Day 1 of Counting Methods and Probability. Fundamental Counting Principle and Permutations

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1 Grade Level: 10 th -11 th Subject Area: Algebra II Day 1 of Counting Methods and Probability Fundamental Counting Principle and Permutations Materials Needed: Pencils, Notebooks, Graphing Calculators, Slide show presentation on smart board. Standards: REVIEW: 7.SP.8 Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation. a. Understand that, just as with simple events, the probability of a compound event is the fraction of outcomes in the sample space for which the compound event occurs. b. Represent sample spaces for compound events using methods such a organized lists, tables and tree diagrams. For an event described in everyday language (e.g., rolling double sixes ), identify the outcomes in the sample space which compose the event. HS.SCP.9 (+) Use permutations and combinations to compute probabilities of compound events and solve problems. Objectives: 1. TLW be able to use a tree diagram and the fundamental counting principle to count the number of ways to perform a task. 2. TLW use the counting principle to find the number of permutations. 3. TLW define and use factorials. 4. TLW use factorials to determine the number of permutations of r objects taken from a group of n! n distinct objects through this formula: npr ( n r)! 5. TLW use factorials to determine the number of distinguishable permutations of n objects where one object is repeated s 1 times, another is repeated s 2 times, and so on, using this formula: n! a. s! s! s! 1 2 k Learning Activity: Introduction and Overview: Beginning a chapter on Counting Methods and Probability Here are a few big ideas we will be covering this chapter: Using permutations and combinations Finding probabilities Constructing binomial distributions

2 Today we will be examining the Fundamental Principle of Counting as well as Permutations. Our most basic method for counting in problems where there are multiple aspects is through the Fundamental Counting Principle. Do you remember drawing trees to represent different options? The Fundamental Counting Principle will shorten the work but illustrate the same idea. Let s say we wanted to make a pizza. To illustrate all the options, we will make what I call a pizza tree! Suppose with our pizza, we are allowed to pick between mozzarella and parmesan for the cheese, and for toppings we can pick one of the following: pepperoni, pineapple, or chicken. How many different pizzas could we make? While this may not be the most realistic pizza problem, it does illustrate how powerful the fundamental counting principle is. Let s draw a tree to represent the options for the pizza. pepperoni mozzarella pineapple Pizza chicken pepperoni parmesan pineapple chicken We can see that when we have two choices for cheese, and three choices for topping, we have a total of six different pizza combinations. Direct Instruction: This will be on the smartboard Fundamental Counting Principle Suppose k items are to be chosen. If there are n 1 ways to choose the first item, n 2 ways to choose the second item, and so on, then there are n1 n2 n k ways to choose all k items.

3 In OTHER WORDS, if one event can occur in m ways and another event can occur in n ways, then the number of ways that both events can occur is m n. If there are three events occur in m, n, and p ways, then the number of ways that all three events can occur is m n p Notice that there were 2 items to be chosen (cheese type and topping type). There are 2 ways to choose the cheese, and 3 ways to choose the topping, so by the fundamental counting principle, we can simply multiply 2 x 3 to get a total of 6 ways to choose both a cheese and a topping. This fundamental counting principle becomes particularly handy when a tree would be difficult to draw, for example, if the pizza parlor allowed the option of thick or thin crust, alfredo sauce or red sauce, three types of cheese to choose from and for toppings you were allowed to pick one of their 14 available toppings. Using the fundamental counting principle, we know that there are: 2 x 2 x 3 x 14 = 168 ways to choose a crust, a sauce, a cheese, and a topping. Guided discovery: Using the fundamental counting principle to derive the permutation formula! Ask 5 out of the 25 students to stand in front of the class. Draw out 5 blanks on the board like this: Ask the class the following questions: o We are going to use our knowledge of the fundamental counting principle to determine how many ways there are to arrange (name the five people) into a line. Each spot in line will have a certain number of options for how we can fill it, and like the different parts of our pizza, we will multiply the different ways to fill each spot. Make sense? o How many ways can we fill spot number one? 5 ways, since there are 5 people who could fill that spot, pick student to stand in the first spot. o Ok, now that we have spot number one filled, we should fill spot number two. How many ways are there to fill spot number two? 4 ways, since 1 person is already standing in spot number one, and cannot be in both spot number one and spot number two. o Notice we only have four ways to fill spot number two. How many ways can we fill spot number 3? Will there be a pattern? 3 ways, and yes there is a pattern, the number of ways to fill a spot will keep decreasing by one until we reach the last spot, where there will be only one way. o Think of this process now in terms of a tree. The base of our tree (I would label it Line of five people ) has five branches (each of our five options). The second layer, coming off each of those five branches, is an additional four branches representing the four people who are available to fill spot number two, and so on. Draw an incomplete tree diagram, showing all the different levels but not filling the whole tree out. o We know we can use the fundamental counting principle to describe this tree, so the total number of ways to arrange these five people in a line is:

