Unlocking Decision-Making Through Probability and Expected Value

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.

Probability
Simulation
Theoretical Probability
Combination
Combination Formula
Fundamental Counting Principle

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.

Explore

Random Points on a Number Line

The applet shows the number line from

0

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.

Number line from 0 to 5 and random points on it

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.

Generating random points on a number line

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 = x^{2}

and

y = 2

for

0 \leq x \leq 2

on an applet that generates ten random points on

y = 2 .

Generating ten random points on a horizontal line

Help Maya answer the following questions.

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, a^{2}),

(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.

\frac{8 + 7 + 9 + 7 + 6 + 6 + 7 + 9 + 6 + 7}{1 0} = 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

10 .

\frac{7 . 2}{1 0}

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.

\frac{a}{1}

The ratio obtained using the simulation data

\frac{7 . 2}{1 0}

is an estimate of the probability obtained using geometric measures

\frac{a}{1} .

With this in mind, the value of

a

can be calculated.

\frac{a}{2} \approx \frac{7 . 2}{1 0}

MultEqn

$LHS \cdot 2 = RHS \cdot 2$

a \approx \frac{1 4 . 4}{1 0}

WriteAsFrac

Write as a fraction

a \approx 1.44

Therefore,

a,

the value of

2,

is about

1.44 .

Notice that this is close to the value of

2 .

2 = 1.414213 \dots

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.

Generating random points inside a 5x5 square and a movable unit square

Determine around what number does the percent take values. How can this number be related to the areas of the squares?

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.

100 random points distributed over 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.

\frac{Sum of N _{U} Values}{Number of N _{U} Values}

SubstituteValues

Substitute values

\frac{3 5 + 3 9 + 3 4 + 3 3 + 3 1 + 2 8 + 2 7 + 3 7 + 3 3 + 3 5}{1 0}

▼

Evaluate

AddTerms

Add terms

\frac{3 3 2}{1 0}

CalcQuot

Calculate quotient

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

100 .

\frac{3 3 . 2}{1 0 0}

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 .

\frac{A _{U}}{1}

The ratio obtained using the simulation data

\frac{3 3 . 2}{1 0 0}

is an estimate of the probability obtained using geometric measures

\frac{A _{U}}{1} .

With this in mind, the value of

A_{U}

can be calculated.

\frac{A _{U}}{1} \approx \frac{3 3 . 2}{1 0 0}

DivByOne

$\frac{a}{1} = a$

A_{U} \approx \frac{3 3 . 2}{1 0 0}

CalcQuot

Calculate quotient

A_{U} \approx 0.332

Therefore, the area under the graph of

y = x^{2}

from

0

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.

Table showing the frequencies of outcomes when a die is rolled 60 times

a Find the number of coins that Vincenzo and his friends obtain at the end of the first trial.

b Simulate the game

10

times. Record the number of coins that they get or give after each trial in a table. Then, calculate the average of the numbers.

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

- 2

and the frequencies of

5

and

6

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, 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

- 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 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.

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 .

\frac{2 7 + ( - 1 ) + 4 1 + 3 4 + 6 + 6 9 + ( - 8 ) + ( - 8 ) + 3 4 + 2 7}{1 0}

AddSubTerms

Add and subtract terms

\frac{2 2 1}{1 0}

CalcQuot

Calculate quotient

22.1

It looks like they will get

22.1

coins on average.

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

\frac{1}{6} .

Therefore, in theory, the frequency of the possible outcomes should be equal to

60 \cdot \frac{1}{6} = 10 .

Table showing the frequencies of outcomes of the experiment when repeated infinitely many times

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

- 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

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.

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.

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.

$E (X) = i = 1 \sum n x_{i} \cdot P (x_{i})$

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.

E X, E (X), E (X), E [X], E (X), E [X]

Expected value of a random variable, or the average of a random variable, tells what is expected in the long run, and therefore better decisions can be made.

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.

Probability of each possible value of random variable X

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

\frac{1}{5}

and the probability of selecting an incorrect one is

\frac{4}{5} .

The following table shows probabilities.

$X$	$1$	$- 0.5$
$P (X)$	$\frac{1}{5}$	$\frac{4}{5}$

From here, the expected value of the random variable

X

can be calculated.

E (X) = x_{1} \cdot P (x_{1}) + x_{2} \cdot P (x_{2})

SubstituteValues

Substitute values

E (X) = 1 \cdot \frac{1}{5} + (- 0.5) \cdot \frac{4}{5}

▼

Evaluate right-hand side

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

E (X) = \frac{1}{5} + \frac{- 2}{5}

AddNeg

$a + (- b) = a - b$

E (X) = \frac{- 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.

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

P (selecting 4 females) = \frac{1}{3 5}

E (X) \approx 2

Hint

a The probability of an event is the ratio of the number of favorable outcomes to the number of possible outcomes.

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.

_{n} C_{r} = \frac{n !}{r ! ( n - r ) !}

By substituting

7

for

n

and

4

for

r,

the number of combinations can be calculated.

