estimating the Area Under the Graph of y=x^2
Using a similar thought process, Maya now attempts to approximate the area under the graph of y=x^2 between 0 and 1. The following applet generates 100 random points inside a unit square.
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 N_U of random points under the graph of y=x^2 after each trial. The number of random points above the graph N_A will also be recorded. Recall that N_A is the total number of random points subtracted by N_U.
The results can be recorded in a table. Here is an example table.
| Trial | N_U | N_A |
|---|---|---|
| 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 N_U values.
Sum ofN_U Values/Number ofN_U Values
SubstituteValues
Substitute values
35+39+34+33+31+28+27+37+33+35/10
33.2
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. 33.2/100 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 A_U to the area of the unit square 1. A_U/1 The ratio obtained using the simulation data 33.2100 is an estimate of the probability obtained using geometric measures A_U1. With this in mind, the value of A_U can be calculated.
Therefore, the area under the graph of y=x^2 from 0 to 1 is about 0.332 square units. This result is close to the exact value of the area.
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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.
Recall the rules of the game. ∙ & If 1, 2, 3, or4appear, pay 2 coins ∙ & If 5 or6appear, 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
When the results are as shown in the table, they will get 27 coins.
b 
Here is an example table.
It looks like they will get 22.1 coins on average.
By repeating the procedure in Part A ten times, the number of coins Vincenzo can get or give after each trial can be recorded.
| Trial | Number of Coins |
|---|---|
| 1 | 27 |
| 2 | - 1 |
| 3 | 41 |
| 4 | 34 |
| 5 | 6 |
| 6 | 69 |
| 7 | - 8 |
| 8 | - 8 |
| 9 | 34 |
| 10 | 27 |
To find the average of these values, add them and divide the sum by 10.
27 + (- 1)+41+34+6+69+(- 8)+(- 8)+34+27/10
AddSubTerms
Add and subtract terms
221/10
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 16. Therefore, in theory, the frequency of the possible outcomes should be equal to 60 * 16=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
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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, &if1,2,3,or4appear, 5, &if5or6appear. 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.
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. Expected Value ofX: 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)^n x_i * P(x_i)
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]
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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.
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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.
X = 1 if the answer is correct - 0.5 if the answer is incorrect
Since there are 5 choices for each question, the probability of selecting the right answer is 15 and the probability of selecting an incorrect one is 45. The following table shows probabilities.
| X | 1 | - 0.5 |
|---|---|---|
| P(X) | 1/5 | 4/5 |
From here, the expected value of the random variable X can be calculated.
E(X) =x_1 * P(x_1)+ x_2 * P(x_2)
SubstituteValues
Substitute values
E(X) = 1 * 1/5 + (- 0.5) * 4/5
▼
Evaluate right-hand side
MoveLeftFacToNum
a*b/c= a* b/c
E(X) = 1/5 + - 2/5
AddNeg
a+(- b)=a-b
E(X) = - 1/5
CalcQuot
Calculate quotient
E(X) = - 0.2
Because the expected value for each guess is -0.2, Jordan would lose 0.2 points for each guess. Therefore, she should not guess.
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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(selecting4females) = 1/35
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.
Note that there is only one way to select 4 females out of the seven. 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. _nC_r=n!/r!(n-r)! By substituting 7 for n and 4 for r, the number of combinations can be calculated.
_nC_r=n!/r!(n-r)!
SubstituteII
n= 7, r= 4
_7C_4=7!/4!( 7- 4)!
▼
Evaluate right-hand side
SubTerm
Subtract term
_7C_4=7!/4!* 3!
Write as a product
_7C_4=7* 6 * 5 * 4!/4!* 3!
CancelCommonFac
Cancel out common factors
_7C_4=7* 6 * 5 * 4!/4!* 3!
SimpQuot
Simplify quotient
_7C_4=7* 6 * 5/3!
