1 Lesson 6.2.3 Counting Outcomes. 2 Lesson 6.2.3 Counting Outcomes California Standards: Statistics, Data Analysis and Probability 3.1 Represent all possible

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Lesson 6.2.3Lesson 6.2.3

Counting OutcomesCounting Outcomes

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Lesson

6.2.3Counting OutcomesCounting Outcomes

California Standards:Statistics, Data Analysis and Probability 3.1Represent all possible outcomes for compound events in an organized way (e.g., tables, grids, tree diagrams) and express the theoretical probability of each outcome.Statistics, Data Analysis and Probability 3.3 Represent probabilities as ratios, proportions, decimals between 0 and 1, and percentages between 0 and 100 and verify that the probabilities computed are reasonable; know that if P is the probability of an event, 1–P is the probability of an event not occurring.

What it means for you:You’ll see how ways of counting outcomes such as grids and tree diagrams can help you work out probabilities of events.

Key words:• probability• outcome• event• grid• table• tree diagram

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Counting OutcomesCounting OutcomesLesson

6.2.3

Last Lesson, you saw how to find probabilities with this formula.

P(event) = Number of favorable outcomesNumber of possible outcomes

But to use it, you need to know how many possible and favorable outcomes there are.

Luckily, you learned two ways to do that a few Lessons ago.

Now you’ll see just how useful tables, grids, and tree diagrams can be.

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Tables and Grids Help to Calculate Probabilities

Lesson

6.2.3

You know that: P(event) = Number of favorable outcomesNumber of possible outcomes

So if you can make an organized list of possible outcomes, it makes it easier to find probabilities.

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Example 1

Lesson

6.2.3

Solution

Solution follows…

Two dice are rolled. Find the probability that the sum of the numbers rolled is: (i) 8, and (ii) greater than 8.

2 3 4 5 6 7

3 4 5 6 7 8

4 5 6 7 8 9

5 6 7 8 9 10

6 7 8 9 10 11

7 8 9 10 11 12

1 2 3 4 5 6

1st die rolled

1

2

3

4

5

62

nd d

ie r

olle

d

(i) This table shows all the possible outcomes for the sum of the numbers rolled on two dice.

All you need to do is count how many possible and favorable outcomes there are, and put them into the formula.

Solution continues…

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Example 1

Lesson

6.2.3

Solution (continued)


2 3 4 5 6 7

3 4 5 6 7 8

4 5 6 7 8 9

5 6 7 8 9 10

6 7 8 9 10 11

7 8 9 10 11 12

1 2 3 4 5 6

1st die rolled

1

2

3

4

5

62

nd d

ie r

olle

d

There are 5 ways to roll 8, and there are 36 possible outcomes altogether.

So, P(rolling 8) =5

36


Favorable outcomes

Possible outcomes

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Example 1

Lesson

6.2.3



2 3 4 5 6 7

3 4 5 6 7 8

4 5 6 7 8 9

5 6 7 8 9 10

6 7 8 9 10 11

7 8 9 10 11 12

1 2 3 4 5 6

1st die rolled

1

2

3

4

5

62

nd d

ie r

olle

d

(ii) The favorable outcomes for the event “rolling a sum greater than 8” are the outcomes 9, 10, 11, and 12.

In total, this makes 10 outcomes.

So, P(rolling over 8) = =5

181036

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Guided Practice

Solution follows…

Lesson

6.2.3

Use the table below to find the probability of each of the following events when rolling two dice:

1. Rolling a sum of 9



4. Rolling a sum less than 6


6. Rolling a sum that is a prime number

2 3 4 5 6 7

3 4 5 6 7 8

4 5 6 7 8 9

5 6 7 8 9 10

6 7 8 9 10 11

7 8 9 10 11 12

1 2 3 4 5 6

1st die rolled

1

2

3

4

5

62

nd d

ie r

olle

d

P(rolling 9) = =4

36

1

9

P(rolling 10) = =3

36

1

12

P(rolling 2) =1

36

P(rolling 17) = 0

P(rolling < 6) = =10

36

5

18

P(rolling prime) = =15

36

5

12

9


Guided Practice

Solution follows…

Lesson

6.2.3

The table lists the number of students at a middle school by grade and gender. One student is chosen at random.

Determine the probability of choosing each type of student listed in Exercises 7–11. Write each probability as a percent.

7. P(girl)

8. P(boy)

9. P(6th grader)

10. P(7th grader)

11. P(8th grader)

Boys Girls Totals

57 51 108Grade 6

42 57 99Grade 7

45 48 93Grade 8

144 156 300Totals

(156 ÷ 300) × 100 = 52%

(144 ÷ 300) × 100 = 48%

(108 ÷ 300) × 100 = 36%

(99 ÷ 300) × 100 = 33%

(93 ÷ 300) × 100 = 31%

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Guided Practice

Solution follows…

Lesson

6.2.3

The table lists the number of students at a middle school by grade and gender. One student is chosen at random.

Determine the probability of choosing each type of student listed in Exercises 12–16. Write each probability as a percent.

12. P(6th grade boy)

13. P(7th grade girl)

14. P(8th grade boy)

15. P(9th grader)

16. P(6th, 7th, or 8th grade student)

Boys Girls Totals

57 51 108Grade 6

42 57 99Grade 7

45 48 93Grade 8

144 156 300Totals

(57 ÷ 300) × 100 = 19%

(57 ÷ 300) × 100 = 19%

(45 ÷ 300) × 100 = 15%

(0) × 100 = 0%

(300 ÷ 300) × 100 = 100%

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Example 2

Lesson

6.2.3

Solution

Solution follows…

A coin is tossed three times. Find the probability of getting tails on exactly two of the three tosses.

