Math 7 Notes - Unit 11 Probability

Math 7 Notes - Unit 11 Probability Probability Syllabus Objective: (7.2)The student will determine the theoretical probability of an event. Syllabus Objective: (7.4)The student will compare theoretical and experimental probabilities of an event. Syllabus Objective: (7.5)The student will represent the probability of an event as a number between 0 and 1. Most of us would like to predict the future; just think of the possibilities if we could! Since we cannot, the best we can do is tell how likely something is to happen. It s helpful to know if something is impossible, likely, unlikely or certain to happen. People like to know if it is a sure thing, or a 50-50 chance or it will never happen. It is more useful if you can use a number to describe the likelihood. Both probability and odds are ways to tell how likely it is that an event will or will not happen. Note to CCSD teachers: The textbook uses words or phrases to represent the probability of an event from impossible to certain. The CRT requires number values to represent probability (from 0 to 1). Probability is the measure of how likely an event is to occur. They are written as fractions or decimals from 0 to 1. Probability may be written as a percent, 0% to 100%. The higher the probability, the more likely an event is to happen. For instance, an event with a probability of 0 will never happen. If you have a probability of 100%, the event will always happen. An event with a probability of 1 2 or 50% has the same chance of happening as not happening. Example: How likely is it that a coin tossed will come up heads? This means that there is as likely a chance of heads as not heads. In other words, a probability of 1 or 0.5 or 50%. 2 Example: The weather report gives a 75% chance of rain for tomorrow. This means that there is a likely chance of rain (75%) and an unlikely chance of no rain (25%). In other words, the probability of rain is 3 or 0.75 or 75%. 4 Math 7, Unit 11: Probability Holt: Chapter 11 Page 1 of 14

Outcomes are the possible results of an experiment. Example: Tossing a coin; the possible outcomes (results) are a head or a tail. Theoretical probability is based on knowing all the equally likely outcomes of an experiment, and it is defined as a ratio of the number of favorable outcomes to the number of possible outcomes. Mathematically, we write: probability = number of favorable outcomes number of possible outcomes or success probability = success + failure Example: Suppose you pick a marble from a hat that contains three red, two yellow and one blue marble. What is the probability you draw a yellow marble? P( yellow marbles ) = # of yellow marbles total number of marbles 2 1 P = or 6 3 The probability found in the above example is an example of theoretical probability. Experimental probability is based on repeated trials of an experiment. Example: In the last thirty days, there were 7 cloudy days. What is the experimental probability that tomorrow will be cloudy? 7 P( cloudy days ) = 30 Math 7, Unit 11: Probability Holt: Chapter 11 Page 2 of 14

Odds Note to CCSD teachers: odds is not taught in the textbook; however, this concept is included in the CCSD benchmarks and Nevada Standards and will be tested on the CRT. Also, teachers should be aware that one can convert from odds to probability, and vice versa. For example, if the odds of winning a game is 2 3, then the probability of winning is 2. If the 5 probability of rain is 7 10, then the odds of rain is 7 3. Syllabus Objective: (7.3) The student will determine the odds of an event. Odds: the ratio of favorable outcomes to the number of unfavorable outcomes, when all outcomes are equally likely. Odds in favor = Odds against = Number of favorable outcomes Number of unfavorable outcomes Number of unfavorable outcomes Number of favorable outcomes Example: Suppose you pick a marble from a hat that contains three red, two yellow and one blue marble. What are the odds (in favor) you draw a yellow marble? 2 of yellow marbles 2 1 P = = = or 1:2 # of marbles not yellow 4 2 Example: If the probability of an event is 4, find the odds. 9 4 9 4 = 5, so the odds are. 5 Math 7, Unit 11: Probability Holt: Chapter 11 Page 3 of 14

