A team of 8 students has finished an investigation on the many uses of parabolas. They must select a single group member to present their findings to the class. They have only a single coin to make chance selections, but they can flip it as many times as necessary to select a student fairly.
Our goal is to propose a method that gives 8 distinct and equally probable outcomes. Let's then think about the number of outcomes when tossing a fair coin. For example, if we toss it once, the number of outcomes is 2,heads or tails.
Number of Tosses
Number of Outcomes
1
2
If we toss the coin twice, then the number of outcomes is 2⋅2, or 4, because there are 2 possible outcomes for each toss. Let's extend this pattern for 3 and 4 tosses. Notice that in each case, the number of outcomes is a product of the number of outcomes for each toss.
Number of Tosses
Number of Outcomes
1
2
2
2⋅2, or 4
3
2⋅2⋅2, or 8
4
2⋅2⋅2⋅2, or 16
We found that tossing a coin 3 times produces 8 distinct outcomes. If we assign each of these outcomes to one member of the team, then we can be sure to select a student fairly.
b We want to find the smallest number of coin tosses needed to select a presenter fairly. In Part A we found that 3 tosses are enough to do so since tossing the coin 3 times produces 8 distinct outcomes.
Number of Tosses
Number of Outcomes
1
2
2
2⋅2, or 4
3
2⋅2⋅2, or 8
4
2⋅2⋅2⋅2, or 16
Let's check if 3 is also the smallest possible number of tosses. We can draw a tree diagram of this method to represent the sample space of all possible outcomes. Keep in mind that we associate each unique outcome with one team member. We can denote those team members as M1,M2, and so on.
We can see that the number of possible outcomes increases with the number of tosses. If we stopped after the second toss, then there would be only 4 distinct outcomes. Since we need 8 distinct outcomes, 3 is the minimal number of tosses.
c Two of the group members realize they will not be in class the day of the presentation, so we now want to propose a fair method that gives 6 distinct and equally probable outcomes. Let's think of a way that we can modify the method from Part A.
One way to redesign our method is to assign 6 out of the 8 outcomes to the group members who are available on the presentation day and leave the other 2 outcomes unclaimed. If we get one of these 2 outcomes, we repeat the process until we get one of the assigned outcomes.
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