Tree diagrams can help you find probabilities too.

This tree diagram shows all the possible outcomes for tossing a coin three times.

HHH HHT HTH HTT THH THT TTH TTTOutcomes

Coin toss 3

Coin toss 2

Coin toss 1


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Example 2

Lesson

6.2.3


A coin is tossed three times. Find the probability of getting tails on exactly two of the three tosses.

There are 8 possible outcomes.

There are 3 outcomes that have exactly two tails: HTT, THT, and TTH.

So, P(exactly 2 tails) =38 HHH HHT HTH HTT THH THT TTH TTT

Coin toss 3

Coin toss 2

Coin toss 1

Favorable outcomes

Possible outcomes

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Guided Practice

Solution follows…

Lesson

6.2.3

Write your answers to Exercises 17–21 as decimals. Use the tree diagram shown to find the probability of each of the following events for tossing a coin three times:

17. Heads on exactly 2 tosses

18. Tails on 2 or more tosses

19. Tails on fewer than 2 tosses

20. The coin landing on the same side all 3 tosses

21. The first and third coins landing on the same side HHH HHT HTH HTT THH THT TTH TTT

Coin toss 3

Coin toss 2

Coin toss 1HHT, HTH, THH 3 ÷ 8 = 0.375

HTT, THT, TTH, TTT 4 ÷ 8 = 0.5

HHH, HHT, HTH, THH 4 ÷ 8 = 0.5

HHH, TTT 2 ÷ 8 = 0.25

HHH, HTH, TTT, THT 4 ÷ 8 = 0.5

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Independent Practice

Solution follows…

Lesson

6.2.3

Exercises 1–4 are about the color spinner shown below.

The spinner is spun once. Find the following probabilities.

1. P(green)

2. P(yellow)

3. P(white)

4. P(not purple)

1

4

1

0

1

4

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Solution follows…

Lesson

6.2.3

Exercise 5 is about the color spinner shown below.

5. The spinner is spun twice. Make a tree diagram or a table that shows all the possible outcomes. Use the abbreviations R for red, Y for yellow, G for green, and B for blue.

R Y G B

R Y G B R Y G B R Y G B R Y G B

1st spin

2nd spin

Outcomes RR RY RG RB YR YY YG YB GR GY GG GB BR BY BG BB

BGYR

BRGRYRRRR

BYGYYYRYY

BGGGYGRGG

BBGBYBRBB

1st spin

2nd s

pin

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Solution follows…

Lesson

6.2.3

Use your diagram from Exercise 5 to answer Exercises 6–7.

16

4

BGYR

BRGRYRRRR

BYGYYYRYY

BGGGYGRGG

BBGBYBRBB

1st spin

2nd s

pin

6. How many possible outcomes are there?

7. How many ways can you get 2 spins of the same color?

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Solution follows…

Lesson

6.2.3

Use your diagram from Exercise 5 to answer Exercises 8–12.

Find the following probabilities.Write your answers as fractions in their lowest terms.

8. P(2 spins the same color)

9. P(one blue and one green in any order)

10. P(blue first, then green)

11. P(exactly one spin is red)

12. P(at least one spin is red)

BGYR

BRGRYRRRR

BYGYYYRYY

BGGGYGRGG

BBGBYBRBB

1st spin

2nd s

pin

1

4

3

8

1

16

7

16

1

8

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Solution follows…

Lesson

6.2.3

Contestants on a game show spin a wheel twice. If the product of the numbers is odd, they win a prize. The numbers on the wheel are 1, 2, 4, 5, and 6.

13. Make a table with a grid that shows all the possible products from any two spins of the wheel.

14. What is the probability that a contestant will win?

15. What is the probability that a contestant will lose?

(4 ÷ 25) × 100 = 16%

(21 ÷ 25) × 100 = 84%

× 1 2 4 5 6

1 1 2 4 5 6

2 2 4 8 10 12

4 4 8 16 20 24

5 5 10 20 25 30

6 6 12 24 30 36

First spin

Sec

on

d s

pin

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Solution follows…

Lesson

6.2.3

The wheel in Exercises 13–15 had two odd numbers and three even numbers on it.

16. Design a new wheel that has an equal number of odd and even numbers on it.

17. Use your new wheel. Determine which probability is greater, P(even product) or P(odd product).

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For example,

Using the wheel shown in Exercise 16, the possible outcomes are: 1 × 1 = 1, 1 × 2 = 2, 2 × 2 = 4 and 2 × 1 = 2.3 out of the 4 outcomes are even, so P(even product) = 0.75.1 out of the 4 outcomes is odd, so P(odd product) = 0.25. P(even product) is greater

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Solution follows…

Lesson

6.2.3

18. If your wheel in Exercise 16 had a different number of integers on it (but half of those integers were still even, with the rest being odd), would your answer to Exercise 17 have been different? Explain your answer.No. To get an odd product, you must multiply two odd numbers together.

Getting two odd numbers will always be the case for exactly a quarter of the possible outcomes (since the possibilities are “odd × odd,” “odd × even,” “even × odd,” and “even × even”) — so your answer to Exercise 17 will always be the same.

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Couting OutcomesCouting OutcomesLesson

6.2.3

Round UpRound Up

The most important use of tree diagrams and tables of outcomes is in finding probabilities.

Look back at Lessons 6.1.1–6.1.3 if you need a reminder about anything to do with listing possible outcomes.

Documents

1 Lesson 6.2.3 Counting Outcomes. 2 Lesson 6.2.3 Counting Outcomes California Standards: Statistics, Data Analysis and Probability 3.1 Represent all possible