Tree Diagrams and The Fundamental Counting Principle Syllabus Objective: (7.6) Accelerated only The student will determine the number of possible outcomes using a variety of counting methods. One method students may use is to determine the number of possible outcomes is to write an organized list. Example: Cassie has a blue sweater, a red sweater, and a purple sweater. She has a white shirt and a tan shirt. How many different ways can she wear a sweater and a shirt together? blue sweater white shirt or BW blue sweater tan shirt or BT red sweater white shirt or RW red sweater tan shirt or RT purple sweater white shirt or PW purple sweater tan shirt or PT 6 ways As more items are added, this method becomes cumbersome. A tree diagram makes it easier to see (count) the number of possible outcomes for experiments when the numbers are small and there are multiple events. To draw a tree diagram, you: 1) begin with a point; then you draw a line for each outcome in the first event. 2) draw lines for subsequent outcomes based on the outcomes from the first event. Example: Draw a tree diagram to show the outcomes for flipping two coins. Start with a point. There are two outcomes for the first coin, a head (H) or a tail (T). Draw lines and label H and T. H T Math 7, Unit 11: Probability Holt: Chapter 11 Page 4 of 14

For either of the outcomes in the first flip, the second coin could be a head or a tail. So the tree diagram would look like this. H T H T H T Now reading down the tree diagram, the possible outcomes are HH, HT, TH, or TT. There are 4 possible outcomes when flipping two coins. Extend the tree diagram for three coins. How many outcomes are there? Example: Draw a tree diagram to determine the number of different outfits that could be worn if you had two pairs of pants and three shirts. Starting with a point, you have 2 pairs of pants. For each pair of pants (P1 and P2), you have three shirts (S1, S2 and S3) to choose from. P1 P2 S1 S2 S3 S1 S2 S3 There are 6 possible outcomes: P1S1, P1S2, P1S3, P2S1, P2S2, and P2S3. Math 7, Unit 11: Probability Holt: Chapter 11 Page 5 of 14

Fundamental Counting Principle Tree diagrams are useful to get a picture of what is occurring, but with a large number of events, the tree can get out of hand in a hurry. A quick way to determine the number of possible outcomes in a tree diagram is to multiply the number of outcomes in each event. With the first example using 2 coins, there are 2 outcomes when you flip the first coin and two outcomes when you flip the second coin. 2 2= 4. In the last example, we choose from 2 pairs of pants, then from three shirts. Notice the total number of outcomes we identified using the tree diagram was 6 and 2 3= 6. Those examples lead us to the following generalization: Fundamental Counting Principle: If one event can occur in m ways, and for each of these ways a second event can happen in n ways, then the number of ways that the two events can occur is m n. Example: How many possible outcomes are there if you roll two cubes with the numbers one through six written on each face? There are 6 outcomes on the first cube, 6 outcomes on the second cube, so using the Fundamental Counting Principal we have 6 6 36 = outcomes. Example: How many possible outcomes are possible for tossing a coin and rolling a cube with the numbers one through six written on each face? There are two things that can happen when tossing a coin. There are six things that can happen when rolling the cube. Using the Fundamental Counting Principle, we have 2 6= 12outcomes. Example: How many possible answers are there to a 10 question True-False test? Using Fundamental Counting Principal, 10 2 2 2 2 2 2 2 2 2 2 = 2 = 1024. Let s extend this to finding the probability of compound events (an event made up of two or more separate events). If the occurrence of one event does not affect the probability of the other, the events are independent. Math 7, Unit 11: Probability Holt: Chapter 11 Page 6 of 14

Probability of Independent Events = P(A) i P(B) Example: An experiment consists of flipping a coin 2 times. What is the probability of flipping heads both times? The flip of a coin does not affect the results of the other flips, so the flips are independent. For each flip, P(H) = 1 1 1 1. So P(H, H) = i or 2 2 2 4 Example: You have 3 colors of t-shirts (red, blue, green) 2 colors of shorts (white, black) from which to choose. What is the probability of randomly choosing a blue shirt with black pants? For the choice of shirt, P(T) = 1 3, for the shorts P(S) = 1 1 1 1 ; So P(T, S) = i or 2 2 3 6 Probability of Dependent Events P(A and B) = P(A) P(B after A) Example: Seven different books are on a shelf in the classroom. If Jewel chooses a book from the shelf to read, and then Cheryl chooses a book from the ones that remain, what is the probability of them choosing Book 1 and Book 2? P(Book 1) = 1 7. The P(Book 2) = 1 6. P(Book 1, then Book 2) 1 1 = 1 7 6 42 First, we need to learn a little notation. Factorials We are going to abbreviate products that start at one number and work their way back to one by using an exclamation point (!). In math, however, that won t mean the number is excited. And we won t call it an exclamation point; we ll call it a factorial. So 5! is read as five factorial. Math 7, Unit 11: Probability Holt: Chapter 11 Page 7 of 14