4 5 x 4 x 3 x 2 x 1. We can use a short hand symbol to describe this kind of multiplication: Display this on smartboard: n Factorial The factorial of n is denoted as n! n! n( n 1)( n 2)( n 3)...1 So instead of writing 5 x 4 x 3 x 2 x 1 we could write 5! How does this relate to permutations? A permutation is an arrangement of all or part of a set of items where order is important. Permutations are expressed by this formula, where n represents the whole set, and r represents the part of the set that is being arranged. Permutations The permutation of n objects taken r at a time is: n! P( n, r) ( n r)! n P Another way of thinking: The number of permutations of r objects taken from a group of n distinct objects is denoted by np r. r Note that 0 factorial (0!) is defined to be 1. (How many ways can we arrange 0 items? One way! It is already done!) Also notice that 0! must be defined as 1 so that P(n,n) will equal the familiar n! For example, let s take a look at lining up five people from our class. Suppose n is 25 (since there are 25 of you) and r is 5 (since I want 5 people to be lined up). According to the formula, there are: 25! ! ,375, 600 distinct (or different) (25 5)! 20! ways to arrange five people from the 25 available. This formula makes sense, since there are 25 ways to pick the first person, 24 ways to pick the second, all the way down to 21 ways to fill the last spot. After all spots are filled, we don t care about the order of the rest of the people. That is why we divide by 20! (the number of ways to order the rest of the class). ***If you are wanting to find the! symbol on your calculator, go into CATALOG, hit A on your keypad to bring it to the top, and then scroll up through the symbols until you find!.*** Suppose we want to find how many different arrangements we can make with the letters in the word BANANA. In this case, we know that the A s and N s are indistinguishable (meaning we can t tell one A from another), therefore we need a slightly different formula to solve this problem.

5 Permutations with Repetition The number of distinguishable permutations of n objects where one object is repeated s 1 times, another is repeated s 2 times, and so on, using this formula: n! s! s! s! 1 2 k For the word BANANA there are 6 letters in all (n=6) and A is repeated 3 times, and N is repeated two times. Therefore the number of distinguishable permutations of BANANA is Work these problems as a class if time allows: 6! ! 3! 2! = 60 3! 2 1 Find the number of distinguishable permutations of the letters in the word CLEVELAND. 9! Answer: 90, 720 2! 2! Find the number of permutations: 6 P 2 6! Answer: 6 P 2 30 (6 2)! For the given configuration, determine how many different license plates are possible if (a) digits and letters can be repeated, and (b) digits and letters cannot be repeated. Use the Fundamental Counting Principle. 4 letters followed by 3 digits a) = ,976,000 b) ,336,000 ****Remind students there will be a quiz on Day 3 on Permutations (with and without repetition) and Combinations, covered on day 2, (formulas will be provided but not labeled)**** Allow students to work quietly on homework during the remaining time period. Assessment:

6 Objectives 1-2: Informal Assessment done by in class observation. Formal Assessment via homework. Objectives 3-5: Informal Assessment done by in class observation. Formal Assessment via homework, quiz, and unit test. Homework assignment: #1 (vocabulary), #3 (tree diagram), #7-9 (Fundamental Counting Principle), #12-14 (License Plate configurations), #27-29 (Factorials), #33-36 (Permutations), #43-46 (Permutations with Repetition) #56 (Error Analysis) Response:

7 Grade Level: 10 th -11 th Subject Area: Algebra II Day 2 of Counting Methods and Probability Combinations Using the Calculator for Permutations and Combinations Materials Needed: Pencils, Notebooks, Graphing Calculators, 25 Calculator Instruction sheets, 25 group worksheets on Combinations and Permutations, Graphing Calculator Simulator on smartboard. Standards: HS.SCP.9 (+) Use permutations and combinations to compute probabilities of compound events and solve problems. Objectives: TLW find the number of combinations of n objects taken r at a time using the following formula: n! C( n, r) r!( n r)! TLW will distinguish between when to use a permutation vs. a combination. Learning Activity: ANSWER ANY QUESTIONS ON HOMEWORK. Grade homework using a homework quiz have students copy their solutions for #7, #14, #35, and #45 on to a blank sheet of paper. Collect the homework quizzes. Review of what we discussed yesterday: Tree Diagrams, Fundamental Counting Principle, Factorials, Permutations, Permutations with Repetition. Have students summarize what they learned about each of those areas. Direct Instruction: *Displayed on smartboard: A combination is a selection of items where order does not matter. Combinations The combinations of n objects taken r at a time is: n! C( n, r) ncr r!( n r)! Another way of thinking: The number of combinations of r objects taken from a group of n distinct objects is denoted by nc r

8 **Notice that a combination, like a permutation, can be notated a number of different ways. nc r and C(n, r) will be the two ways I will write combinations for tests or quizzes. Example: I am forming a math club on campus and I will be selecting 7 of the students in this class to be part of this club (no, the students do not have a choice as to whether or not they want to be in the club ). There are 25 people in the class, which means that n= 25. I am choosing 7 out of the 25 to be in my club, so that means r= 7. Let s plug these numbers into the formula to see if the formula makes sense: 25! ! C(25, 7) 7!(25 7)! 7! 18! 480, 700 Once again, since repetition is not allowed (I can t pick the same person twice to be in my club), the options there are 25 options to pick the first person, and 19 options to pick the last person. What s different about the combinations formula in comparison to the permutations formula? Why do you think that is? Remember, the order of those that are selected into our groups no longer matters (unlike permutations). Notice that since order no longer matters, any rearrangement (or permutation) of the same seven people makes it the same group. How many different ways can we arrange 7 people? 7! That is why we divide by r!; since the order in which we pick the objects no longer matters. Have students use the formula to solve for C(5,3) and C(14,5): C(5,3) 10 C(14,5) 2002 Decide whether to multiply or add combinations! Are we choosing one thing OR another? The word OR is a signal that we must add the two events. Are we doing one thing AND another? The word AND is a signal that we must multiply the outcomes of the two events. A story of Miss Lange s date night with Mr. Combination (see what I did there?). Combinations Example: PIZZA! Story: Mr. Combination and I decided that we would decide to order a pizza since there was a special going on at Bruno s pizza. How many different pizza varieties can be created if there are two choices of crust, thick or thin, and two out of three possible toppings are selected? Notice we pick a crust AND two out of three toppings? Solution The combinations of the 2 types of crusts is C (2,1). The combination of 2 out of 3 toppings is C (3,2) So the number of pizza varieties that can be created is: 2! 3! C(2,1) C(3,2) 1!(2 1)! 2!(3 2)!