_{n} C_{r} = \frac{n !}{r ! ( n - r ) !}

SubstituteII

$n = 7$ , $r = 4$

_{7} C_{4} = \frac{7 !}{4 ! ( 7 - 4 ) !}

▼

Evaluate right-hand side

SubTerm

Subtract term

_{7} C_{4} = \frac{7 !}{4 ! \cdot 3 !}

Write as a product

_{7} C_{4} = \frac{7 \cdot 6 \cdot 5 \cdot 4 !}{4 ! \cdot 3 !}

CancelCommonFac

Cancel out common factors

_{7} C_{4} = \frac{7 \cdot 6 \cdot 5 \cdot 4 !}{4 ! \cdot 3 !}

SimpQuot

Simplify quotient

_{7} C_{4} = \frac{7 \cdot 6 \cdot 5}{3 !}

Write as a product

_{7} C_{4} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1}

Multiply

_{7} C_{4} = \frac{2 1 0}{6}

CalcQuot

Calculate quotient

_{7} C_{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

35 .

P (selecting 4 females) = \frac{1}{3 5}

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 ⇓_{4} C_{1} and ⇓ \cdot 3 Males ⇓_{3} C_{3}

Here,

_{4} C_{1}

represents the number of possible ways

1

female is selected out of

4

females and

_{3} C_{3}

represents the number of possible ways

3

males are selected out of

3

males. Compute the value of the product.

_{4} C_{1} \cdot_{3} C_{3}

▼

Evaluate

Rewrite

Rewrite $_{4} C_{1}$ as $\frac{4 !}{1 ! ( 4 - 1 ) !}$

\frac{4 !}{1 ! ( 4 - 1 ) !} \cdot_{3} C_{3}

Rewrite

Rewrite $_{3} C_{3}$ as $\frac{3 !}{3 ! ( 3 - 3 ) !}$

\frac{4 !}{1 ! ( 4 - 1 ) !} \cdot \frac{3 !}{3 ! ( 3 - 3 ) !}

SubTerm

Subtract term

\frac{4 !}{1 ! \cdot 3 !} \cdot \frac{3 !}{3 ! \cdot 0 !}

MultFrac

Multiply fractions

\frac{4 ! \cdot 3 !}{1 ! \cdot 3 ! \cdot 3 ! \cdot 0 !}

Write as a product

\frac{4 \cdot 3 ! \cdot 3 !}{1 ! \cdot 3 ! \cdot 3 ! \cdot 0 !}

CancelCommonFac

Cancel out common factors

\frac{4 \cdot 3 ! \cdot 3 !}{1 ! \cdot 3 ! \cdot 3 ! \cdot 0 !}

SimpQuot

Simplify quotient

\frac{4}{1 ! \cdot 0 !}

$1! = 1$

\frac{4}{1 \cdot 0 !}

$0! = 1$

\frac{4}{1 \cdot 1}

MultByOne

$a \cdot 1 = a$

\frac{4}{1}

DivByOne

$\frac{a}{1} = a$

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
$_{4} C_{1} \cdot_{3} C_{3}$	$_{4} C_{2} \cdot_{3} C_{2}$	$_{4} C_{3} \cdot_{3} C_{1}$	$_{4} C_{4} \cdot_{3} C_{0}$
$4 \cdot 1$	$6 \cdot 3$	$4 \cdot 3$	$1 \cdot 1$
$4$	$18$	$12$	$1$

Considering the above table, the probabilities can be determined.

$X$	$1$	$2$	$3$	$4$
$P (X)$	$\frac{4}{3 5}$	$\frac{1 8}{3 5}$	$\frac{1 2}{3 5}$	$\frac{1}{3 5}$

From here, the expected value of the random variable

X

can be computed.

E (X) = x_{1} \cdot P (x_{1}) + x_{2} \cdot P (x_{2}) + x_{3} \cdot P (x_{3}) + x_{4} \cdot P (x_{4})

SubstituteValues

Substitute values

E (X) = 1 \cdot \frac{4}{3 5} + 2 \cdot \frac{1 8}{3 5} + 3 \cdot \frac{1 2}{3 5} + 4 \cdot \frac{1}{3 5}

▼

Evaluate right-hand side

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

E (X) = \frac{4}{3 5} + \frac{3 6}{3 5} + \frac{3 6}{3 5} + \frac{4}{3 5}

AddFrac

Add fractions

E (X) = \frac{8 0}{3 5}

CalcQuot

Calculate quotient

E (X) = 2.285714 \dots

RoundInt

Round to nearest integer

E (X) \approx 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.

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.

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} \cdot P (x_{1}) + x_{2} \cdot P (x_{2}) + x_{3} \cdot P (x_{3})

SubstituteValues

Substitute values

E (X) = - 25 \cdot 0.4 + 10 \cdot 0.2 + 50 \cdot 0.4

▼

Evaluate right-hand side

Multiply

E (X) = - 10 + 2 + 20

AddSubTerms

Add and subtract terms

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} \cdot P (y_{1}) + y_{2} \cdot P (y_{2}) + y_{3} \cdot P (y_{3})

SubstituteValues

Substitute values

E (Y) = - 25 \cdot 0.1 + 10 \cdot 0.75 + 50 \cdot 0.15

▼

Evaluate right-hand side

Multiply

E (Y) = - 2.5 + 7.5 + 7.5

AddSubTerms

Add and subtract terms

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) 12.5 > E (X) > 12

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Catch-Up and Review

Answer

Hint

Solution

Answer

Hint

Solution

Answer

Hint

Solution

Answer

Hint

Solution

Answer

Hint

Solution

Hint

Solution

The Expected Value of Gain or Loss for Coin Q

The Expected Value of Gain or Loss for Coin R

Conclusion

The Expected Value of Gain or Loss for Coin $Q$

The Expected Value of Gain or Loss for Coin $R$