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12.5: The Counting Principle

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Introduction

Alicia’s Outfit

Remember Alicia’s outfit? Well, when Alicia was working on figuring out the number of possible outfits that she could create, we used a tree diagram. A tree diagram is very useful because it provides us with a visual display of the data.

From the tree diagram, we could count all of the possible outcomes. There are 12 possible outfits for Alicia to choose from.

What about if we didn’t want to draw a tree diagram? Is there another way that we could have thought about figuring out the number of outfits possible?

This lesson is all about The Counting Principle. The Counting Principle makes figuring outcomes possible by using a mathematical method, not a tree diagram. Use this lesson to learn about The Counting Principle and we will apply it to Alicia’s outfits at the end of the lesson.

What You Will Learn

In this lesson, you will learn how to do the following things:

  • Recognize the number of possible outcomes of a series of events as the product of the number of possible outcomes for each event.
  • Use the Counting Principle to find all possible outcomes of a series of events involving two or more choices or results.
  • Find probabilities of specified outcomes using the Counting Principle.

Teaching Time

I. Recognize the Number of Possible Outcomes of a Series of Events as the Product of the Number of Possible Outcomes for Each Event

In the last lesson, you learned about tree diagrams. Tree diagrams provide you with a visual way of seeing all of the possible outcomes for a set of particular events.

What if there was a simpler way?

Sometimes, you don’t want to have to draw a tree diagram to figure out all of the possible outcomes for a series of events. When this happens, you can use another principle to figure out the possible outcomes.

Let’s look at an example.

Example

Molly’s All Star Farm Breakfast features 3 choices of eggs–scrambled, fried, or omelet–plus a choice of bacon or sausage. You can use a tree diagram to find that there are 6 different choices, or outcomes, for the breakfast.

What if we wanted to look at this in another way?

We could look at the number of possible breakfast options in terms of outcomes.

For the first choice there are 3 different outcomes. For the second choice there are 2 different outcomes.

3 \ \text{outcomes} \cdot 2 \ \text{outcomes} = 6 \ \text{outcomes}

Yes they are. That is because this is an example of a new principle for figuring out the total possible outcomes of a series of events. We call it the Counting Principle.

Counting Principle: The number of choices or outcomes for two independent events, A and B, taken together, is the product of the total number of outcomes for each event.

Total outcomes for A and B = (\text{number of outcomes for} \ A) \cdot (\text{number of outcomes for} \ B)

II. Use the Counting Principle to Find all Possible Outcomes of a Series of Events Involving Two or More Choices or Results

Now that you know what the Counting Principle is, you can practice using it. The Counting Principle will work for 2, 3 even 4 or more different events. Just follow the procedure of using multiplication to find the number of possible outcomes.

Let’s apply the Counting Principle to an example.

Example

For buying gum you have the following choices:

  • 3 flavor choices–spearmint, peppermint, cinnamon
  • 2 sugarless choices–sugarless or non-sugarless
  • 2 bubble choices–bubble gum or regular

To find the number of gum choices you have you could make a tree diagram, or you could simply use the Counting Principle:

3 \ \text{choices} \cdot 2 \ \text{choices} \cdot 2 \ \text{choices} = 12 \ \text{choices}

To check the answer, you can write out all of the possible options for gum:

& \text{spear-sugarless-bubble} && \text{pepper-sugarless-bubble} && \text{cinnamon-sugarless-bubble}\\& \text{spear-sugarless-regular} && \text{pepper-sugarless-regular} && \text{cinnamon-sugarless-}\\& \text{spear-non-bubble} && \text{pepper-non-bubble} && \text{regular}\\& \text{spear-non-regular} && \text{pepper-non-regular} && \text{cinnamon-non-bubble}\\& && && \text{cinnamon-non-regular}

You can see that the Counting Principle worked out fine for the solution.

Example

You’re buying a sweater and have the following choices.

  • 5 different color choices–black, yellow, blue, red, green
  • 3 different material choices–wool, cotton, fleece
  • 4 different style choices–v-neck, crew, button-down, turtle

To find the number of sweater choices you have, you could make a tree diagram, or you could simply use the Counting Principle:

5 \ \text{choices} \cdot 3 \ \text{choices} \cdot 4 \ \text{choices} = 60 \ \text{choices}

12H. Lesson Exercises

Use the Counting Principle to count outcomes.

  1. Omar is buying a skateboard. He has 5 different skateboard decks to choose from and 4 different wheel choices. How many different skateboard choices does Omar have?
  2. Ice Stone ice cream shop has 3 different sundae sizes: baby, large, and grand. You can choose from 6 different ice cream flavors and add 4 different toppings. How many sundae choices are there?
  3. Gina tosses a number cube 2 times. How many different outcomes are there?

Take a few minutes to check your answers with a partner.

III. Find Probabilities of Specified Outcomes Using the Counting Principle

You can use the Counting Principle to help find probabilities of events. For example, suppose you wanted to know the probability of the arrow landing on the same color on both spinners. Keep in mind that for any probability you can use this ratio.

P (\text{event}) = \frac{favorable \ outcomes}{total \ outcomes}

Here, you can use the Counting Principle to find the number of total outcomes for the two spins with the two spinners. There are four outcomes on one spinner and three outcomes on the other spinner.

\text{Total outcomes} &= 4 \ \text{outcomes} \cdot 3 \ \text{outcomes}\\&= 12 \ \text{outcomes}

Now list those 12 outcomes and mark the outcomes that are the same color for both spins.

