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Alex, Mel, and Chelsea Play a Game


Alignments to Content Standards: S-CP.B.9

Task

Alex, Mel, and Chelsea play a game that has 6 rounds. In each round there is a single winner, and the outcomes of the rounds are independent. For each round the probability that Alex wins is $\frac{1}{2}$, and Mel is twice as likely to win as Chelsea. What is the probability that Alex wins three rounds, Mel wins two rounds, and Chelsea wins one round?

IM Commentary

This task combines the concept of independent events with computational tools for counting combinations, requiring fluent understanding of probability in a series of independent events. 

After setting up the problem, students will need to know that probabilities for independent events are multiplied and then take into account all of the different scenarios which result in the desired outcome (Alex winning 3 rounds, Mel 2, and Chelsea 1). The solution provided uses the binomial coefficient notation for the number of choices, but this is not required.

This task was adapted from problem #11 on the 2012 American Mathematics Competition (AMC) 12A Test.

For the 2012 AMC 12A, which was taken by 72,238 students, the multiple choice answers for the problem had the following distribution:

Choice Answer Percentage of Answers
(A) ${\frac{5}{72}}$ 9.35
(B)* ${\frac{5}{36}}$ 20.51
(C) ${\frac{1}{6}}$ 12.83
(D) ${\frac{1}{3}}$ 6.16
(E) ${1}$. 19.48
Omit -- 31.63

Of the 72,238 students: 28,268, or 39%, were in 12th grade; 34,124 or 47%, were in 11th grade; 4,615, or 6%, were in 10th grade; and the remainder were below 10th grade.

The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable during exploration of a task depend on how instruction unfolds in the classroom. While it is possible that tasks may be connected to several practices, only one practice connection will be discussed in depth. Possible secondary practice connections may be discussed but not in the same degree of detail.

This task helps illustrate Mathematical Practice Standard 1, Make sense of problems and persevere in solving them.  As students begin, they may ask themselves questions such as, “What are the probabilities for each of these independent events?” “What are all the possible scenarios?” and “What operation do I use when I am dealing with independent events?” They will find an entry point that makes sense to them and determine a solution pathway, knowing along the way they may take a wrong turn.  Students will realize that they must complete three separate steps (detailed in the solution set) in order to solve this task and understand the need for each of the steps. Lastly, students will consider how to best explain their reasoning to their peers while preparing themselves to make sense of the reasoning of others since they recognize their solution path through the task may not be the only correct one.

 

 

Solution

There are three separate steps to this problem. We need to calculate the probabilities that Mel and Chelsea win a single round. Next, we must compute the probability of any given scenario where Alex wins 3 rounds, Mel wins 2, and Chelsea wins 1 round. Then we have to find out how many ways there are for Alex to win three rounds, Mel 2, and Chelsea 1. With all of this information we can then calculate the probability that Alex wins 3 rounds, Mel wins 2, and Chelsea wins 1.

We are given that Mel is twice as likely to win a round as Chelsea. Their combined probability of winning a round is $\frac{1}{2}$ since this number, combined with Alex's probability of winning a round, must add up to 1. If we let $x$ denote Chelsea's probability of winning a given round then $2x$ is Mel's probability of winning a round and $$ 2x + x = \frac{1}{2} $$ so $x = \frac{1}{6}$. So the probability that Mel wins a round is $\frac{2}{6} = \frac{1}{3}$ and the probability that Chelsea wins a round is $\frac{1}{6}$.

Suppose we let AAAMMC denote the situation where Alex wins the first three rounds, Mel wins the fourth and fifth, and Chelsea wins the sixth. To find the likelihood that this happens, we use the information from the previous paragraph about the probability that each individual wins a single round. So the probability that Alex wins the first round is $\frac{1}{2}$. Similarly the probability that Alex wins round 2 is $\frac{1}{2}$ as is the probability that he wins round 3. The probability that Mel wins round 4 is $\frac{1}{3}$ and so is the probability that Mel wins round 5. Finally the probability that Chelsea wins round 6 is $\frac{1}{6}$. Since the different rounds are independent events, this means that the probability of an outcome such as AAAMMC (with three wins for Alex, two wins for Mel and one win for Chelsea) is $$ \left(\frac{1}{2}\right)^3 \times \left(\frac{1}{3}\right)^2 \times \left(\frac{1}{6}\right) $$ This is the same as the probability of, say, MMAAAC, where Mel wins the first two rounds, Alex the next three, and Chelsea the final one. This brings us to our last step, finding out how many ways there are for Alex to win 3 rounds, Mel 2 and Chelsea 1.

Since there are six rounds, the number of ways Alex can win 3 of them is the combinatorial symbol

$$ \left( \begin{array} &6 \\ 3 \end{array} \right) $$
which is equal to 20. Once the three games which Alex has won have been chosen, Mel must win two of the other three and there are
$$ \left( \begin{array} &3 \\ 2 \end{array} \right) $$
for this to happen: this is equal to 3. The remaining game, won by Chelsea, has no further choices.

Putting together all of this information, the probability that Alex wins three games, Mel two, and Chelsea one is $$ \left(\frac{1}{2}\right)^3 \times \left(\frac{1}{3}\right)^2 \times \left(\frac{1}{6}\right) \times 20 \times 3 = \frac{5}{36}. $$