Download Section 11.1 The Fundamental Counting Principle

Section 11.1
The Fundamental Counting Principle
Objective
1. Use the Fundamental Counting Principle to
determine the number of possible outcomes in a
given situation.
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Section 11.1
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Fundamental Counting Principle
• Definition
If you can choose one item from a group of M items and a
second item from a group of N items, then the total
number of two-item choices is M∙N.
• Tree Diagram
A representation of all possible
choices. This tree diagram shows
that there are 2·3 = 6 different
Outfits from 2 pairs of jeans and
three T-shirts.
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Section 11.1
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Example 1
Applying the Fundamental Counting
Principle
The Greasy Spoon Restaurant offers 6 appetizers and 14
main courses. In how many ways can a person order a
two-course meal?
Solution:
Choosing from one of 6 appetizers and one of 14 main
courses, the total number of two-course meals is:
6 ∙14 = 84
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Section 11.1
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The Fundamental Counting Principle with
More than Two Groups of Items
• Definition
The number of ways in which
a series of successive things
can occur is found by
multiplying the number of
ways in which each thing can
occur.
• The number of possible outfits
from 2 pair of jeans, 3 T-shirts
and 2 pairs of sneakers are:
2·3·2 = 12
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Section 11.1
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Example 2
Applying the Fundamental Principle of Counting with
More Than Two Groups of Items
Options in Planning a Course Schedule
Next semester, you are planning to take three courses –
math, English and humanities. There are 8 sections of math,
5 of English, and 4 of humanities that you find suitable.
Assuming no scheduling conflicts, how many different threecourse schedules are possible?
Solution:
This situation involves making choices with three groups of
items.
Math
English
Humanities
{8 choices}
{5 choices}
{4 choices}
There are 8∙5∙4 = 160 different three-course schedules.
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Section 11.1
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Example 3
A Multiple-Choice Test
•
You are taking a multiple-choice test that has ten
questions. Each of the questions has four answer choices,
with one correct answer per question. If you select one of
these four choices for each question and leave nothing
blank, in how man ways can you answer the questions?
• Solution:
This situation involves making choices with ten questions:
Question 1
{4 choices}
Question 2
{4 choices}
Question 3 ∙ ∙ ∙ Question 9
{4 choices}
{4 choices}
Question 10
{4 choices}
The number of different ways you can answer the questions is:
4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 = 410 = 1,048,576
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Section 11.1
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Example 4
Telephone Numbers in the United States
• Telephone numbers in the United States begin with threedigit area codes followed by seven-digit local telephone
numbers. Area codes and local telephone numbers cannot
begin with 0 or 1. How many different telephone numbers
are possible?
• Solution:
This situation involves making choices with ten groups of
items. Here are the choices for each of the ten groups of
items: Area Code
Local Telephone Number
8 10 10
8 10 10 10 10 10 10
The total number of different telephone numbers is:
8 ∙ 10 ∙ 10 ∙ 8 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 = 6,400,000,000
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Section 11.2
Permutations
Objectives
1. Use the Fundamental Counting Principle
to count permutations.
2. Evaluate factorial expressions.
3. Use the permutations formula.
4. Find the number of permutations of
duplicate items.
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Section 11.2
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Permutations
• Permutation is an ordered arrangement of items that
occurs when:
– No item is used more than once.
– The order of arrangement makes a difference.
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Section 11.2
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Example 1
Counting Permutations
• You are in charge of planning a concert tour with U2, ‘N
Sync, Aerosmith and the Rolling Stones. You decide
that the Rolling Stones will perform last. How many
different ways can you put together the concert?
• Solution:
You can now choose any one of the three groups, U2, ‘N
sync or Aerosmith, as the opening act. That leaves two
groups left to choose from for the second group and only
one group left for the third group.
Using the Fundamental Counting Principle there are:
3∙2∙1∙1 = 6 different ways to arrange the concert.
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Factorial Notation
• The product 7∙6∙5∙4∙3∙2∙1 is called factorial and is written 7!
• Definition:
If n is a positive integer, the notation n! (read “n factorial”) is
the product of all positive integers from n down through 1.
n! = n(n - 1)(n - 2)···(3)(2)(1)
0! (zero factorial), by definition, is 1.
0! = 1
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Section 11.2
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Example 2
Using Factorial Notation
• Evaluate without using your calculator:
8! 8  7  6  5∕ !

 8  7  6  336
∕
5!
5!
26! 26  25  24  23  22  21
∕ !

 26  25  24  23  22  7,893,600
∕ !
21!
21
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A Formula for Permutations
• Permutations of n things taken r at a time.
n!
P
n r
(n  r )!
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Example 3
Using the Formula for Permutations
• You and 19 of your friends have decided to form a
business. The group needs to choose three officers– a
CEO, an operating manager, and a treasure. In how many
ways can those offices be filled?
• Solution:
Your group is choosing r = 3 officers from a group of n = 20
people. The order matters because each officer has
different responsibilities:
20!
20! 20 19 18 17
∕ !
P  20 P3 


 20 19 18  6840
n r
(20  3)! 17!
17∕ !
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A Formula for Permutations of Duplicate
Items
• Permutations of Duplicate Items
The number of permutations of n items, where p items are
identical, q items are identical, r items are identical, and
so on, is given by:
n!
p!q!r!...
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Section 11.2
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Example 4
Using the Formula for Permutations of Duplicate
Items
• In how many distinct ways can the letters of the word
MISSISSIPPI be arranged?
• Solution:
The word contains 11 letters (n = 11) where four Is are
identical (p = 4), four Ss are identical (q = 4) and 2 Ps are
identical (r = 2). The number of distinct permutations are:
n!
11! 1110  9  8  7  6  5  4!