Write as a product
_7C_4=7* 6 * 5/3*2*1
Multiply
Multiply
_7C_4=210/6
CalcQuot
Calculate quotient
_7C_4=35
There are 35 ways of selecting 4 people out of 7. This is the number of possible outcomes. Therefore, the probability of selecting 4 females is given by the ratio of 1 to 35. P(selecting4females) = 1/35
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 selected 2 if two females are selected 3 if three females are selected 4 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 & and& 3 Males ⇓ & ⇓ & ⇓ _4C_1 & * & _3C_3 Here, _4C_1 represents the number of possible ways 1 female is selected out of 4 females and _3C_3 represents the number of possible ways 3 males are selected out of 3 males. Compute the value of the product.
_4C_1 * _3C_3
▼
Evaluate
Rewrite
Rewrite _4C_1 as 4!/1!(4-1)!
4!/1!(4-1)! * _3C_3
Rewrite
Rewrite _3C_3 as 3!/3!(3-3)!
4!/1!(4-1)! * 3!/3!(3-3)!
SubTerm
Subtract term
4!/1!* 3! * 3!/3!* 0!
MultFrac
Multiply fractions
4! * 3!/1!* 3! * 3! * 0!
Write as a product
4 * 3! * 3!/1!* 3! * 3! * 0!
CancelCommonFac
Cancel out common factors
4 * 3! * 3!/1!* 3! * 3! * 0!
SimpQuot
Simplify quotient
4/1!* 0!
1!=1
4/1 * 0!
0!=1
4/1 * 1
MultByOne
a * 1=a
4/1
DivByOne
a/1=a
4
Therefore, there are 4 possible ways to select 1 female and 3 males. Similarly, the rest of the possible cases can be found.
| 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 |
| _4C_1 * _3C_3 | _4C_2 * _3C_2 | _4C_3 * _3C_1 | _4C_4 * _3C_0 |
| 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) | 4/35 | 18/35 | 12/35 | 1/35 |
From here, the expected value of the random variable X can be computed.
E(X) = x_1 * P(x_1)+ x_2 * P(x_2)+ x_3 * P(x_3)+ x_4 * P(x_4)
SubstituteValues
Substitute values
E(X) = 1 * 4/35 + 2 * 18/35 + 3 * 12/35 + 4 * 1/35
▼
Evaluate right-hand side
MoveLeftFacToNum
a*b/c= a* b/c
E(X) = 4/35 + 36/35 + 36/35 + 4/35
AddFrac
Add fractions
E(X) = 80/35
CalcQuot
Calculate quotient
E(X) = 2.285714 ...
RoundInt
Round to nearest integer
E(X) ≈ 2
This equation means that when 4 people are randomly selected from a group of 3 males and 4 females, approximately 2 are expected to be female. Therefore, selecting only four females, in this case, is unlikely.
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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 |
Which cryptocurrency — Coin Q or Coin R — should Kevin choose? Justify his selection.
{"type":"choice","form":{"alts":["Coin Q","Coin R"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":1}
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 |
To calculate the expected value, multiply each X-value by its probability.
E(X) = x_1 * P(x_1) + x_2 * P(x_2) + x_3 * P(x_3)
SubstituteValues
Substitute values
E(X) = - 25 * 0.4 + 10 * 0.2 + 50 * 0.4
▼
Evaluate right-hand side
E(X) = 12
Kevin can expect to gain $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 |
From here, the expected value of the random variable Y can be calculated.
E(Y) = y_1 * P(y_1) + y_2 * P(y_2) + y_3 * P(y_3)
SubstituteValues
Substitute values
E(Y) = - 25 * 0.1 + 10 * 0.75 + 50 * 0.15
▼
Evaluate right-hand side
E(Y) = 12.5
Kevin can expect to gain $12.5 from Coin R.
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) & > E(X) 12.5 & > 12
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Examining Decisions
Exercises
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{"type":"choice","form":{"alts":["LaShay","Her classmates","Neither"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}
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The statement no more than 3
is every outcome that sum to either 1, 2 or 3. However, since the sum of two dice is at least 2, a player can only win when the sum of the dice is 2 or 3. Let's draw the sample space and count all of the outcomes where the sum is either 2 or 3.