Examples: 5! = 5 4 3 2 1 3! = 3 2 1 1! = 1 = 120 = 6 4! = 4 3 2 1 2! = 2 1 = 24 =2 Permutations Syllabus Objective: (7.1 )The student will determine the number of permutations using a variety of counting methods. Example: If a different person must be selected for each position, in how many ways can we choose the president, vice president, and secretary from a group of seven members if the first person chosen is the president, the second the vice president, and the third is the secretary? We have a total of 7 people taken three at a time. Using the Fundamental Counting Principle, the first person can be chosen 7 ways, the next 6, and the third 5, we have 7 6 5 or 210 ways of choosing the officers. Another way of doing the same problem is by developing a formula. Let s see what that might look like and define it. Permutation is an arrangement of objects in which the order matters without repetition. Order typically matters when there is position or awards like first place, second place, or when someone is named president or vice president. The notation used to represent the number of permutations of a set of n objects taken r at a time is P. In the last example, we would use the notation 7 P 3 to represent picking three people out of the seven. We could then use the Fundamental Counting principle to determine the number of permutations. 7P 3= 7 6 5 = 210 Generalizing this algebraically, we could develop the formula for a permutation. n r n P r n! = ( n r)! Math 7, Unit 11: Probability Holt: Chapter 11 Page 8 of 14

Using the formula for the last example would give us: P = 7 3 7! 7 3! ( ) 7! = 4! 7 6 5 4 3 2 1 = 4 3 2 1 = 7 6 5 = 210 Example: There are 5 runners in a race. How many different permutations are possible for the places in which the runners finish? By formula, we have a permutation of 5 runners being taken 5 at a time. 5! P = (5 5)! 5! = 0! 5 5 Note: 0! is defined as 1 = 5! or 5 4 3 2 1 = 120 Notice we could have just as easily used the Fundamental Counting Principle to solve this problem. In using a permutation or the Fundamental Counting Principle, order matters. A permutation does not allow repetition. For instance, in finding the number of arrangements of license plates, the digits can be re-used. In other words, someone might have the license plate 333 333. To determine the possible number of license plates, I could not use the permutation formula because of the repetitions. I would have to use the Fundamental Counting Principle. Since there are 10 ways to choose each digit on the license plate, the number of plates would be determined by 10 10 10 10 10 10 = 1,000,000. Math 7, Unit 11: Probability Holt: Chapter 11 Page 9 of 14

Combinations Note: Combinations are NOT a Nevada State Standard at the 7 th grade level. These notes are included here for teacher development only. Combination is an arrangement of objects in which the order does not matter without repetition. This is different from a permutation because the order does not matter. If you change the order, you don t change the group and you do not make a new combination. So, a dime, nickel and penny is the same combination of coins as a penny, dime and nickel. Example: Bob has four golf shirts. He wants to take two of them on his golf outing. How many different combinations of two shirts can he take? One way I could solve this problem is to create a table. 1 2 3 4 1 1, 2 1, 3 1, 4 2 2, 1 2, 3 2, 4 3 3, 1 3, 2 3, 4 4 4, 1 4, 2 4, 3 Then, since choosing Shirt 1 and Shirt 2 is the same as choosing Shirt 2 and Shirt 1, we can eliminate duplicates (because order does not matter). 1 2 3 4 1 1, 2 1, 3 1, 4 2 2, 1 2, 3 2, 4 3 3, 1 3, 2 3, 4 4 4, 1 4, 2 4, 3 So, there are 6 combinations of choosing two shirts. Math 7, Unit 11: Probability Holt: Chapter 11 Page 10 of 14