9 Next, we went to a movie store. The movie store advertises that it houses 100 movies, 32 of which are comedies, 17 are horror, 22 are family, and 29 are dramas. How many different sets of exactly 2 comedies, AND 3 dramas can we rent? Solution: You can choose 2 out of the 32 comedies and 3 out of the 29 dramas, so the number of possible movie rentals is: 32! 29! ! C C 2!(32 2)! 3!(29 3)! ! = 1,812,384 movie rentals! ! ! How many different sets of exactly 2 comedies OR 3 dramas can we rent? 32! 29! ! C C + 2!(32 2)! 3!(29 3)! ! = 4150 movie rentals! How many different sets of at most 3 movies can we rent? ! ! Solution: we can rent 0, 1, 2, or 3 movies. Because there are 100 movies available, the number of possible sets of movies is: C(100,0) C(100,1) C(100, 2) C(100,3) , ,751 possible sets PERMUTATION OR COMBINATION? Because distinguishing between a permutation and combination can be tricky, I want to provide you a lot of practice doing so. First we will cover some examples together as a class, and then I will divide you into groups of three (with one group of 4) to work on a worksheet I have prepared Discuss the difference between a permutation and a combination. A permutation is an arrangement of all or part of a set of items where order is important. A combination is an arrangement of all or part of a set of items where order is not important. For example, given the objects a, b, c 3P 2 = 6 {ab, ac, bc, ba, ca, cb} 3C 2 = 3 {ab, ac, bc} Example: Explain whether the following is a combination or a permutation and find the solution. How many twoletter arrangements can be formed from the letters CAT? Solution

10 Order makes a difference in this problem because CA and AC are not the same arrangement. The 3! solution using permutations: P(3, 2) 6.There are 6 possible arrangements (3 2)! {CA, AC, CT, TC, AT, TA} Example: Explain whether the following is a combination or a permutation and find the solution. How many twoman crews can be selected from the set of three males: {Charlie, Anton, Tom}? Solution For this problem, order does not make a difference. The two-man crew of Charlie/Anton is the same as 3! the two-man crew of Anton/Charlie. The solution is found using combinations: C(3, 2) 3. 2!(3 2)! There are only 3 different two-man crews {Charlie/Anton, Charlie/Tom, Anton/Tom} Explain whether each situation is a combination or a permutation and find the solution. (questions will be displayed on smartboard) How many 4-letter passwords can be made from the first 10 letters of the alphabet if no letter is used more than once in the same password? Solution: Permutation; P(10,4) 5,040 How many sets of officers of 3 people can be selected from a club that has 22 members? Solution: Combination; C(22,3) 1,540 Guided Discovery Problems (link to bigger ideas) 1. Give an example to show why 3 P 2 = 3 P 3. Consider the objects a, b, c. 3P 2 =6 {ab, ac, bc, ba, ca, cb} 3P 3 =6 {abc, acb, bac, bca, cba, cab} Or when we think of this problem in regards to using the formula: 3! P 6 (3 2)! 3 2 3! P 6 (3 3)! 3 3

11 Hand out Calculator Instruction sheets, walk through examples on the page using the graphing calculator simulator on the smartboard for students to follow along. At this point remind students that they will be having a QUIZ tomorrow at the end of class, covering permutations (with and without repetition) and combinations and one easy question on tomorrow s lesson dealing with the Binomial Theorem and Pascal s Triangles. Break students off into their groups, handing each student a group worksheet on Combinations and Permutations. Students are to work on these worksheets for the remainder of the class period. If students are unable to complete the worksheet during class, they are to take it home and finish it themselves, along with the few homework problems assigned from the book. Assessment: Objectives 1-2: Informal Assessment done by in class observation. Formal Assessment via homework assignment and Unit test Homework assignment: 1.) Complete worksheets for class tomorrow. 2.) page 694, #3-5 (Basic Combinations, use formula for #3, can use calculator for 4 and 5), #11-12 (Error Analysis), #15-17 (Possible 5 card hands taken from standard 52 deck). Instructions for calculator included below. Group worksheet not included. Response:

12 Calculator Instructions for Permutations and Combinations with Examples PERMUTATIONS Example: How many possible permutations of 2 cards can be chosen from a deck of 5 cards? Input 5. Press [MATH], arrow left to highlight PRB, then press [2] to select the npr function. Input 2 and press [ENTER]. There are 20 possible permutations of choosing 2 cards from a deck of 5 cards. COMBINATIONS Example: How many possible combinations of 2 cards can be chosen from a deck of 5 cards? Input 5. Press [MATH], arrow left to highlight PRB, then press [3] to select the ncr function. Input 2 and press [ENTER]. There are 10 possible combinations of choosing 2 cards from a deck of 5 cards

