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| 12 Theory slides |
| 8 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
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.
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 |
Trial | NL | NR |
---|---|---|
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 |
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.
Help Maya estimate the area under the curve.
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 |
Trial | NU | NA |
---|---|---|
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 |
Substitute values
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.
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 |
Multiply
Add and subtract terms
Trial | Number of Coins |
---|---|
1 | 27 |
2 | -1 |
3 | 41 |
4 | 34 |
5 | 6 |
6 | 69 |
7 | -8 |
8 | -8 |
9 | 34 |
10 | 27 |
Add and subtract terms
Calculate quotient
Under these conditions, the number of coins Vincenzo can get will be equal to 20.
Multiply
Add and subtract terms
E(X)=i=1∑nxi⋅P(xi)
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.
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.
X | 1 | -0.5 |
---|---|---|
P(X) | 51 | 54 |
Substitute values
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.
n=7, r=4
Subtract term
Write as a product
Cancel out common factors
Simplify quotient
Write as a product
Multiply
Calculate quotient
Rewrite 4C1 as 1!(4−1)!4!
Rewrite 3C3 as 3!(3−3)!3!
Subtract term
Multiply fractions
Write as a product
Cancel out common factors
Simplify quotient
1!=1
0!=1
a⋅1=a
1a=a
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 |
4C1⋅3C3 | 4C2⋅3C2 | 4C3⋅3C1 | 4C4⋅3C0 |
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 |
Substitute values
a⋅cb=ca⋅b
Add fractions
Calculate quotient
Round to nearest integer
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 |
Substitute values
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 |
Substitute values
LaShay invites her classmates to play a game she has created. She instructs them to bet one dollar and roll a pair of dice. If the sum of the dice shows no more than 3, LaShay will give $10.
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?
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?
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.
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
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.
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
For the following spinner, landing on -16 and -8 are equally likely outcomes.
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.
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 |
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.
Assume that each spinner is divided into equal portions.
We will first determine the probabilities of the sectors and then calculate the expected values of the spinners.
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.
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