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{{ printedBook.courseTrack.name }} {{ printedBook.name }} 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.**

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 <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.8777699999999999em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\">Q<\/span><\/span><\/span><\/span>","Coin <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.00773em;\">R<\/span><\/span><\/span><\/span>"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":1}

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?

When ten points are distributed randomly on a line segment, half of the points are expected to be on each half of the segment.

### Answer

### Hint

### Solution

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 $(a,2).$

$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… $

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 $0≤x≤2$ on an applet that generates ten random points on $y=2.$

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 $2 .$

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 $1.44$

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?

a Let $a$ represent the value of $2 .$ Then, the intersection point of the graphs is the point $(a,a_{2}),$ or $(a,2).$

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.

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.

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

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.
### Answer

### Hint

### Solution

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 $100.$
$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 $A_{U}$ to the area of the unit square $1.$
$1A_{U} $
The ratio obtained using the simulation data $10033.2 $ is an estimate of the probability obtained using geometric measures $1A_{U} .$ 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 $1.$

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

a Count the number of points under the curve.

b Calculate the average of the numbers found.

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.

$Number ofN_{U}ValuesSum ofN_{U}Values $

SubstituteValues

Substitute values

$1035+39+34+33+31+28+27+37+33+35 $

$33.2$

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?

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:** $20$

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?

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.

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.

Here is an example table.

Trial | Number of Coins |
---|---|

$1$ | $27$ |

$2$ | $-1$ |

$3$ | $41$ |

$4$ | $34$ |

$5$ | $6$ |

$6$ | $69$ |

$7$ | $-8$ |

$8$ | $-8$ |

$9$ | $34$ |

$10$ | $27$ |

$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

$20$

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, call it $X.$ $X={-25 if1,2,3,or4appearif5or6appear $ The variable $X$ has a special name in probability.

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. A continuous random variable, on the other hand, takes any value within a continuous interval. Continuous random variables are usually measurements such as height, weight, and time.

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] $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.

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.

No, it is not advantageous to guess. See solution.

Define a random variable for the situation. Then, find the expected value of it.

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={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 $ |

$E(X)=x_{1}⋅P(x_{1})+x_{2}⋅P(x_{2})$

SubstituteValues

Substitute values

$E(X)=1⋅51 +(-0.5)⋅54 $

Evaluate right-hand side

MoveLeftFacToNum

$a⋅cb =ca⋅b $

$E(X)=51 +5-2 $

AddNeg

$a+(-b)=a−b$

$E(X)=5-1 $

CalcQuot

Calculate quotient

$E(X)=-0.2$

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.

a $P(selecting4females)=351 $

b $E(X)≈2$

b Start by defining a random variable for the possible number of females.

a Recall that the probability of an event $A$ is the ratio of the number of favorable outcomes to the number of possible outcomes.

$_{n}C_{r}=r!(n−r)!n! $

SubstituteII

$n=7$, $r=4$

$_{7}C_{4}=4!(7−4)!7! $

Evaluate right-hand side

SubTerm

Subtract term

$_{7}C_{4}=4!⋅3!7! $

Write as a product

$_{7}C_{4}=4!⋅3!7⋅6⋅5⋅4! $

CancelCommonFac

Cancel out common factors

$_{7}C_{4}=4!⋅3!7⋅6⋅5⋅4! $

SimpQuot

Simplify quotient

$_{7}C_{4}=3!7⋅6⋅5 $

Write as a product

$_{7}C_{4}=3⋅2⋅17⋅6⋅5 $

Multiply

Multiply

$_{7}C_{4}=6210 $

CalcQuot

Calculate quotient

$_{7}C_{4}=35$

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.

$_{4}C_{1}⋅_{3}C_{3}$

Evaluate

Rewrite

Rewrite $_{4}C_{1}$ as $1!(4−1)!4! $

$1!(4−1)!4! ⋅_{3}C_{3}$

Rewrite

Rewrite $_{3}C_{3}$ 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 |

$_{4}C_{1}⋅_{3}C_{3}$ | $_{4}C_{2}⋅_{3}C_{2}$ | $_{4}C_{3}⋅_{3}C_{1}$ | $_{4}C_{4}⋅_{3}C_{0}$ |

$4⋅1$ | $6⋅3$ | $4⋅3$ | $1⋅1$ |

$4$ | $18$ | $12$ | $1$ |

Considering the above table, the probabilities can be determined.

$X$ | $1$ | $2$ | $3$ | $4$ |
---|---|---|---|---|

$P(X)$ | $354 $ | $3518 $ | $3512 $ | $351 $ |

$E(X)=x_{1}⋅P(x_{1})+x_{2}⋅P(x_{2})+x_{3}⋅P(x_{3})+x_{4}⋅P(x_{4})$

SubstituteValues

Substitute values

$E(X)=1⋅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

$E(X)≈2$

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 $R$ | $0.10$ | $0.75$ | $0.15$ |

Start by calculating the expected value of the gain or loss for each cryptocurrency.

The expected value of gain or loss for each cryptocurrency will be calculated one at a time.

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

$E(X)=12$

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

$E(Y)=12.5$

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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