Probability for Kids

Using Model-Eliciting Activities to Investigate Probability Concepts (Grades 4-6)

Scott Chamberlin

• Publisher: Routledge
• Year: 2021
• ISBN:
  9781000495324
• Edition: 1
• Topic: Education
• Subtopic: Education General

Probability for Kids features real-world probability scenarios for students in grades 4-6. Students will encounter problems in which they read about students their age selling magazines for a school fund raiser, concerned about their homeroom assignments, and trying to decode the combination to a safe that their grandfather abandoned, among others, all of which maximizes learning so students gain a deep understanding of concepts in probability. This book will help teachers, parents, and other educators to employ best practices in implementing challenging math activities based on standards. Problem solvers who complete all six activities in the book will understand the six basic principles of probability and be high school ready for discussions in probability. Grades 4-6

Chapter 1
Likelihood

The Junior High School Problem

DOI: 10.4324/9781003237280-2

For some students, the Junior High School Problem is the introductory model-eliciting activity that they will do in probability. Consequently, the content and context are rather important. This problem, written specifically with advanced grades 4–6 students in mind, but aptly challenging for middle grade students up to grade 9, should sync well with their interests, as many of them will be making the transition from upper elementary to junior high or high school in the coming months or years. Identifying a focus in probability for high-ability students was difficult in the respect that the problem needed to be sufficiently, but not overly, challenging. That is, on either end of the developmental spectrum, a problem may pose barriers for students. If the problem is too difficult for the students, then they may become frustrated and disengaged. If, on the other hand, the problem is too pedestrian for them, they may not take it seriously and produce less-than-satisfactory solutions. The hope with the Junior High School Problem is that the context and content meet expectations and provide ideal challenge for upper elementary high-ability students and middle grade general population students.

The focus of the Junior High School Problem is to get students to quantify the probability of an event transpiring. More specifically, this MEA is about compound probability because two events must occur, that is, both friends being placed in the same classroom. At its most basic definition, probability is the likelihood of an event happening and it is represented by the number of favorable outcomes (numerator) relative to the number of potential outcomes (denominator). This is one of the reasons that highly elementary events are used in introductory probability activities, such as flipping a coin (restricted to 2 outcomes), drawing a card from a fair deck (52 outcomes), and/or rolling a die (6 outcomes). In this book, problems are not centered on such (un)common events, as they are ubiquitous in probability chapters, books, and curricular materials. Instead, much more realistic problems are presented in hopes of stimulating students’ interest in the problems.

In the Junior High School Problem, two friends are trying to identify the likelihood that they will get in the same class when they leave their respective elementary schools. Several elementary schools feed into a larger junior high school and the friends are hoping to identify what their chance is of being placed in the same classroom. The students have known each other for years and they are faced with attending a much larger school, junior high, than which they currently attend (elementary school). If they could be placed together, they would at least start junior high with one friend that they already know.

Links to Standards

The Junior High School Problem links to the CCSS-M (NGA & CCSSO, 2010) and the Principles and Standards for School Mathematics (NCTM, 2000).

Regarding the CCSS-M, problem solvers can be expected to engage in mathematical practices 1, 2, 4, 6, and 7, and depending on the manner in which the problem is implemented and whether the facilitator has problem solvers discuss solutions, mathematical practices 3, 5, and 8 may be utilized. Certainly, MP4 is a practice in which students rarely engage and MEAs are designed with the intent of getting young problem solvers to create mathematical models. In addition, MP1 is likely a practice in which few students have engaged because it may be common for textbook problem-solving tasks to require very little time to solve. Anecdotal data suggests that MEAs often require upper elementary students approximately 45–60 minutes to solve. Consequently, perseverance can be a necessary skill that high-ability students need to refine in order to realize success with MEAs.

Regarding the Principles and Standards for School Mathematics, problem solvers will generally meet several expectations. For instance, problem solvers will typically select and use appropriate statistical methods to analyze data, they may also develop and evaluate inferences based on data, and in this problem, they will most certainly understand and apply basic concepts of probability (grade bands 3–5, 6–8, and 9–12 expectations).

Potential Student Responses

Most of the MEAs in this book have at least five solutions. This problem-solving task, however, only has four because the problem is rather straightforward. As this problem may be the introduction for many high-ability students in grades 4–6, the intent is to get them to consider what probability is and how it can be quantified. To add a level of complexity, problem solvers are asked to consider two events in relation to one another in an attempt to quantify the probability of both events happening. This concept is referred to as compound probability. Among the responses are several initial attempts, with the last one considered the most accurate and sophisticated. This is the case with all chapters in that the solutions are listed from most simplistic to most sophisticated. That is, Solution 1 may or may not satisfy what teachers are seeking with respect to student understanding. Responses listed are potential responses and represent the most common responses that a teacher can anticipate. If the first and second responses created by a group of problem solvers are considered unacceptable, then teachers should utilize questions to redirect students’ thinking in an attempt to seek a refined and more advanced solution than those presented.

Solution 1

The first solution is a strong explanation, but it is confined to only the first homeroom. Consequently, many teachers might desire a more comprehensive mathematical model than this one and encourage problem solvers to explain what could happen in homerooms after the first one. The first solution reflects the idea that there is a 1 in 100 chance of the two friends being placed in the same class. According to the problem solvers, this is the case because there are 10 homerooms and each boy has a 1 in 10 chance of being selected for the first homeroom.