13 Day 3 of Counting Methods and Probability Binomial Theorem, Pascal s Triangle and Combinations Grade Level: 10 th -11 th Subject Area: Algebra II Materials Needed: Pencils, Notebooks, Graphing Calculators, Extra Calculator Instruction Sheets, Slideshow for Pascal s Triangle pictures and it s connections. Standards: HS.SCP.9 (+) Use permutations and combinations to compute probabilities of compound events and solve problems. HS.AAPR. 5 (+) Know and apply the Binomial Theorem gives the expansion of (x + y) n in powers of x and y for a positive integer n, where x and y are any numbers, with coefficients determined for example by Pascal s Triangle. The Binomial Theorem can be proved by mathematical induction or by a combinatorial argument. (CCSS) Objectives: 1 TLW complete additional rows of Pascal s Triangle 2 TLW explore the connections between the entries of Pascal s Triangle and combinations and binomial expansions. 3 TLW use the entries of Pascal s Triangle to find the values for specific combinations. 4 TLW use the entries of Pascal s Triangle to complete binomial expansions. 5 TLW Know and apply the Binomial Theorem to expand binomials. Learning Activity: ANSWER ANY QUESTIONS ON PREVIOUS HOMEWORK ASSIGNMENT. Have students pass in their group worksheets on combinations and permutations as well as the textbook assignment. Announce that there will be a quiz on combinations and permutations at the end of the class period, as well as a simple question about today s topic: binomial theorem, Pascal s triangle, and combinations. Hook: Start drawing out the entries for Pascal s Triangle down to the fifth row. Introduce the drawing as Pascal s Triangle and ask if any of the students could take a guess as to what the 6 th row would be. Allow students to work together to solve for the next row. If students believe they have figured out the pattern, instruct them to write the next row of Pascal s Triangle in their notebooks ?????? Drawing of Pascal s Triangle

14 Hints: There is a pattern. The next row relies on the row before it. Can you spot the pattern? Notice that all the outside entries are 1 s. Do you think that will be true for the next row? Why does each row get bigger by one entry? Notice how the rows are spaced, with some numbers falling in between two numbers of the row above it. Is there any pattern to that? Try using addition. After giving all of the hints, reveal the next row in the triangle! Write this in for the next line: Did anyone find that to be the next row? If so, can you explain to us how you got those numbers? The trick to finding the entries of the next row is to take the two numbers above the entry and add them together. That is why the outsides were always 1 (0+1=1), and why the second entry for the row was 5 (1+4=5) and so on. This is just one of the patterns associated with Pascal s Triangle, created by the famous French mathematician Blaise Pascal. We will look at two very important connections with Pascal s triangle: It s connection with combinations and with expanding binomials ((x+y) n ). Combinations and Pascal s Triangle I want you to take a close look at these two pictures (you can also find a similar picture on page 692 in your textbook). This picture claims that the entries for Pascal s triangle have a relationship with C(n,r), with the top row being C(0,0) = 1. Notice how as we go down in rows, n increases, and as we go across, r increases. n increases r increases Using your calculator, I want you to find the values of the C(4,0) through C(4,4) to verify that they match. If you didn t bring your calculator instruction sheets for combinations and permutations to class today, I have extras up on my desk. How might this information be useful? Let s say we didn t know the pattern for finding the next row, how could we use combinations to figure out the next row in the triangle? We could let n be equal to 5, and find the values for C(5,0) all the way to C(5,5).

15 More realistic use: Let s say we wanted to write out all the values for C(5,0) to C(5,5). Instead of going through the process of solving via the formula or using our calculator, we could quickly find the next row in the triangle, and those entries would be the solutions! On smartboard: Suppose there are 6 members in student council, and we must choose two of them to present an idea to the principal. Use Pascal s triangle to find the number of combinations of 2 members that can be chosen as representatives. Solution: First, we would find the values for the 6 th row of Pascal s triangle. Once that is complete, we would need to find the entry that corresponds with C(6,2). What is that value? C(6,2) = 15. There are 15 combinations of members to present an idea to the principal. This is the first connection we ve explored with Pascal s Triangle, and the second one is even more powerful! If I were to ask you to expand (x+y) 2, how would you expand it? Most people would use double distribution, or the FOIL method. What if I wanted you to expand (x+y) 7? Or (x+y) 15? We will need a better system than just FOIL or distribution! This is where Pascal s Triangle comes in! Take a look at this picture: As you can see, the values that are highlighted are the coefficients when the binomial is expanded. AND they are the SAME values for Pascal s Triangle and the values for Combinations! How can we use this information to help us expand binomials? Certainly we don t want to have to draw the fifteenth row in order to find the coefficients of (x+y) 15. Instead, we use a shortcut! Using the knowledge we have of combinations!

16 On Smartboard: Binomial Theorem For any positive integer n, the binomial expansion of (a+b) n is: ( a b) n C a n b C a n b C a n b C a b n n 0 n 1 n 2 n n Notice that each term in the expansion of (a+b) n nr r has the form ncra b where r is an integer from 0 to n. Compare this theorem with picture of the expanded binomials. Notice that the exponents for a and b always add up to n, starting with the exponent for a being equal to n, and ending with the exponent of b being equal to n. This is important to note as you try expanding these on your own. Try expanding these binomials using the formula: You may use the values from the triangle for the coefficients, but I want you to pay particular attention to the exponents. (Write these problems on the board, then solve together as a class) (x+3) 5 (a+2b) 4 What happens if we want to expand (a-2b) 4?( Make note that we can use a substitution by letting c= - 2b, using the binomial theorem, and then plug -2b back in for c at the end. Students do not have to write out all the steps for substitution, but they may find it helpful.) Isn t this powerful information?! You will have more opportunities to practice expanding binomials in your homework, but now it is time to take the quiz. I will hand out the quiz and write your homework assignment up on the board. Assessment: Objectives 1-2: Informal Assessment done by in class observation. Objectives 3-5: Informal Assessment done by in class observation. Formal Assessment via homework, quiz, and unit test. Homework assignment: #20-23 (Using rows of Pascal s Triangle to write binomial expansions allow students to look up a completed Pascal s Triangle or draw out a triangle for rows not covered in class) #25-33 odd (use binomial theorem to write expansion) #49 & 50 (Review of Combinations- Word Problems) Quiz on the following page. Response:

17 Name Subject Quiz Date n! r!( n r)! 1 2 Formulas Bank n! s! s! s! k n! ( n r)! 1. Find C(6,4) using the appropriate formula. Show all your steps. You may check your work with a calculator. 2. A Spanish club is electing a president, a vice president, and secretary. The club has 9 members many different ways can the 3 offices be filled? (HINT: Does order matter? Meaning, is there a difference in the positions of president, vice president and secretary?) Show all your steps, you may check your work with your calculator. 3. OHIO has 4 letters with O repeated twice. Find the number of distinguishable permutations of letters in OHIO.

18 4. Permutation or combination? Examine the following scenario and determine if it is solved using the permutation formula or combination formula, explain your reasoning. Solve for the answer. A photographer lines up the 15 members of a family in a single line in order to take a photograph. How many different ways can the photographer arrange the family members for the picture? 5. List areas of math that have connection to Pascal s Triangle (Discussed in class today)

19 Name Subject Quiz Answer Key Date n! r!( n r)! 1 2 Formulas Bank n! s! s! s! k n! ( n r)! 1. Find C(6,4) using the appropriate formula. Show all your steps. You may check your work with a calculator. 6! 6! C(6, 4) 360 (6 4)! 2! 2. A Spanish club is electing a president, a vice president, and secretary. The club has 9 members many different ways can the 3 offices be filled? (HINT: Does order matter? Meaning, is there a difference in the positions of president, vice president and secretary?) Show all your steps. You may check your work with a calculator. 9! 9! P(9,3) 504 (9 3)! 6! 3. OHIO has 4 letters with O repeated twice. Find the number of distinguishable permutations of letters in OHIO. 4! 12 distinguishable permutations of letters in OHIO 2!

20 4. Permutation or combination? Examine the following scenario and determine if it is solved using the permutation formula or combination formula, explain your reasoning. Solve for the answer. A photographer lines up the 15 members of a family in a single line in order to take a photograph. How many different ways can the photographer arrange the family members for the picture? This is a permutation, since we are arranging people into a line, therefore order matters. 15! P(15,15) 15! 1,307, 674,368, 000 ways (15 15)! 5. List areas of math that have connection to Pascal s Triangle (Discussed in class today) Binomial Expansion and the Binomial Theorem (anything that includes binomial ) and Combinations

21 Day 4 of Counting Methods and Probability Define and Use Probability Grade Level: 10 th -11 th Subject Area: Algebra II Materials Needed: Pencils, Notebooks, Graphing Calculators, 25 M&M Fun Size packages, 25 M&M probability worksheets, ** Card Facts Handouts for those who want them** Standards: REVIEW: 7.SP.5 Understand that the probability of a chance event is a number between 0 and 1 that expresses the likelihood of the event occurring. Larger numbers indicate greater likelihood. A probability near 0 indicates an unlikely event, a probability around ½ indicates an event that is neither unlikely nor likely, and a probability near 1 indicates a likely event. Objectives: TLW make decisions based on the probability of an event (ranged from 0 to 1). (i.e. Is this event likely to happen?) TLW understand the difference between theoretical probability and experimental probability. TLW determine the odds in favor of or odds against an event Learning Activity: ANSWER ANY QUESTIONS ON HOMEWORK. Collect the homework. Hand back quizzes, explain difficult problems. Direct Instruction: Definitions: Probability: The probability of an event is the likelihood of that event. It is given by a number on a continuous scale from 0 to 1, where 0 means that the event is impossible and 1 means that the event will definitely occur. Probabilities can be written as fractions, decimals, or percents, so for an event that is just as likely to occur as to not occur, the probability, denoted P, is 1/2= 0.5 = 50% Probability experiment: an occurrence whose outcome is uncertain. Often called a random event, because it occurs at random. Displayed on Smartboard Theoretical Probability of an Event

