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{{ printedBook.courseTrack.name }} {{ printedBook.name }} Examining Decisions

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.

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 to 5. Each point on the number line represents a random number. Move the interval to see how the percentage of the random numbers inside the interval changes. Create some cases where ten points are randomly distributed along the number line. Does an interval which contains half of the points also come out to be about half of the range of the number line? Example Using Random Points to Estimate the Value of 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 using probability concepts. To do so, Maya draws the graphs of y=x2 and y=2 for 0x2 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 Answer a Example Table: Trial Number of points to the left of Number of points to the right of 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 Then, the intersection point of the graphs is the point (a,a2), 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 be the number of points to the left and be the number of points to the right of (a,2). Here, an example simulation is shown. Trial 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 values. 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. 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. The ratio obtained using the simulation data is an estimate of the probability obtained using geometric measures With this in mind, the value of a can be calculated. Therefore, a, the value of is about 1.44. Notice that this is close to the value of 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. 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=x2 Using a similar thought process, Maya now attempts to approximate the area under the graph of y=x2 between 0 and 1. The following applet generates 100 random points inside a unit square. 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 of random points under the graph of y=x2 after each trial. The number of random points above the graph will also be recorded. Recall that is the total number of random points subtracted by The results can be recorded in a table. Here is an example table. Trial 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 values. Evaluate 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. 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 to the area of the unit square 1. The ratio obtained using the simulation data is an estimate of the probability obtained using geometric measures With this in mind, the value of can be calculated. Therefore, the area under the graph of y=x2 from 0 to 1 is about 0.332 square units. This result is close to the exact value of the area. Example Rolling a Die 60 Times Vincenzo and his friends decide to play a game. They roll a die up to 60 times. They have to pay 2 coins if 1, 2, 3, or 4 appears. They get 5 coins if 5 or 6 appears. a Find the number of coins that Vincenzo and his friends obtain at the end of the first trial. 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 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. 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 -22182216+45+60 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. 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 Therefore, in theory, the frequency of the possible outcomes should be equal to 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 -20202020+50+50 20 Discussion Random Variable and Expected Value Consider an experiment involving the rolling of a die 60 times in which the outcome of each roll determines the number of coins that will be received or lost. If 5 or 6 appears, 5 coins are received. Otherwise, 2 coins are lost. This situation can be represented by a variable, call it X. 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. 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. 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. 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. 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. 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. 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. Since there are 5 choices for each question, the probability of selecting the right answer is and the probability of selecting an incorrect one is The following table shows probabilities.  X P(X) 1 -0.5 From here, the expected value of the random variable X can be calculated. E(X)=x1P(x1)+x2P(x2) Evaluate right-hand side 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 a b 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. 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. By substituting 7 for n and 4 for r, the number of combinations can be calculated. Evaluate right-hand side Write as a product Write as a product 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. 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. 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. Here, 4C1 represents the number of possible ways 1 female is selected out of 4 females and 3C3 represents the number of possible ways 3 males are selected out of 3 males. Compute the value of the product. 4C13C3 Evaluate Write as a product 1!=1 0!=1 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 4C13C3 4C23C2 4C33C1 4C43C0 41 63 43 11 4 18 12 1 Considering the above table, the probabilities can be determined.  X P(X) 1 2 3 4 From here, the expected value of the random variable X can be computed. E(X)=x1P(x1)+x2P(x2)+x3P(x3)+x4P(x4) Evaluate right-hand side 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 P(X) -25 10 50 0.4 0.2 0.4
To calculate the expected value, multiply each X-value by its probability.
E(X)=x1P(x1)+x2P(x2)+x3P(x3)
E(X)=-250.4+100.2+500.4
Evaluate right-hand side
E(X)=-10+2+20
E(X)=12

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.