Here are a few recommended readings before getting started with this lesson.
A local $3-$on$-3$ beach volleyball tournament was hosted in South Beach. Heichi, Izabella, and Kevin pulled off a huge upset and won the tournament. The reward is a bicycle that flies as wind gusts propel the rider anywhere they so please to go. Everyone wants it, but, unfortunately, there is only one. Therefore, each player suggests a method to decide who gets to keep the highly coveted bicycle.
Considering the three suggestions, which method is fair?
Try to determine which game leads to a fair decision. In order to do this, complete the following steps.
A fair decision is a decision for which all events are equally likely to happen, meaning that the probabilities of the events are the same. Fair decisions can be made by using several methods, such as selecting names from a hat, flipping a coin, rolling a die, drawing a card from a standard deck of cards, using a random number generator, and so on.
Suppose that the captains of two baseball teams $A$ and $B,$ have to decide which team bats first. To do so, the captains will roll one die each, simultaneously.
One example of a fair method and one example of an unfair method will be presented for this situation.
The fairness of the mentioned methods will be explained one at a time.
This method suggests rolling two dice and calculating the sum of the numbers obtained. Since rolling each die has $6$ possible outcomes, the number of all possible combinations is $36.$
Recall that if the sum of the numbers is less than $7,$ then Team $A$ bats first. For a clearer understanding, highlight all the outcomes which satisfy this event.
After counting, there is a total of $15$ favorable outcomes. Using this information, the probability of Team A batting first can be calculated.$Favorable outcomes=15$, $Total outcomes=36$
$ba =b/3a/3 $
Calculate quotient
Round to ${\textstyle 2 \, \ifnumequal{2}{1}{\text{decimal}}{\text{decimals}}}$
Again, there are $15$ favorable outcomes for the event. Therefore, the probability that Team $B$ bats first is also about $0.42.$ Since two events, Team $A$ batting first and Team $B$ batting first, have equal probabilities, this method is fair.
This method suggests rolling two dice and multiplying the numbers obtained. Since rolling each die has $6$ possible outcomes, there are $36$ possible products.
According to this method, if the product of the numbers is even, then Team $A$ bats first. Hence, start by identifying the favorable outcomes which are the even numbers in the table.
There are $27$ favorable outcomes for Team $A$ out of $36$ possible outcomes. With this information, the probability of Team $A$ batting first can be calculated.$Favorable outcomes=9$, $Total outcomes=36$
$ba =b/9a/9 $
Calculate quotient
$Favorable outcomes=9$, $Total outcomes=36$
$ba =b/9a/9 $
Calculate quotient
In the last daring coin flip, Emily won and devoured the red velvet cake. The girls enjoyed the game so much that they decided to continue playing these games of probability just for fun. This time they will use a standard deck of cards.
Dominika will first remove all of the clubs and diamonds from the deck. Emily will then draw two cards in succession without replacing them. If she draws two cards with the same suit, then Emily wins. If she draws two face cards, then Dominika wins. For any other outcome, they draw again.
Try to determine if the game leads to a fair decision. To do so, complete the following steps.
Note that the event of drawing two hearts and the event of drawing two spades are mutually exclusive, therefore, the probability on the right-hand side equals the sum of the individual probabilities.
Since there are $26$ cards in total, the probability of getting a heart on the first draw is $2613 .$ If the first card is drawn is a heart, then only $12$ hearts remain in the deck, and because the cards are not being replaced there are $25$ remaining cards in total. Hence, the probability of drawing a second heart is $2512 .$ $P(♡first)P(♡second∣♡first) =2613 =2512 $ By the Multiplication Rule of Probability for dependent events, the probability of drawing two hearts equals the product of the two probabilities written above.Substitute values
$ba =b/13a/13 $
Multiply fractions
$ba =b/2a/2 $
Substitute values
Add fractions
Calculate quotient
Substitute values
$ba =b/2a/2 $
$ba =b/5a/5 $
Multiply fractions
Use a calculator
Round to ${\textstyle 3 \, \ifnumequal{3}{1}{\text{decimal}}{\text{decimals}}}$
$P(Emily wins)P(Dominika wins) =0.48≈0.046 $
As can be seen, the probability of Emily winning is greater than the probability of Dominika winning. $0.48>0.046 $ Oh snaps, the game is not fair! Also, note that if two face cards of the same suit are drawn, both Emily and Dominika will win, which is ambiguous. This also says that the game is not well set.
There are opportunities to make fair decisions in different contextual situations by the use of simulations.
