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Chapter 12 Section 7 - Slide 1 Copyright © 2009 Pearson Education, Inc. AND.

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Presentation on theme: "Chapter 12 Section 7 - Slide 1 Copyright © 2009 Pearson Education, Inc. AND."— Presentation transcript:

1 Chapter 12 Section 7 - Slide 1 Copyright © 2009 Pearson Education, Inc. AND

2 Copyright © 2009 Pearson Education, Inc. Chapter 12 Section 7 - Slide 2 Chapter 12 Probability

3 Chapter 12 Section 7 - Slide 3 Copyright © 2009 Pearson Education, Inc. WHAT YOU WILL LEARN Conditional probability The counting principle, permutations

4 Copyright © 2009 Pearson Education, Inc. Chapter 12 Section 7 - Slide 4 Section 7 Conditional Probability

5 Chapter 12 Section 7 - Slide 5 Copyright © 2009 Pearson Education, Inc. Conditional Probability In general, the probability of event E 2 occurring, given that an event E 1 has happened (or will happen; the time relationship does not matter), is called a conditional probability and is written P(E 2 |E 1 ).

6 Chapter 12 Section 7 - Slide 6 Copyright © 2009 Pearson Education, Inc. Example Given a family of two children, and assuming that boys and girls are equally likely, find the probability that the family has a) two girls. b) two girls if you know that at least one of the children is a girl. c) two girls given that the older child is a girl.

7 Chapter 12 Section 7 - Slide 7 Copyright © 2009 Pearson Education, Inc. Solutions a)two girls There are four possible outcomes BB, BG, GB, and GG. b) two girls if you know that at least one of the children is a girl

8 Chapter 12 Section 7 - Slide 8 Copyright © 2009 Pearson Education, Inc. Solutions (continued) two girls given that the older child is a girl

9 Chapter 12 Section 7 - Slide 9 Copyright © 2009 Pearson Education, Inc. Conditional Probability For any two events, E 1 and E 2,

10 Chapter 12 Section 7 - Slide 10 Copyright © 2009 Pearson Education, Inc. Example Use the results of the taste test given at a local mall. If one person from the sample is selected at random, find the probability the person selected: 257127130Total 1327260Women 1255570Men TotalPrefers Wintergreen Prefers Peppermint

11 Chapter 12 Section 7 - Slide 11 Copyright © 2009 Pearson Education, Inc. Example (continued) a) prefers peppermint b) is a woman

12 Chapter 12 Section 7 - Slide 12 Copyright © 2009 Pearson Education, Inc. Example (continued) c) prefers peppermint, given that a woman is selected d) is a man, given that the person prefers wintergreen

13 Copyright © 2009 Pearson Education, Inc. Chapter 12 Section 7 - Slide 13 Section 8 The Counting Principle and Permutations

14 Chapter 12 Section 7 - Slide 14 Copyright © 2009 Pearson Education, Inc. Counting Principle If a first experiment can be performed in M distinct ways and a second experiment can be performed in N distinct ways, then the two experiments in that specific order can be performed in M N distinct ways.

15 Chapter 12 Section 7 - Slide 15 Copyright © 2009 Pearson Education, Inc. Example A password to logon to a computer system is to consist of 3 letters followed by 3 digits. Determine how many different passwords are possible if: a) repetition of letters and digits is permitted b) repetition of letters and digits is not permitted c) the first letter must be a vowel (a, e, I, o, u), the first digit cannot be 0, and repetition of letters and digits is not permitted.

16 Chapter 12 Section 7 - Slide 16 Copyright © 2009 Pearson Education, Inc. Solutions a. repetition of letters and digits is permitted.  There are 26 letters and 10 digits. We have 6 positions to fill.

17 Chapter 12 Section 7 - Slide 17 Copyright © 2009 Pearson Education, Inc. Solution b. repetition of letters and digits is not permitted.

18 Chapter 12 Section 7 - Slide 18 Copyright © 2009 Pearson Education, Inc. Solution c. the first letter must be a vowel (a, e, i, o, u), the first digit cannot be 0, and repetition of letters and digits is not permitted.

19 Chapter 12 Section 7 - Slide 19 Copyright © 2009 Pearson Education, Inc. Permutations A permutation is any ordered arrangement of a given set of objects. Number of Permutations The number of permutations of n distinct items is n factorial, symbolized n!, where n! = n(n  1)(n  2) (3)(2)(1)

20 Chapter 12 Section 7 - Slide 20 Copyright © 2009 Pearson Education, Inc. Example How many ways can 6 different stuffed animals be arranged in a line on a shelf? 6! = 6 5 4 3 2 1 = 720 The 6 stuffed animals can be arranged in 720 different ways.

21 Chapter 12 Section 7 - Slide 21 Copyright © 2009 Pearson Education, Inc. Example Consider the six numbers 1, 2, 3, 4, 5 and 6. In how many distinct ways can three numbers be selected and arranged if repetition is not allowed? 6 5 4 = 120 Thus, there are 120 different possible ordered arrangements, or permutations.

22 Chapter 12 Section 7 - Slide 22 Copyright © 2009 Pearson Education, Inc. Permutation Formula The number of permutations possible when r objects are selected from n objects is found by the permutation formula

23 Chapter 12 Section 7 - Slide 23 Copyright © 2009 Pearson Education, Inc. Example The swimming coach has 8 swimmers who can compete in a “new” 100m relay (butterfly, backstroke, free style), he must select 3 swimmers, one for each leg of the relay in the event. In how many ways could he select the 3 swimmers?

24 Chapter 12 Section 7 - Slide 24 Copyright © 2009 Pearson Education, Inc. Permutations of Duplicate Objects The number of distinct permutations of n objects where n 1 of the objects are identical, n 2 of the objects are identical, …, n r of the objects are identical is found by the formula

25 Chapter 12 Section 7 - Slide 25 Copyright © 2009 Pearson Education, Inc. Example In how many different ways can the letters of the word “CINCINNATI” be arranged? Of the 10 letters, 2 are C’s, 3 are N’s, and 3 are I’s.


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