22 When all outcomes are equally likely, the theoretical probability that an event A will occur is: Number of outcomes in event A PA ( ) Total number of outcomes (sample space) Often, the theoretical probability of an event is often simply called the probability of the event. Notice that all outcomes must be equally likely. Think about rolling a die, the likelihood of each of the six outcomes is the same. If we wanted to find the probability of rolling an odd number (notated P(odd number)) that would be equal to 3/6 or ½ since there are 3 favorable outcomes (1,3,5) out of the possible six outcomes. This is a theoretical probability, meaning that this is what should happen. In theory, half the rolls should yield an odd number, but an actual experiment might show differently. Experimental Probability of an Event When an experiment is performed that consists of a certain number of trials, the experimental probability of an event A is given by: Number of trials where A occurs PA ( ) Total number of trials This probability is determined based on actual trials. The more trials, the more accurate the results for the probability. For example: Have you ever made decisions based off a coin toss? Best two out of three? We know that the theoretical probability of having the coin be heads up is 50%, however, there have been times where the odds appear to not be that way (like the six times in a row you betted it would be heads, and each time it was tails). According to our experimental probability, if 6 out of 6 times the results came back tails, then the experimental probability of getting a tails is 100% (6/6) and the probability of getting a tails is 0% (0/6). We know that those values aren t accurate reflections of the real probability, and with more trials we might find that our experimental probability approaches the theoretical probability of 50%. Today s activity will focus on finding theoretical probabilities and then experimenting to see if the experimental probabilities match the theoretical probabilities. I have set up different probability stations where you will be responsible for 1) finding the theoretical probability of your particular scenario 2) recording the outcomes of the trials to determine the experimental probability. Determining Odds: Odds in Favor of or Odds Against an Event When all outcomes are equally likely, the odds in favor of an event A and the odds against an event A are defined as follows:

23 Number of outcomes in A Odds in favor of event A = Number of outcomes not in A Odds against event A= Number of outcomes not in A Number of outcomes in A You can write the odds in favor of or against an event in the form a/b or in the form a:b. M&M s Activity: Theoretical and Experimental Probability Have you ever wondered if, when you reach into an M&M Fun Size bag, if the probability of picking an Orange M&M is the same as the probability of picking a blue M&M? Did you know that it is NOT the same? Did you know that the M&M manufacturing company actually intentionally creates less orange M&M s than blue M&M s? According to some research that I have done, in a regular package of Milk Chocolate M&M s should contain : 24% blue, 14% brown, 16% green, 20% orange, 13% red, and 14% yellow M&M s (Write percentages on the board). That means, in any given package, the probability of drawing a yellow is 14%. I want to see if these theoretical probabilities are the same for all packages! So today, I have brought in enough Fun Size packages of the milk chocolate M&M s so that each of you will find the experimental probability of drawing each color. Hand out M&M Probability worksheets, and display these instructions on the smartboard: 1. Record the theoretical probability for each color. 24% blue, 14% brown, 16% green, 20% orange, 13% red, and 14% yellow 2. Open up your M&M Package, but DO NOT EAT ANY M&M S! 3. Record your total number of M&M s. This will be your total number of outcomes for the experimental probability. Number of outcomes in event A PA ( ) Total number of outcomes (sample space) 4. Record how many M&M s there are of each color (number of outcomes in the event of each color), and divide each number by the total number of M&M s. Record your result as a decimal to two decimal places in the section marked Theoretical Probability-Personal.

24 5. When you are finished, write your theoretical probabilities for each color on the smart board. We will average the probabilities and see if they approximate the values provided to us by the M&M manufacturers. ****If you have time: On the backside of your sheet there is a similar table, except this time there is a section for Experimental Probability. With all your M&M s out on our your desk, close your eyes and pick an M&M. Open your eyes, and the put a tally mark beside each color. Repeat this process 20 times, and then calculate the Experimental Probability. Does this match your personal theoretical probabilities from the first page? Write assignment up on the board while students are working. If students would like a card facts sheet (Answers questions like: How many cards are in a suit? How many are face cards? How many of each type of card), instruct them to pick up a sheet on the way out. Assessment: Objectives 1-3: Informal Assessment done by in class observation. Formal Assessment via homework, quiz, and unit test. Homework Assignment: pg 702 #11-14 (Finding probability- cards) #20-23 (Finding Odds), (Error Analysis) Response:

25 M&M Probability Sheet Color Theoretical Probability Provided Theoretical Probability Personal Class Red Orange Yellow Green Blue Brown

26 M&M Probability Sheet Color Tallies Experimental Probability Red Orange Yellow Green Blue Brown

27 Grade Level: 10 th -11 th Subject Area: Algebra II Day 5 & 6 of Counting Methods and Probability Finding Probabilities of Disjoint and Overlapping Events Materials Needed: Pencils, Notebooks, Graphing Calculators, smartboard slideshow Standards: HS.SCP.7 Apply the Addition Rule, P(A or B) = P(A) + P(B) P(A and B), and interpret the answer in terms of the model. HS.SCP. 1 Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes, or as unions, intersections, or complements of other events ( or, and, not ). Objectives: 1.) TLW use the rules of Probability of Compound Events to find the probability of compound events 2.) TLW use the Probability of the Complement of an Event to find the probability of an event. 3.) TLW use the ideas of set theory to better understand the probability of compound events and finding the complement of an event. Day 5 Learning Activity: Begin by answering homework questions and There are times when we want to find the probability of compound events (two events occurring at the same time). For example, when picking a card out of the deck, what is the probability that we could draw a 7 or draw a spade? We would use the rules of finding the probability of compound events. Probability of Compound Events If A and B are any two events, then the probability of A or B is: P(A or B) = P(A) + P(B) P(A and B) Inclusive Events If A and B are disjoint events, then the probability of A or B is: P(A or B)= P(A) + P(B)Mutually Exclusive Events First thing we need to discuss: What are Inclusive Events? Inclusive Events are events such that both events share some outcomes, or when thinking of sets, it is two sets that have an intersection. Think of them as being Overlapping (Draw two circles on the board that intersect to represent inclusive events). An example of this: The event that you draw a 7 and the event