The problem solvers therefore suggested that the probability of both boys being placed in the same homeroom is 1/10 for Maurice and 1/10 for Marvin or (1/10)². Other problem solvers may represent this as 1/10 multiplied by 1/10. Nevertheless, the probability of the event happening is 1 in 100, 1/100, or it can be expressed as 1%. These are not great odds and some problem solvers might suggest that with a much smaller school and fewer homerooms, the likelihood of being placed together would be increased.

Solution 2

The second solution is only somewhat more complex than the first solution. This solution represents the model expressed in the first solution—1 in 100— but with this model, the problem solvers added the caveat that if either of the boys were selected for one homeroom and the other was not, the conditions of the problem could not be met. The second solution provided an example to substantiate this claim. Solution 2 suggests that if Marvin was placed in the first homeroom and Maurice was not, then compound probability could not be met in this case because the problem statement prompted individuals for the likelihood of both boys being placed in the same homeroom. So, if Marvin were placed in the first homeroom and Maurice in the sixth homeroom, they would not be together. This solution does not specify the precise probability of this event transpiring, but does acknowledge that it was far more likely that the boys were separated than together in one homeroom.

Solution 3

With this solution, the problem solvers realize that the probability of being placed together could be quantified quite precisely if certain conditions were met. Each homeroom could be given an actual probability if the boys were not separated in the drawing. Figure 1.1 shows the probability for each homeroom, assuming no separation in the drawing.

Figure 1.1. Probability for each homeroom.

Solution 3 also shows that the precise percentage (probability) can be quantified by ascertaining the numerator, 1, and dividing it by the number of homerooms remaining and squaring that number. That eventual decimal could be represented as a percentage, as was done in Figure 1.1. Problem solvers stipulate that if the boys were separated by the draw, then the conditions of the problem could not be met. They are not only able to quantify the exact, rounded in some cases, percentage, but they did it for each homeroom. They also realize that as homerooms were removed or already drawn, the probability of the event happening increased greatly. Interestingly enough, they note that even with two homerooms remaining, the probability was still not greater than 50% that the boys would end up in the same homeroom.

Solution 4

The fourth solution is the most accurate, sophisticated, and insightful. This solution identifies the same information as Solution 3, but also shows how to compute the inverse for each event (i.e., the likelihood that the two best friends could be separated in the drawing at any moment) by subtracting the percentage in the second column from 100% or 1/1. This information is represented in the chart prepared in Figure 1.2. This solution also shows that probabilities can/should be represented on a scale that ranges from 0 to 1 or 0% to 100% with 0 being no chance at all and 1 or 100% being complete certainty of an event transpiring.

Figure 1.2. Probability and inverse probability of each homeroom.

Instructor Questions

Imperative to the success of MEAs is anticipating instances in which problem solvers may reach an impasse and not be able to successfully construct a suitable mathematical model. These questions might help problem solvers clarify thinking, change directions altogether, and/or come to a refined solution. Appropriate use and timing of questions can help problem solvers attain considerable insight as to the solution of the problem and help them create strong mathematical models.

Three caveats are issued with the instructor questions in each chapter. First, not all questions will be used with the MEA. Second, the list is intended to be comprehensive, but it does not represent all possible questions. Third, the questions are not arranged in any particular order and sometimes an alteration of the wording in the questions can change them dramatically—for better or worse.

1. Can you provide an example of chance in your life?
2. Can you name a situation in which the chance of something happening was 100% or nearly so? Can you name a situation in which the chance of something happening was 0% or nearly so?
3. Does it have applications to this problem?
4. What have you done so far and what does not seem clear to you?
5. Does your final mathematical model and the number at which you arrived seem logical?
6. Can you use your mathematical model in future, similar situations?
7. How many events must happen for the two friends to be placed in the same homeroom?
8. Is there some relationship between the two events?
9. Does the placement of students in homerooms have any biases? For example, can parents of students call the principal’s office and get two people placed together in one homeroom or not placed in the same homeroom?
10. If you were to guess what the likelihood of being placed in the same homeroom with your friend with 20 homerooms are possible, what would you guess the likelihood to be?
11. If you were to guess what the likelihood of being placed in the same homeroom with your friend with 10 homerooms possible, what would you guess the likelihood to be?
12. If you were to guess what the likelihood of being placed in the same homeroom with your friend with only four homerooms possible, what would you guess the likelihood to be?
13. If you were to guess what the likelihood of being placed in the same homeroom with your friend with only two homerooms possible, what would you guess the likelihood to be?
14. In which of the four previous scenarios do you have the best chance of being placed in the same homeroom with a friend? Provide reasoning for your selection.
15. Mathematically speaking, how do you represent or write the possibility of something happening (such as in this problem)?
16. Have you considered the likelihood that the friends would not be placed in the same homeroom?
17. Does that answer help you identify the opposite likelihood (i.e., that they would be placed together)?
18. When representing probability (or likelihood) as a fraction, is there meaning attached to the numerator and the denominator of the fraction?
19. Have you ever done a problem such as this one?
20. If so, can you apply any of the methods used to th...