We have 36 possible outcomes and three of them have a sum of either 2 or 3. With this information, we can calculate the probability of rolling a sum of no more than 3 with two dice. P(no more than 3) & =3/36, or 1/12
We will first define a random variable that shows the possible gains a player has in the game. X = $10, if the sum is no more than 3. $0, otherwise. From Part A, we know that a player has a probability of 112 to win $10. Therefore, 1112 is the probability of a player winning nothing. Then, the expected value of X can be found as follows.
On average, a player will get about $0.83. However, since the game costs one dollar to play, a player will actually lose $0.17 per game. 0.83-1 = - 0.17 On the other hand, a player's loss will be LaShay's gain. Therefore, LaShay will make a profit of $0.17 per game. Hey, who said house rules are fair?
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Hint & Answer
S
Solution
Prêmio Casino in Vegas is designing a game where a player has to spin the following wheel twice. If a player lands on the same color on both spins, the player will win a cash prize.
It costs $5 to play the game. To entice players to play, they have to make it somewhat fair. If Prêmio Casino wants to offer the prize in a whole dollar amount, what is the greatest amount they can offer to make the game profitable for themselves?
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The casino has to make the game somewhat fair in order to entice players to try their luck. At the same time, they want to make money for themselves, which means the game must be set in a way that the expected value for a player is still negative.
Determining Probabilities
Notice that we have three sectors, two have central angles of 135^(∘) and one has a central angle of 90^(∘). By dividing these central angles by 360^(∘), we get the probability of landing on each of the colors. P(red)&=135^(∘)/360^(∘)=3/8 [1em] P(blue)&=135^(∘)/360^(∘)=3/8 [1em] P(green)&=90^(∘)/360^(∘)=1/4
Illustrating the Sample Space
We have three possible outcomes for each spin — red, green, or blue. Since we must make two spins, there is a total of nine possible outcomes. Let's make a tree diagram to show the sample space and mark the paths which result in a win with their related probabilities.
As we can see, there are three paths that results in a win. By the Multiplication Rule of Probability, we can find the probability of each of these outcomes. P(two red)&=3/8* 3/8 = 9/64 [1em] P(two blue)&=3/8* 3/8 = 9/64 [1em] P(two green)&=1/4* 1/4 = 1/16 If we add these probabilities, we get the total probability of winning.
The probability of winning is 1132.
Expected Value
Let A be the amount of the prize. We can define a random variable that shows the possible gains of a player as follows.
We have found that a player has a probability of 1132 to win $A. By the Complement Rule, the probability of its complement is 1 minus 1132. 1-11/32 = 21/32 Therefore, 2132 is the probability of a player winning nothing. Then, the expected value of X can be found as follows.
On average, a player will get $ 11A32. Recall that the game costs $5 to play. For the game to be fair, the difference between the initial cost and the expected value should be zero. E(X)-5 = 0 We have found the expected value of X in terms of the prize. Let's substitute it into the above equation and solve.
If the prize is more than this number, the casino would, on average, lose money. Therefore, we must round down to $14. When the prize is $14, the expected value of X becomes less than 5, which is the cost for playing the game. E(X)- 5 & = 11* 14/32 - 5 [0.8em] & = 4.8125 -5 & = - 0.1875
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Hint & Answer
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Solution
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Let X be a random variable whose values depend on the outcomes of the spinner. We are given that x=8. Therefore, the random variable X can be written as follows. X = 32 &if it stops on blue sector 8 &if it stops on purple sector - 16 &if it stops on red sector - 8 &if it stops on orange sector To find the expected value of X, we need to determine the probability of spinning each sector.
Determining Probabilities
The probability of spinning each sector is the ratio of the central angle to 360^(∘). Let's first determine each of the central angles.
We see that the blue sector is a semicircle which means it has a central angle of 180^(∘). Since both red and orange sectors have the same probability of occurring, the measures of their central angles must be equal. 180^(∘)+ 90^(∘)+ m∠ θ + m∠ θ=360^(∘) ⇓ m∠ θ=45^(∘) Let's add the central angles to the spinner.
