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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Lesson Settings & Tools
| | 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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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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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?
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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. 
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 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.
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 ...
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).
| 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.
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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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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.
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.
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.
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.
| 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
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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.
( - 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 |
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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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
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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.
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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.
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.
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]Pop Quiz
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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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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 |
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
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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.
_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
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.
_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
| 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 |
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
Closure
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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 |
{"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 |
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
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) = 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
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
Examining Decisions
Exercise 2.1
Kriz is creating a game that requires the contestant to roll two dice. The contestant can win the following amount of prize money.
| Sum of the Roll | Prize Money |
|---|---|
| 10 | $10 |
| 11 | $40 |
| 12 | $100 |
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We will start by defining a random variable whose values depend on the outcomes of a roll of two dice. Considering the table given in the prompt, we can define the random variable and outcomes as follows.
X = 100 &if the sum is12 40 &if the sum is11 10 &if the sum is10 0 &otherwise To find the expected value of X, we need to determine the probabilities of rolling the different outcomes.
Determining Probabilities
To calculate the probabilities of rolling a sum of 10, 11 and 12, we will begin by drawing a sample space and count all of the outcomes where the sum is 10, 11 and 12.
We have 36 possible outcomes. Three of the outcomes add to 10, two add to 11 and one adds to 12. With this information, we can calculate the probability of rolling these outcomes. P(sum of 10)=3/36 [1em] P(sum of 11)=2/36 [1em] P(sum of 12)=1/36 The probability of the remaining outcomes is, therefore, 3036.
Calculating the Expected Value
Now, we can calculate the expected value of X by multiplying the probability of each event by their respective prize.
This is the amount of money that a player will win on average. For the game to be completely fair, Kriz should charge roughly that amount. However, for Kriz turn the game in their own favor, and charge a whole number of dollars, Kriz would have to round up to $6. E(X) ≈ 5.83 ⇒ Charge = 6 It is also noteworthy that this charge makes the expected value for playing the game negative. E(X) - Charge & = 5.83-6 & = - 0.17
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A wheel of fortune is divided into a blue and a green field.
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To determine the expected value of the game, we have to determine the probability of losing 10 million dollars. In other words, find the probability of landing on the green region three times in a row. To do so, we must first find the probability of stopping on the blue region one time.
Probability of Landing on the Blue Region
Let's call the event of the wheel stopping on the blue and green region as B and G, respectively. We know that the probability of spinning blue three times in a row is 34.3 %. Let's illustrate all possible outcomes and the given probability in a tree diagram.
According to the Multiplication Rule of Probability, the probability of several events happening is the product of their individual probabilities. Therefore, the probability of landing on the blue region three times in a row is the product of the probabilities of landing on the blue region one time. P(three blue)=P(B)* P(B)* P(B) Since we know that P(three blue) is 34.3 %, or 0.343, we can solve for P(B) by substituting this into the equation.
The spinner stops on the blue region with probability 0.7, or 70 %.
Probability of Landing on the Green Region Three Times in a Row
If the probability of the wheel showing blue is 70 %, then the probability of the wheel showing green is the complement of this. Therefore, by the Complement Rule, its probability can be calculated as follows. 1-0.7=0.3, or 30 % Let's show the probability of landing on the green region in the tree diagram.
Again, we can use the Multiplication Rule of Probability to determine the probability of losing 10 million dollars, or the probability of landing on green three times in a row.
Expected Value
By adding the products of winning and losing with the given price and punishment, we can determine the expected value of the game.
The expected value is $73 000.
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Mark designs a dartboard painted in three colors. He assigns a different point value to each zone as shown below.
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We can start by defining a random variable to represent the possible points Mark can get. X = 18 &if the blue region is hit 63 &if the white region is hit 90 &if the red region is hit We are asked to find the average of the points Mark can score, in other words, the expected value of X. To find it, we first need to determine the areas of the colored regions. Then, using the geometric probability, we will calculate the probability of hitting each colored part of the board.
Determining Areas
We see that the dart board has a diameter of 24 inches, which means it has a radius of 12 inches. With this information, we can find the area of the dartboard, A_(DB).
Let's also determine the area of the red bullseye A_R, which has a diameter of 8 inches and therefore a radius of 4 inches.
We must also determine the blue and white regions. If we disregard the bullseye, we see that they are equally large as all sectors have the same central angle. Since there are two of each color, they each make a semicircle. Therefore, we can determine their area by dividing the area of the dart board by 2. 144π/2=72π However, they do not make a whole semicircle as the bullseye covers part of them. Half of the bullseye covers part of the white regions, and the other half covers the blue regions. If we rearrange the sectors, we can see that more clearly.
Therefore, we can find the area of the blue and white region by subtracting half the area of the bullseye from each of them. A_B&=144π/2=72π-8π=64π [1em] A_W&=144π/2=72π-8π=64π
Determining Probabilities
By the geometric probability, we know that the probability of hitting a colored region is the ratio of the area of the region to the area of the dartboard. Let's find the probabilities.
| Region | Blue Region | White Region | Red Region |
|---|---|---|---|
| Probability | 64π/144π=4/9 | 64π/144π=4/9 | 16π/144π=1/9 |
Expected Value
Using the above probabilities, the probability distribution of the random variable X can be shown as follows.
| X | 18 | 63 | 90 |
|---|---|---|---|
| P(X) | 4/9 | 4/9 | 1/9 |
Finally, we can find the expected value. To do so, we will multiply every possible X-value by its probability and add them.
Therefore, the player would score 46 points on average.
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Examining Decisions
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