chapter 8: further topics in algebra
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Chapter 8: Further Topics in Algebra. 8.1Sequences and Series 8.2Arithmetic Sequences and Series 8.3Geometric Sequences and Series 8.4The Binomial Theorem 8.5Mathematical Induction 8.6Counting Theory 8.7Probability. 8.7 Probability. Basic Concepts - PowerPoint PPT PresentationTRANSCRIPT
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Copyright © 2007 Pearson Education, Inc. Slide 8-1
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Chapter 8: Further Topics in Algebra
8.1 Sequences and Series
8.2 Arithmetic Sequences and Series
8.3 Geometric Sequences and Series
8.4 The Binomial Theorem
8.5 Mathematical Induction
8.6 Counting Theory
8.7 Probability
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Copyright © 2007 Pearson Education, Inc. Slide 8-3
8.7 Probability
Basic Concepts
• An experiment has one or more outcomes. The outcome of rolling a die is a number from 1 to 6.
• The sample space is the set of all possible outcomes for an experiment. The sample space for a dice roll is {1, 2, 3, 4, 5, 6}.
• Any subset of the sample space is called an event. The event of rolling an even number with one roll of a die is {2, 4, 6}.
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Copyright © 2007 Pearson Education, Inc. Slide 8-4
8.7 Probability
Probability of an Event E
In a sample space with equally likely outcomes, the probability of an event E, written P(E), is the ratio of the number of outcomes in sample space S that belong to E, n(E), to the total number of outcomes in sample space S, n(S). That is,
( )( ) .
( )
n EP E
n S
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Copyright © 2007 Pearson Education, Inc. Slide 8-5
8.7 Finding Probabilities of Events
Example A single die is rolled. Give the probabilityof each event.
(a) E3 : the number showing is even
(b) E4 : the number showing is greater than 4
(c) E5 : the number showing is less than 7
(d) E6 : the number showing is 7
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Copyright © 2007 Pearson Education, Inc. Slide 8-6
8.7 Finding Probabilities of Events
Solution The sample space S is {1, 2, 3, 4, 5, 6} so
n(S) = 6.
(a) E3 = {2, 4, 6} so
(b) E4= {5, 6} so
33
( ) 3 1( ) .
( ) 6 2
n EP E
n S
44
( ) 2 1( ) .
( ) 6 3
n EP E
n S
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Copyright © 2007 Pearson Education, Inc. Slide 8-7
8.7 Finding Probabilities of Events
Solution
(c) E5 = {1, 2, 3, 4, 5, 6} so
(b) E6 = Ø so
55
( ) 6( ) 1 .
( ) 6
n EP E
n S
66
( ) 0( ) 0 .
( ) 6
n EP E
n S
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Copyright © 2007 Pearson Education, Inc. Slide 8-8
8.7 Probability
• For an event E, P(E) is between 0 and 1 inclusive.
• An event that is certain to occur always has probability 1.
• The probability of an impossible event is always 0.
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Copyright © 2007 Pearson Education, Inc. Slide 8-9
8.7 Complements and Venn Diagrams
• The set of all outcomes in a sample space that do not belong to event E is called the complement of E, written E´. If S = {1, 2, 3, 4, 5, 6} and E = {2, 4, 6} then E´ = {1, 3, 5}.
• ' , 'E E S E E
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Copyright © 2007 Pearson Education, Inc. Slide 8-10
8.7 Complements and Venn Diagrams
• Probability concepts can be illustrated with Venn diagrams. The rectangle represents the sample space in an experiment. The area inside the circle represents event E; and the area inside the rectangle but outside the circle, represents event E´.
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Copyright © 2007 Pearson Education, Inc. Slide 8-11
8.7 Using the Complement
Example A card is drawn from a well-shuffled
deck, find the probability of event E, the card is
an ace, and event E´.
Solution There are 4 aces in the deck of 52
cards and 48 cards that are not aces. Therefore
( ) 4 1 ( ') 48 12
( ) ( ') .( ) 52 13 ( ) 52 13
n E n EP E P E
n S n S
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Copyright © 2007 Pearson Education, Inc. Slide 8-12
8.7 Odds
The odds in favor of an event E are expressed as the
ratio of P(E) to P(E´) or as the fraction
( ).
( ')
P E
P E
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Copyright © 2007 Pearson Education, Inc. Slide 8-13
8.7 Finding Odds in Favor of an Event
Example A shirt is selected at random from a dark
closet containing 6 blue shirts and 4 shirts that are
not blue. Find the odds in favor of a blue shirt
being selected.
Solution E is the event “blue shirt is selected”.
6 3 4 2
( ) , ( ') .10 5 10 5
P E P E
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Copyright © 2007 Pearson Education, Inc. Slide 8-14
8.7 Finding Odds in Favor of an Event
Solution The odds in favor of a blue shirt are
or 3 to 2.
33 2 35( ) to ( ') to
25 5 25
P E P E
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Copyright © 2007 Pearson Education, Inc. Slide 8-15
8.7 Probability
Probability of the Union of Two Events
For any events E and F,
( or ) ( ) ( ) ( ) ( ) .P E F P E F P E P F P E F
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Copyright © 2007 Pearson Education, Inc. Slide 8-16
8.7 Finding Probabilities of Unions
Example One card is drawn from a well-shuffled
deck of 52 cards. What is the probability of each
event?
(a) The card is an ace or a spade.
(b) The card is a 3 or a king.
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Copyright © 2007 Pearson Education, Inc. Slide 8-17
8.7 Finding Probabilities of Unions
Solution (a) P(ace or space) = P(ace) + P(spade)
– P(ace and spade)
(b) P(3 or K) = P(3) + P(K) – P(3 and K)
4 13 1 16 4.
52 52 52 52 13
4 4 8 20 .
52 52 52 13
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Copyright © 2007 Pearson Education, Inc. Slide 8-18
8.7 Probability
Properties of Probability
1.
2. P(a certain event) = 1;
3. P(an impossible event) = 0;
4.
5. ( or ) ( ) ( ) ( ) ( ) .P E F P E F P E P F P E F
0 ( ) 1;P E
( ') 1 ( );P E P E
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Copyright © 2007 Pearson Education, Inc. Slide 8-19
8.7 Binomial Probability
An experiment that consists of
• repeated independent trials,• only two outcomes, success and
failure, in each trial,
is called a binomial experiment.
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Copyright © 2007 Pearson Education, Inc. Slide 8-20
8.7 Binomial Probability
Let the probability of success in one trial be p.
Then the probability of failure is 1 – p.
The probability of r successes in n trials is given by
(1 ) .r n rnp p
r
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Copyright © 2007 Pearson Education, Inc. Slide 8-21
8.7 Finding Binomial Probabilities
Example An experiment consists of rolling a die 10
times. Find the probability that exactly 4 tosses result
in a 3.
Solution Here , n = 10 and r = 4. The required probability is
4 10 4 4 610 1 1 1 51 210 .054 .
4 6 6 6 6
1
6p