counting methods lesson 11-5 pg. # 426-428. ca content standards statistics, data analysis, and...

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Counting Methods Lesson 11-5 Lesson 11-5 Pg. # 426-428 Pg. # 426-428

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Page 1: Counting Methods Lesson 11-5 Pg. # 426-428. CA Content Standards Statistics, Data Analysis, and Probability 3.1 ***: I represent all possible outcomes

Counting Methods

Lesson 11-5Lesson 11-5

Pg. # 426-428Pg. # 426-428

Page 2: Counting Methods Lesson 11-5 Pg. # 426-428. CA Content Standards Statistics, Data Analysis, and Probability 3.1 ***: I represent all possible outcomes

CA Content StandardsStatistics, Data Analysis, and Probability 3.1 ***:

I represent all possible outcomes of compound events in an organized way and express the theoretical probability of each outcome.

Statistics, Data Analysis, and Probability 3.3***:

I represent probabilities as ratios and percentages between 0 and 100.

Page 3: Counting Methods Lesson 11-5 Pg. # 426-428. CA Content Standards Statistics, Data Analysis, and Probability 3.1 ***: I represent all possible outcomes

Vocabulary: TREE DIAGRAM

A diagram used to organize outcomes of an experiment to make them easier to count.

Page 4: Counting Methods Lesson 11-5 Pg. # 426-428. CA Content Standards Statistics, Data Analysis, and Probability 3.1 ***: I represent all possible outcomes

Vocabulary: COUNTING PRINCIPLEIf one choice can be made in m ways and a second choice can be made in n ways, then the two choices can be made together in m x n ways.

Page 5: Counting Methods Lesson 11-5 Pg. # 426-428. CA Content Standards Statistics, Data Analysis, and Probability 3.1 ***: I represent all possible outcomes

Objective

Use a tree diagram or the counting principle to find the total number of outcomes for an event.

Math Link: You can use what you know about making tables and tree diagrams to find all possible outcomes.

Page 6: Counting Methods Lesson 11-5 Pg. # 426-428. CA Content Standards Statistics, Data Analysis, and Probability 3.1 ***: I represent all possible outcomes

Example 1. The cooking club is having a sandwich fundraiser. A sandwich consists of one choice of bread and one choice of filling. How many different kinds of sandwiches can be made with the following ingredients?

Bread: Pita, Tortilla

Filling: Chicken, Beef, Vegetable

Page 7: Counting Methods Lesson 11-5 Pg. # 426-428. CA Content Standards Statistics, Data Analysis, and Probability 3.1 ***: I represent all possible outcomes

You can use a tree diagram…To represent the different kinds of sandwiches.

Sandwich Fundraiser

ChickenOutcome 1

BeefOutcome 2

VegetableOutcome 3

Pita

ChickenOutcome 4

BeefOutcome 5

VegetableOutcome 6

Tortilla

Sandwich

Menu includes: pc, pb, pv, tc, tb, tv

Page 8: Counting Methods Lesson 11-5 Pg. # 426-428. CA Content Standards Statistics, Data Analysis, and Probability 3.1 ***: I represent all possible outcomes

Example 2. One of each type of sandwich is prepared. You choose one without looking. Find the probability of getting a beef sandwich.

The menu includes 6 kinds of sandwiches: pc, pb, pv, tc, tb, tv2 kinds of sandwiches have beef: pb, tb

So…

P(beef) = 2/6 = 1/3 = 33%

The probability of getting a beef sandwich is about 33%.

Page 9: Counting Methods Lesson 11-5 Pg. # 426-428. CA Content Standards Statistics, Data Analysis, and Probability 3.1 ***: I represent all possible outcomes

Example 3. If there are three kinds of breads and four kinds of fillings at the fundraiser, how many different kinds of sandwiches can be made?

Use the counting principle.

Possible Possible Kinds ofBreads FillingsSandwiches3 x 4 = 12

Twelve different kinds of sandwiches could be made.

If there are m possible outcomes for the first event and n possible outcomes for the second event, then there are m x n possible

outcomes.

Page 10: Counting Methods Lesson 11-5 Pg. # 426-428. CA Content Standards Statistics, Data Analysis, and Probability 3.1 ***: I represent all possible outcomes

Moral of the Story

Use tree diagrams to organize possible outcomes. Also, if there are m possible outcomes for the first event and n possible outcomes for the second event, then there are m x n total possible outcomes.