28 that you draw a spade are inclusive events. This is because we can draw a 7 that is a spade, in other words, the two events share at least one outcome. That shared outcome is what lies in the intersection of our circles. Mutually exclusive events are events that DO NOT share anything in common. If we were to think of this in terms of sets, it is two sets that have no intersection (Draw two circles on the board that do not intersect). Think of them as being Disjoint An example of this: The event that we draw a spade and the event that we draw a diamond. We cannot draw both a spade and a diamond at the same time. These events are mutually exclusive, they share no common outcomes. Today we will cover some examples of finding probability using permutations, combinations, and the rules for finding probability of compound events. Tomorrow I will set up various scenarios (with cards, dice, and a twister spinner) in which you will first have to determine the theoretical probability, and then perform trials to come up with an experimental probability. Example 1: Suppose I roll a regular six-sided die. Find the probability of rolling a number greater than or equal to 4 or rolling an even number. This is a compound event, since I am looking for the probability of one event or the other. Are these events inclusive or mutually exclusive? (inclusive, because the event of even numbers contains 4 and 6 which is included in the event of greater than or equal to 4). First I must find the probability of rolling a number greater or equal to four. Going back to my theoretical probability formula, what is the total number of outcomes? (6) How many outcomes are in the event of rolling a 4 or greater? (3). P(rolling a 4 or greater)= 3/6 or ½ We know that the P(rolling an even number) is also 3/6 or ½. What is the P(rolling a 4 or greater AND rolling an even number)? The total number of outcomes is the same (6), but how many outcomes are both 4 or greater and even? (2, 4 and 6 ). Based on the rule for Probability of Compound Events that are Inclusive, we add P(A)+P(B) and subtract off P(A and B). Therefore, the probability of rolling a number greater than equal to four or an even number is the following: 1/2 +1/2 1/3= 2/3 There is a 2/3 chance that you will role a number greater than equal to four or an even number. Example 2: A card is randomly selected from a standard deck of 52 cards. What is the probability that it is a face card or a diamond? Have students guide you through the solution, here are the general steps:

29 Let event A be selecting a face card and event B be selecting a diamond. A has 12 outcomes and B has 13 outcomes. Of these, 3 outcomes are common to A and B. So the probability of selecting a face card or a diamond is: P(A or B) = P(A) + P(B)- P(A and B) = 12/ /52-3/52. Example 3: Out of the 25 people in this class, suppose 11 students are either involved in band or on the football team. There are 7 people in this class who are involved in band, and 7 people who are on the football team. What is the probability that a randomly selected student is both involved in band AND on the football team? P(A)= 7/25, P(B) 7/25 and P(A or B)=11/25. Find P(A and B). Write the general formula: P(A or B) = P(A) + P(B)- P(A and B) Substitute in the known probabilities: 11/25= 7/25 +7/25 P(A and B) Solve for P(A and B). P(A and B)= 14/25-11/25 = 3/25 Run through these additional example problems if there is time (if not, cover them beginning of Day 6): A card is randomly selected from a standard deck of 52 cards. Find the probability of the given event: 1) Selecting a two or a king. 2.) Selecting an even number or a red face card. Assessment Day 5: Objective 1: Informal Assessment done by in class observation. Formal Assessment via homework, quiz, and unit test. Homework assignment: pg 710, 3-5 (Disjoint Events), 9-14 (Overlapping Events), (Choosing Cards), (Error Analysis), 43, 49 (Story Problems) Response: Day 6 Learning Activity: ANSWER ANY QUESTIONS ON HOMEWORK. Grade homework based on homework quiz. Have each student take out a half sheet of paper and copy their solutions onto the paper for problems #4, 12, 26, 43. Collect homework quizzes. Review topics from yesterday: Probability of Compound Events-Inclusive and Mutually Exclusive Events (Overlapping and Disjoint). Review rules.

30 Transition: Yesterday, we focused on ways to find the probability of a specific event. We did this by finding how many outcomes of an event are in A, and dividing that by the total number of outcomes. Sometimes, it is easier to find the number of outcomes that are NOT in A, rather than those that are in A. Example 1: A pair of dice is rolled. What is the probability of NOT rolling doubles? We could potentially try to how many of the 36 possible outcomes are NOT doubles, however, it would be easier to simply find the number of outcomes that are doubles, and then subtracting off that probability. This is called finding the complement of an event. Displayed on smartboard: Probability of the Complement of an Event The probability of the complement of A is P A 1 P( A) In this case, let A be the event that we roll doubles. We are looking for the probability that we do not roll doubles ( P A ). Therefore, if we can find the probability that we do roll doubles, we can subtract that from 1, and find the probability that we do not roll doubles. Does that make sense? What is the probability that we roll doubles? To break this problem down, what is the probability that we roll double one s? There is a 1/6 probability that we will roll a 1 for the first die, and a 1/6 probability that we will roll a 1 for the second die. The probability that both would occur is 1/6 x 1/6 or 1/36. That is the probability that we would roll double one s, how many possible double numbers can we roll? (there are 6 numbers). We could roll double one s, OR double two s, OR double six s. In order to find the probability of rolling any doubles, we need to add up the probabilities of rolling double digits for 1-6, therefore the probability of rolling doubles is (1/36) x 6 or (1/6). The probability that we will NOT role doubles is 1- (1/6) = 5/6 Set Theory and Complement of an Event I want to introduce you to a few of the same ideas we have covered yesterday and today (working with compound events) but using set theory. Do you remember how I drew circles on the board, two that intersected to represent inclusive events and two circles that did not intersection to represent disjoint events? We will be looking set theory to help us understand the concept of a complement better.