Now, we can determine probabilities of spinning each sector.
| Sector | Blue | Purple | Red | Orange |
|---|---|---|---|---|
| Probability | 180^(∘)/360^(∘)=1/2 | 90^(∘)/360^(∘)=1/4 | 45^(∘)/360^(∘)=1/8 | 45^(∘)/360^(∘)=1/8 |
Expected Value
Let's make a table to represent the probability distribution of the random variable X.
| X | 32 | 8 | - 16 | - 8 |
|---|---|---|---|---|
| P(X) | 1/2 | 1/4 | 1/8 | 1/8 |
To find E(X), we will find the sum of the products of the probability of spinning a certain sector by the value shown on the sector.
When the value of x is 8, a player will get 15 points on average.
We see that only the value on the purple sector became - 12. We can define another random variable for it. Y = 32 &if it stops on blue sector - 12 &if it stops on purple sector - 16 &if it stops on red sector - 8 &if it stops on orange sector Since none of the areas of the sectors changed, each sector will have the same probability found in Part A. Therefore, the probability distribution of the random variable Y can be represented by the following table.
| Y | 32 | - 12 | - 16 | - 8 |
|---|---|---|---|---|
| P(Y) | 1/2 | 1/4 | 1/8 | 1/8 |
Now, we can calculate E(Y).
When the value of x is - 12, a player will get 10 points on average.
This time we are asked to find the value of x that makes the expected value 0. Again, we will start by defining a random variable. Z = 32 &if it stops on blue sector x &if it stops on purple sector - 16 &if it stops on red sector - 8 &if it stops on orange sector The probabilities of the possible values of Z are also the same as the probabilities found in Part A because no sector's area has changed.
| Z | 32 | x | - 16 | - 8 |
|---|---|---|---|---|
| P(Z) | 1/2 | 1/4 | 1/8 | 1/8 |
By using the expected value formula and substituting E(Z)=0, we can find the value of x.
To get 0 points on average, the value of x should be - 52.
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{"type":"choice","form":{"alts":["Left spinner","Right spinner","Both give the same value"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":1}
{"type":"choice","form":{"alts":["Left spinner","Right spinner","Both give the same value"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}
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We will first determine the probabilities of the sectors and then calculate the expected values of the spinners.
Determining Probabilities
To determine the probability of spinning a sector, we divide the central angle of each sector by 360^(∘).
Since the sectors of the left spinner measures 90^(∘), each sector of it has the same probability p_l of occurring. p_l= 90^(∘)/360^(∘) = 1/4 Similarly, each sector of the right spinner has the same probability p_r of occurring. p_r = 180^(∘)/360^(∘) = 1/2 Now, the expected values of the spinner can be calculated.
Expected Values of the Spinners
Let's recall the probabilities of the sectors of the left spinner and their related values.
| Left Spinner | ||||
|---|---|---|---|---|
| Values on Sectors | 20 | 0 | 0 | 0 |
| Probability | 1/4 | 1/4 | 1/4 | 1/4 |
The expected value of a spinner is the sum of the products of the value on each sector by the probability of spinning the sector. EV_l & = 20 * 1/4 + 0 * 1/4 + 0 * 1/4 + 0 * 1/4 & = 5 The expected value of the left spinner is 5. Similarly, using the table for the right spinner, its expected value can be found.
| Right Spinner | ||
|---|---|---|
| Values on Sectors | 8 | 4 |
| Probability | 1/2 | 1/2 |
Let's calculate it. EV_r & = 8 * 1/2 + 4 * 1/2 & = 6 Since the expected value of the right spinner is greater, the right spinner should be chosen. EV_r =6 > 5= EV_l
If the sector valued at 20 was changed to 40, the expected value of the spinner will change. We can use the probabilities of the sectors found in Part A because the measures of the sectors did not change.
| Left Spinner | ||||
|---|---|---|---|---|
| Values on Sectors | 40 | 0 | 0 | 0 |
| Probability | 1/4 | 1/4 | 1/4 | 1/4 |
With this new value, let's calculate the expected value. EV_l& = 40 * 1/4 + 0 * 1/4 + 0 * 1/4 + 0 * 1/4 & = 10 Since the expected value from the left spinner is now 10, the left spinner should be chosen if we want to choose the spinner with the greatest expected value. EV_l =10 > 6= EV_r
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