A simulation is a model that imitates a real-life process or situation. In particular, a simulation can be used as a probability model to make predictions about real-life events. Also, the experimental probability can be estimated by simulating the events. $P(event)=Number of trialsNumber of times event occurs $
In general, simulations are used when actual trials of some experiment are impossible or unreasonable to conduct.For example, Mark and Tadeo want to eat dinner at different restaurants and they are not able to come together on a decision. It so happens that they both really like basketball. They then came up with an idea to make a free throw to determine the person who will choose the restaurant.
Mark only recently began practicing free throws, so the probability that Mark makes a free throw is $0.5$ — this is the success rate. Unfortunately, there is no basketball court nearby, so they decide to simulate the free throws using a random generator. $Made Free Throw:Missed Free Throw: 10 $ If the random number generator produces $1$ — meaning that Mark made a free throw — then they will go to the restaurant that Mark wants. Otherwise, they will go to the restaurant that Tadeo wants.
A company with $m$ employees wants to award a prize to $n$ employee(s) at random — it has been a good year. Propose a fair way to choose the random winners for the given values of $m$ and $n.$
Method I: Assign each person a number, write the numbers on slips of paper, and then draw one of them out of a hat.
Method II: Assign each person a number from $1$ to $8$ and then roll a fair octahedron die.
Method III: Flipp three unbiased coins and assign each employee one of $8$ possible outcomes.
Method I: Assign each person a number, write the numbers on pieces of paper, and then draw two pieces out of a hat.
Method II: Eight spades ranging from $2$ to $9$ are taken from a deck. The numbers are assigned to employees and then two cards are drawn without being replaced.
One possible method is to assign a number to each employee. Then the numbers should be written on pieces of paper and put in a hat. After shaking the hat, one piece of paper is randomly picked and the employee with the matching number wins the prize. This method has eight possible outcomes. ${1,2,3,4,5,6,7,8} $ Note that each number has an equal chance of being chosen. That means each employee has the same probability of winning the prize, which is $81 .$ Therefore, this method is fair.
Another possible method is to use an octahedron die — an eight-sided die. Each employee is assigned a number from $1$ to $8,$ and then a fair octahedron die is rolled. All possible outcomes of rolling a fair octahedron die can be listed as follows. ${1,2,3,4,5,6,7,8} $ Similar to the previous method, since the die is fair, each outcome is equally likely to happen. Therefore, this method also results in a fair decision.
The third method is to use three unbiased coins. By flipping three unbiased coins, $8$ possible outcomes can be obtained. Note that the order in which heads or tails appear matters, meaning that ${H,H,T}$ and ${H,T,H}$ are two different outcomes. ${H,H,H},{H,H,T},{H,T,H},{T,H,H},{T,T,H},{T,H,T},{H,T,T},{T,T,T} $ If each of these outcomes is assigned to one employee, then a fair decision can be made, since each outcome is equally likely. The employee with the matching outcome wins the prize.
$n=8$, $r=2$
Subtract term
Write as a product
Split into factors
Cross out common factors
Cancel out common factors
Multiply
$1a =a$
To determine who will win the prize, first flip a coin and then roll a six-sided die. If the coin lands on heads, the tens digit will be $0.$ If it lands on tails, the tens digit will be $1.$ The second digit will be determined by rolling a die. Here is the list of all possible outcomes of this event. ${01,02,03,04,05,06,11,12,13,14,15,16} $ Note that there is a total of $12$ possible outcomes. The probability of each outcome is $121 .$ Since each possible outcome is equally likely, this method results in a fair decision.
Start by assigning a number from $1$ to $246$ to each employee and then use a technology-based random number generator to select one number. As an example, a graphing calculator can be used. First, push the $MATH $ button. Then, scroll to the right to the PRB menu and choose the fifth option called randInt(.
The function randInt$(a,b,c)$ outputs $c$ integers in the range from $a$ to $b,$ inclusive. Since $5$ integers from $1$ to $246$ are required in this case, evaluate randInt$(1,246,5)$ in the calculator.
The random number generator provides each number from $1$ to $246$ an equal chance of being chosen, so each employee has an equal probability of winning the prize. Therefore, this method is fair.
A chess club of $4$ members decided gift a hand-carved wooden chess set to one of its members. The club wants each member to have a chance of winning this gift based on how many games they have won this week.
Members | Number of Games Won |
---|---|
Ali | $2$ |
Maya | $3$ |
Diego | $1$ |
Dylan | $0$ |
Sample Method: Assign integers from $1$ to $6$ to each member according to the number of games they won. Then roll a die to determine the winner.
Why it is Fair: The outcomes of rolling a die are equally likely. Additionally, the probability of winning the gift for each member is proportional to the number of games they have won.