31 Picture will be on the smartboard. Label left circle as A and right circle as B Top example: This picture is a representation of the probability of event A OR B occurring. When both circles are completely shaded (or when we have all the outcomes from A plus all the outcomes from B) we can this the union of two sets, written as A B (write union and A Bnext to the picture.) The second set of circles represents an intersection of sets A and B. Everything that falls in this category must be both an outcome of A and an outcome of B at the same time. This is written as A B (write intersection and A B next to the picture) The third set of circles is the one I really would like us to concentrate on. This is suppose to represent the complement of a set A, and is written as A. However, there is just one thing wrong with this picture. Yes, A isn t shaded, but what should be shaded that is not? Look at your textbook on page 715 and describe to me the difference between the two pictures (Everything in the background should be shaded).this background area is what we call the universe and it includes ALL possible outcomes (it s like the sample space!) If we shaded the whole box, that would mean that we wanted to include EVERY possible outcome, those that are in A, and B, and outside of those too! Can anyone tell me why the background should be shaded? (Because we are looking for all outcomes that are not in A, which include outcomes outside of B as well). (Using smartboard marker, fill in the rest of the background and write complement of A and A ) Example 2: The numbers 1-50 are put into a hat, what is the probability that the number drawn is not divisible by 3? The complement of the event not divisible by three is the event divisible by three. To calculate the probability of not divisible by three we will find the probability of the complement and subtract it off from 1. To find the numbers from 1-50 that are divisible by three, take 50 divided by 3 and round down (16). The probability of drawing a number divisible by three is 16/50. The probability that the probability of the number drawn is NOT divisible by three is 1-16/50= 34/50. Example 3:

32 Here is a tricky one: Suppose I am forming a whiteboard cleaning committee (yes, I need people to clean my whiteboards, a lot of people ). Out of all the possible committees I could form (of all different sizes) what is the probability that my committee has at least three people? Ok, let s first try to think of the total number of committees I could form. How I like to think of it as, there are 2 options for each person: either they are on the committee, or they are not. Therefore the total number of committees that could be formed is 2 25 (since there are 25 people, 2 options for each person). This is my sample space, or total number of outcomes (which is my denominator) Does this make sense so far? What is the complement of having at least 3 people? (having 2 or less people). Therefore, I need to find the probability that the committee I choose has 2 or less people. How many ways can I choose a committee of 2 people? ( C(25,2)) And 1 person? (C(25,1)) Don t forget about 0 people! (C(25,0)) Therefore, the probability of having 2 or less people on my committee is as follows: C(25, 2) C(25,1) C(25, 0) ,554, 432 The probability of me forming a committee of at least three people is 1-326/33,554,432= Summary: In conclusion, we can see that finding the complement of an event can be a powerful tool for finding the probability of an event. If there is time: At this point in time, I would like you to take out a piece of paper and jot down some thoughts in regards to these questions: 1. Is there any part of this unit that I am having difficulty understanding? (Fundamental Counting Principle, Combinations, Permutations, Probability of Compound Events, Using the Complement, etc.) 2. What is my favorite part of the unit? (Activities, concepts, etc.) 3. Is there a topic in this unit that I would like to know more about? While students are reflecting, write out homework assignment on the board: #2, 16-19, 35-38, and 1-3 on page 716. Remind students that if they have any questions on homework, they are welcome to stop by before or after school or to me with their questions. Have students turn in their reflections as a Ticket to Leave Assessment:

33 Objectives 1-3: Informal Assessment done by in class observation. Objectives 2-3: Informal Assessment done by in class observation. Formal Assessment via homework, quiz, and unit test. Homework assignment: #2 (conceptual writing), #16-19 (Finding the Complement of A), #35-38 (Some are done regularly, others are easier to do by finding the complement), and #1-3 on page 716 (Operations on Sets). Response:

34 Grade Level: 10 th -11 th Day 7&8 of Counting Methods and Probability Finding Probabilities of Independent and Dependent Events Subject Area: Algebra II Materials Needed: Pencils, Notebooks, Graphing Calculators, Slideshow on Smartboard for Conditional Probability (Finding Probability of Dependent and Independent Events), 25 Opinion Surveys, 25 Opinion Survey Instruction sheets Standards: HS.SCP.4 Construct and interpret two-way frequency tables of data when two categories are associated with each object being classified. Use the two-way table as a sample space to decide if events are independent and to approximate conditional probabilities. For example, collect data from a random sample of students in your school on their favorite subject among math, science, and English. Estimate the probability that a randomly selected student from your school will favor science given that the student is in tenth grade. Do the same for other subjects and compare the results. HS.SCP.6 Find the conditional probability of A given B as the fraction of B s outcomes that also belong to A, and interpret the answer in terms of the model. HS.SCP.8 (+) Apply the general Multiplication Rule in a uniform probability model, P(A and B) = P(A)P(B A) = P(B)P(A B), and interpret the answer in terms of the model. Objectives: TLW make two-frequency tables of data as a sample space to decide if events are independent and to figure out conditional probabilities. TLW apply the general multiplication rule to solve for any one of the variables given the other three: P(A and B), P(A) or P(B), P(B A) or P(A B). Day 7 Learning Activity: ANSWER ANY QUESTIONS ON HOMEWORK. Collect the homework. Definitions: Independent and Dependent Events Independent Events: Two events are independent if the occurrence of one has no effect on the occurrence of the other. The probability of two independent events occurring is the product of their individual probabilities. Display on smartboard:

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