& \text{red-red} && \text{blue-red} && \text{yellow-red} && \text{green-red}\\& \text{red-blue} && \text{blue-blue} && \text{yellow-blue} && \text{green-blue}\\& \text{red-green} && \text{blue-green} && \text{yellow-green} && \text{green-green}

Since there are 3 outcomes that have the same color for both spins:

P (\text{same}) = \frac{3}{12}=\frac{1}{4}

The probability of both spinners landing on the same color is \frac{1}{4}.

You can also apply the Counting Principle to a variety of different probability problems.

Example

Anna flips a coin 3 times in a row. What is the probability that she will get heads all 3 times?

Step 1: Rather than draw a tree diagram, to find the number of total outcomes you can simply multiply the number of outcomes for each flip

\text{Total outcomes} &= 2 \ \text{outcomes} \cdot 2 \ \text{outcomes} \cdot 2 \ \text{outcomes}\\&= 8 \ \text{outcomes}

Step 2: Now list all 8 outcomes and find the number of ways Anna can get heads all 3 times. Clearly, there is only one arrangement in which all 3 flips result in heads.

& \text{heads-heads-heads} && \text{tails-heads-heads}\\& \text{heads-heads-tails} && \text{tails-heads-tails}\\& \text{heads-tails-heads} && \text{tails-tails-heads}\\& \text{heads-tails-tails} && \text{tails-tails-tails}

Step 3: Find the ratio of favorable outcomes to total outcomes:

P (\text{red-red-red}) = \frac{1}{8}

12I. Lesson Exercises

Use the Counting Principle to find each probability.

  1. Abra flips a coin 2 times. What is the probability that both flips will match?
  2. Abra flips a coin 2 times. What is the probability that both flips will NOT match?
  3. Abra flips a coin 3 times. What is the probability that all 3 flips will match?

Check each answer with a friend.

Real–Life Example Completed

Alicia’s Outfit

Here is the original problem once again. Reread it and then we will look at applying The Counting Principle to this problem.

Remember Alicia’s outfit? Well, when Alicia was working on figuring out the number of possible outfits that she could create, we used a tree diagram. A tree diagram is very useful because it provides us with a visual display of the data.

From the tree diagram, we could count all of the possible outcomes. There are 12 possible outfits for Alicia to choose from.

What about if we didn’t want to draw a tree diagram? Is there another way that we could have thought about figuring out the number of outfits possible?

Think back to the lesson about The Counting Principle.

Total Outcomes = (Number of outcomes)(Number of outcomes)

With Alicia’s problem, there are three possible numbers of outcomes. We have the shirts that she selected as options, there are two of them. We have the skirts that she selected as options, there are three of those. We have the shoes that she selected as options, there are two of those.

3 \times 2 \times 2 = 12 \ \text{possible outfits}

You can see that we ended up with the same answer as we did with the tree diagram. The Counting Principle definitely works.

Vocabulary

Here are the vocabulary words that are found in this lesson.

Probability
the ratio of favorable outcomes to total possible outcomes.
The Counting Principle
the product of the outcomes of a series of events gives the total number of outcomes.

Technology Integration

Khan Academy, Probability Part 2

James Sousa, The Counting Principle

Other Videos:

  1. http://www.5min.com/Video/Application-of-the-Fundamental-Counting-Principle-of-Probability-275652517 – This video shows how to use the Counting Principle on a variety of different problems from a worksheet.

Time to Practice

Directions: Use the Counting Principle to solve each problem.

1. The Cubs have 3 games left to play this year. How many different outcomes can there be for the three games?

2. Svetlana tosses a coin 4 times in a row. How many outcomes are there for the 4 tosses?

3. For a new tennis racquet, Danny can choose from 8 different brands, 3 different head sizes, and 4 different grip sizes. How many different racquet choices does Danny have?

4. Gina tosses a number cube 3 times. How many different outcomes are possible? 5. Gina tosses a number cube. Buster flips a coin. How many different outcomes are possible for the two events?

6. Buster flips a coin. Daoud chooses a card from a deck of 52 cards. How many different outcomes are possible for the two events?

7. Rex spins a spinner that has red, blue, and yellow sections two times. How many different outcomes are possible?

8. Daoud chooses a card from a deck of 52 cards, replaces the card in the deck, then chooses a second card. How many different outcomes are possible?

9. Patsy’s Pizza features 3 different pizza types, 14 different toppings, and 2 different sizes. How many different pizzas can you order?

10. Spud has a 2-letter password for his computer using letters only. If there are 26 different letters in the alphabet, how many different passwords are possible?

11. Doreen has a 3-digit password for her computer using digits only. If there are 10 different digits (including zero), how many different passwords are possible?

12. Sebastian has a 3-letter password for his computer using vowels (A, E, O, I, U) only. How many different passwords are possible?

Directions: Find probabilities using the Counting Principle.

13. Abra flips a coin 3 times. What is the probability that all 3 flips will NOT match?

14. Abra flips a coin 3 times. What is the probability that heads will come up exactly 2 times?

15. Billy spins the spinner twice. What is the probability that blue will come up both times?

16. Billy spins the spinner twice. What is the probability that blue will come up at least one time?

17. Billy spins the spinner twice. What is the probability that blue will come up exactly one time?

18. Billy spins the spinner twice. What is the probability that both spins will match?

19. Cindy tosses a number cube two times. What is the probability that both tosses will match?

20. Cindy tosses a number cube two times. What is the probability that both tosses will NOT match?

21. Cindy tosses a number cube two times. What is the probability that the sum of the two tosses will be 7?

22. Cindy tosses a number cube two times. What is the probability that the sum of the two tosses will be greater than 7?  

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