Find a method of assigning outcomes so that the chance of winning is proportional to the number of games won.
A method of choosing the winner of the gift will be fair if members who won more games will have a better chance of winning the chess set. First, calculate the total number of games won by all the members. $2+3+1+0=6 $ Since there are $6$ games in total, integers from $1$ to $6$ can be assigned to the members according to the number of games they won.
Members | Games Won | Numbers Assigned |
---|---|---|
Ali | $2$ | $1and2$ |
Maya | $3$ | $3,4,and5$ |
Diego | $1$ | $6$ |
Dylan | $0$ | None |
Next, a standard six-sided die can be rolled to decide who wins the set. Recall that there are six possible outcomes of rolling a six-sided die and all outcomes are equally likely. $Possible Outcomes:{1,2,3,4,5,6} $ Therefore, the probability of winning the set for each member equals the quotient of the number of favorable outcomes to the number of possible outcomes. For example, for Ali to win, there are two favorable outcomes, $1$ and $2.$ Hence, the probability $P(A)$ that Ali wins the set can be calculated as follows. $P(A)=Number of Possible OutcomesNumber of Favorable Outcomes ⇓P(A)=62 $ Using the same procedure, the probability of winning the set can be calculated for each member.
Members | Games Won | Numbers Assigned | Probability of Winning the Chess Set |
---|---|---|---|
Ali | $2$ | $1and2$ | $62 ≈0.33$ |
Maya | $3$ | $3,4,and5$ | $63 =0.5$ |
Diego | $1$ | $6$ | $61 ≈0.17$ |
Dylan | $0$ | $None$ | $60 =0$ |
Notice that the probability of winning the gift for each member is proportional to the number of games they have won this week. Therefore, rolling a die can be considered as a fair method to decide who gets the gift.
Let $X$ be a random number between $0$ and $1$ produced by a random number generator. In the applet, a random number is represented by the point on the number line.
At the beginning of the lesson it has been said that the team of Heichi, Izabella, and Kevin won the local $3-$on$-3$ beach volleyball tournament and there is a bicycle in the prize package. All players want it, but there is only one. Therefore, each player proposed a method to decide who gets to keep the new bicycle.
Which player's method is fair?Begin by identifying all possible outcomes of the experiments in each method. Then, calculate the probabilities to determine the fair method(s).
Each method will be examined one at a time.
Heichi suggests rolling two dice and multiplying the numbers obtained. Since rolling each die has $6$ possible outcomes, there are $36$ possible products.
According to Heichi's method, if the product of the numbers is odd, then Izabella keeps the bicycle. Hence, the outcomes which are the odd numbers should be identified in the table.
There are $9$ such outcomes, which is the number of favorable outcomes for Izabella out of $36$ possible outcomes. With this information, the probability of Izabella getting the bicycle can be calculated.$Favorable Outcomes=9$, $Total Outcomes=36$
$ba =b/9a/9 $
Calculate quotient
$Favorable Outcomes=10$, $Total Outcomes=36$
$ba =b/2a/2 $
Calculate quotient
Round to ${\textstyle 2 \, \ifnumequal{2}{1}{\text{decimal}}{\text{decimals}}}$
$Favorable Outcomes=17$, $Total Outcomes=36$
Calculate quotient
Round to ${\textstyle 2 \, \ifnumequal{2}{1}{\text{decimal}}{\text{decimals}}}$
Izabella suggested writing each player's name $5$ times on slips of paper and putting all $15$ pieces in a bowl. Then one slip is randomly picked, which determines who gets the bicycle. To investigate if this method is fair, begin by calculating the probability of keeping the bicycle for each player, then analyze the results.
Favorable Outcomes | Total Outcomes | Probability | |
---|---|---|---|
Heichi wins | $5$ | $15$ | $155 ≈0.33$ |
Izabella wins | $5$ | $15$ | $155 ≈0.33$ |
Kevin wins | $5$ | $15$ | $155 ≈0.33$ |
It was obtained that each player is equally likely to get the bicycle. Therefore, Izabella's method is fair.
Kevin suggested flipping a coin two times. To analyze his method, make a table of all possible outcomes.
According to Kevin's method, if two heads show, then Heichi gets the bicycle. Note that there is only $1$ outcome that two heads show out of $4$ possible outcomes. Using this information, the probability that Heichi wins can be calculated.$Favorable Outcomes=1$, $Total Outcomes=4$
Calculate quotient
$Favorable Outcomes=2$, $Total Outcomes=4$
$ba =b/2a/2 $
Calculate quotient