Another way we could solve this problem is using the Fundamental Counting Principle. Bob could choose his first shirt four ways, his second shirt three ways, so 4 3, for a total of 12 ways. But, hold on a minute. Let s say those shirts each had a different color. By using the Fundamental Counting Principle, that would suggest picking the blue shirt, then the yellow shirt is different from picking the yellow first, then the blue. We don t want that to happen since the order does not matter. To find the number of combinations, in other words, eliminating the order of the two shirts, we would divide the 12 permutations by 2! or 2 1. There would be six different shirt combinations Bob could take on his outing. In essence, a combination is nothing more than a permutation that is being divided by the different orderings of that permutation. The notation we will use will follow that of a permutation, n C r = Pr r! n Example: From among 12 students trying out for the basketball team, how many ways can 7 students be selected? Does the order matter? Is this a permutation or combination? Well, if you were going out for the team and a list was printed, would it matter if you were listed first or last? All you would care about is that your name is on the list. The order is not important, therefore this would be a combination problem of 12 students taken 7 at a time. C 12 7 = P 7! 12 7 Use the combination formula. Remember 12! P = (12 7)! 12 7 12! 5! 12! 12 11 10 9 8 7 6 5/ 4/ 3/ 2/ 1/ = Knowing = 7! 5! 5/ 4/ 3/ 2/ 1/ 3 2 12 11 10 9 8 7 6 = 7 6 5 4 3 2 1 = 792 Math 7, Unit 11: Probability Holt: Chapter 11 Page 11 of 14

There would be 792 different teams that can be chosen. In summary, when order matters and there is no repetition, use a permutation. If order matters and there is repetition, then use the Fundamental Counting Principle. If order does not matter, use a combination. In just about all cases, you can use the Fundamental Counting Principle to determine the size of the sample space. The formulas for permutation and combination just allow us to compute the answers quickly. However, if you read a problem and have trouble determining if it s a permutation or combination, then do it by the Fundamental Counting Principle. Venn Diagrams Syllabus Objective: (7.7 Accelerated only)the student will use set theory to find probabilities. Let s look at a Venn Diagram made up of three sets in which the regions are labeled. We ll describe each region. A 1 4 2 3 5 6 B U 7 C 8 Region 5 is in all three circles. So any elements in region 5 would belong to all three sets. What about Region 2? Those are the elements in A and B, but not C. How might you describe Region 6? Those are elements in B and C, but not in A. Try Region 4. The elements in A and C, but not B. Math 7, Unit 11: Probability Holt: Chapter 11 Page 12 of 14

Let s look at some more regions. Region 1 describes the elements in A only. What about region 3? Those elements are only in B. Region 7 then would be the elements in C only. Region 8 would describe elements that are not members of any of the sets, but belong to the universal set. It s important that you become familiar with how each of those regions might be described. Being able to describe those regions would allow you to solve some problems. Let s try a problem. Example: A survey was taken of 650 university students. It was reported that 240 were taking math, 290 were taking biology, and 270 were enrolled in chemistry. Of those students, 80 were taking biology and math, 70 were taking math and chemistry, 60 were taking biology and chemistry, and 50 were taking all three classes. How many students took math only? A Venn Diagram has been drawn to help show the problem. M 1 3 2 5 2 1 B U C 1 The entire circle marked M represents the students taking Math. The entire circle marked B represents the students taking Biology. The entire circle marked C represents the students taking Chemistry. To determine how many students took only math, begin with the information given and fill in the diagram. 80 were taking biology and math 70 were taking math and chemistry 60 were taking biology and chemistry Math 7, Unit 11: Probability Holt: Chapter 11 Page 13 of 14

50 were taking all three classes. U M 70 80 50 60 B C Since I am trying to determine the number of students who took math only, I compute how many students are in the entire math circle marked M. Adding 80 + 50 + 70 = 200. The problem stated there were a total of 240 students taking Math, so I compute 240 200 = 40. U 40 80 M 70 50 60 B C Other questions you can answer: 1. How many took Biology only? 2. How many took Chemistry only? 3. How many students took exactly two of the courses? Answers: 1. 290 (80 + 50 + 60) = 100 2. 270 (70 + 50 + 60) = 90 3. 80 + 70 + 60 = 210 Math 7, Unit 11: Probability Holt: Chapter 11 Page 14 of 14