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Activity 6 - 2

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Activity 6 - 2. Choices. Objectives. Apply the multiplication principle of counting Determine the sample space for a probability distribution Display a sample space with a tree diagram Determine complementary probabilities Use Venn Diagrams to illustrate relationships between events. - PowerPoint PPT Presentation
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Activity 6 - 2 Choices
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Page 1: Activity 6 - 2

Activity 6 - 2

Choices

Page 2: Activity 6 - 2

Objectives

• Apply the multiplication principle of counting

• Determine the sample space for a probability distribution

• Display a sample space with a tree diagram

• Determine complementary probabilities

• Use Venn Diagrams to illustrate relationships between events

Page 3: Activity 6 - 2

Vocabulary• Tree Diagram – a way of listing all possible events in

a sample space

• Complement of event – the probability that the event does not happen

• Venn diagram – logical depictions of sample spaces

Page 4: Activity 6 - 2

Activity

Suppose you are looking for a new car and have narrowed you decision down to a Mustang, but can’t decide on the exact color, transmission, engine, or options package. There are three sizes of engines (3.0, 3.8 and 4.6 liters), two transmissions (standard and automatic), five colors you like (black, silver, red, yellow, and green), and three option packages (GL, Sport, XL). With all these possible choices, you want to know how many different Mustangs there are from which you must choose. 

1. From the choices given above, how many different Mustang combinations are possible?

3 2 5 3 = 90 different Mustangs

Page 5: Activity 6 - 2

Multiplication Principle of Counting

If you can do one task in n number of ways and a second task in m number of ways, then both tasks can be done in n m number of ways.

This can be applied to as many decisions and choices that you have (like 4 in the Activity).

Page 6: Activity 6 - 2

Example 1

How many different dinner combinations can we have if you have a choice of 3 appetizers, 2 salads, 4 entrees, and 5 deserts?

3 2 4 5 = 120 different combinations

Page 7: Activity 6 - 2

Tree Diagrams

Tree Diagram makes the enumeration of possible outcomes easier to see and determine

Running the tree out details an individual outcome

Event 1

Y

N

Event 2

Y

N

Y

N

Event 3

Y

N

Y

N

N

Y

Y

N

HTT

HTH

HHT

HHH

TTT

TTH

THT

THH

Outcomes

Page 8: Activity 6 - 2

Example 2

Given a survey with 4 “yes or no” type questions, list all possible outcomes using a tree diagram. Divide them into events (number of yes answers) regardless of order.

Page 9: Activity 6 - 2

Example 2 cont

Q 1

Y

Q 2

Y

N

Q 3

Y

N

Y

N

YNNNYNNYYNYNYNYYYYNNYYNYYYYNYYYY

OutcomesQ 4

NYNYNYNY

N

Y

NY

N

Y

N

NYNYNYNY

NNNNNNNYNNYNNNYYNYNNNYNYNYYNNYYY

Page 10: Activity 6 - 2

Example 2 contYNNN 1YNNY 2YNYN 2YNYY 3YYNN 2YYNY 3YYYN 3YYYY 4NNNN 0NNNY 1NNYN 1NNYY 2NYNN 1NYNY 2NYYN 2NYYY 3

OutcomesNumber of Yes’s

0 1 2 3 4

1 4 6 4 1

Page 11: Activity 6 - 2

Venn Diagrams

• Event A is defined to be rolling an odd number on a six sided dice

• The complement of Event A (~A) would be even numbers

• Event A and ~A is all of the sample spaceP(A) + P(~A) = 1

A ~A

Page 12: Activity 6 - 2

At least Probabilities

P(at least one) = 1 – P(complement of “at least one”)

= 1 – P(none)

0 1, 2, 3, ….

Page 13: Activity 6 - 2

Example 3

P(rolling at least 4 on a six-sided die) = ??

= 1 - P(less than 4)

= 1 – ( P(1) + P(2) + P(3) )

= 1 – ( 1/6 + 1/6 + 1/6 ) = 0.5

Page 14: Activity 6 - 2

Example 3 cont

Draw a Venn Diagram representing example 3

4, 5, 6

1, 2, 3

Page 15: Activity 6 - 2

Example 4

P(rolling at least 3 on two six-sided die) = ??

= 1 - P(less than 3)

= 1 – ( P(2) )

= 1 – ( 1/36 ) = 35/36 = 0.9722

Page 16: Activity 6 - 2

Summary and Homework

• Summary– Multiplication principle of counting allows us to

calculate the total combinations of all choicesTotal = m (choices) n (choices)

– Tree diagrams displays all outcomes for a event or a series of events

– The complement of a event is everything outside that event; P(event) + P(~event) = 1

– P(at least one) = 1 – P(none)

• Homework– page 719-723; problems 1-3, 5, 6, 9


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