 43,650
p!q!r! 4!4!2!
4!4  3  2 1 2 1
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Section 11.2
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Section 11.3
Combinations
Objectives
1. Distinguish between permutation and
combination problems.
2. Solve problems involving combinations using
the combinations formula.
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Section 11.3
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Combinations
A combination of items occurs when
• The items are selected from the same group.
• No item is used more than once.
• The order of items makes no difference.
Note:
• Permutation problems involve situations in which order
matters.
• Combination problems involve situations in which the
order of items makes no difference.
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Comparing Combinations and Permutations
Given the letters: A,B,C,D: We can compare how many
permutations and how many combinations are possible if
we chose 3 letters at a time:
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Example 1
Distinguishing Between Permutations and
Combinations
Determine which involve permutations and which involve
combinations.
• Six student are running for student government
president, vice-president and treasure. The student with
the greatest number of votes becomes the president, the
second highest vote-getter becomes vice-president, and
the student who gets the third largest number of votes
will be treasurer. How many different outcomes are
possible?
• Solution:
Order matters since the number of the three highest
votes determine the office.
Permutations
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Example 1
Distinguishing Between Permutations and
Combinations (continued)
• Six people are on the board of supervisors for your
neighborhood part. A three-person committee is needed
to study the possibility of expanding the park. How many
different committees could be formed from the six
people?
• Solution:
The order in which the three people are selected does
not matter since they are not filling different roles.
Combinations
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A Formula for Combinations
The number of possible combinations if r items are taken
from n items is:
n!
n Cr 
(n  r )! r!
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Example 2
Using the Formula for Combinations
• How many three-person committees could be formed
from 8 people?
• Solution: We are selecting 3 people ( r = 3) from 8
people ( n = 8)
8!
8! 8  7  6  5∕ !
C r  8 C3 


 56
n
(8  3)!3! 5!3! 5∕ !3  2 1
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Example 3
Using the Formula for Combinations and the
Fundamental Counting Principle
• The U.S Senate of the 104th Congress
consisted of 54 Republicans and 46
Democrats.
How many committees can be formed
if each committee must have 3
Republicans and 2 Democrats?
• Solution:
The order in which members are selected does not
matter so this is a problem of combinations.
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Example 3 continued
First, select 3 Republicans out of 54 Republicans:
54!
54! 54  53  52  51! 54  53  52
Cr  54 C3 



 24,804
n
(54  3)!3! 51!3!
51!3  2 1
3  2 1
Secondly, select 2 Democrats out of 46 Democrats:
Cr  46 C2 
n
46!
46! 46  45  44! 46  45  44



 1,035
(46  2)!2! 44!2!
44!2 1
2 1
Use the Fundamental Counting Principle to find the number
of committees that can be formed.
54
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C3 46 C2  24,804  1,035  25,672,140
Section 11.3
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Section 11.4
Fundamentals of Probability
Objectives
1. Compute theoretical probability.
2. Compute empirical probability.
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Probability
• Probabilities are assigned
values from 0 to 1.
• The closer the probability
of a given event is to 1, the
more likely it is that the
event will occur.
• The closer the probability
of a given event is to 0, the
less likely that the event
will occur.
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Theoretical Probability
• Experiment is any occurrence for which the outcome is
uncertain.
• Sample space is the set of all possible outcomes of an
experiment , denoted by S.
• Event, denoted by E is any subset of a sample space.
• Sum of the theoretical probabilities of all possible
outcomes is 1.
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Computing Theoretical Probability
• If an event E has n(E) equally likely outcomes and its
sample space S has n(S) equally-likely outcomes, the
theoretical probability of event E, denoted by P(E), is:
• P(E) = number of outcomes in event E = n(E)
total number of possible outcomes n(S)
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Example 1
Computing Theoretical Probability
•
A die is rolled once. The sample space is S = {1,2,3,4,5,6}
with n(S) = 6. Find the probability of rolling
a. a 3
b. an even number
• Solution to a:
There is only one way to roll a 3 so n(E) = 1.
P(3) = number of outcomes that result in 3 = n(E) = 1
total number of possible outcomes
n(S) 6
• Solution to b:
Rolling an even number describes the event E = {2,4,6}.
This event can occur in 3 ways: n(E) = 3.
P(even number) =
number of outcomes that result in even number = n(E) = 3 =1
total number of possible outcomes
n(S) 6 2
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Example 2
Probability and a Deck of 52 Cards
•
You are dealt one card from a standard 52-card deck.
Find the probability of being dealt a
a. King
Solution:
P(3) = number of outcomes that result in a king = n(E) = 4 =1
total number of possible outcomes
n(S) 52 13
b. Heart
Solution:
P(heart)
= number of outcomes that result in a heart = n(E) = 13=1
total number of possible outcomes
n(S) 52 4
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Empirical Probability
• Applies to situations in which we observe how frequently
an event occurs.
• Computing Empirical Probability
The empirical probability of event E is:
• P(E) = observed number of times E occurs = n(E)
total number of observed occurrences n(S)
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Example 4
Computing Empirical Probability
If one person is randomly selected from the population
described above, find the probability that the person is
female.
Solution. The probability of selecting a female is the
observed number of females, 110.1(million), divided by the
total number of U.S. adults, 212.5 (million).
P (selecting a female) =
number females_ = 110.1 ≈ 0.52
total number of adults 212.5
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Section 11.5
Probability with the Fundamental counting
Principle, Permutations, and Combinations
Objectives
1. Compute probabilities with permutations.
2. Compute probabilities with combinations,
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Example 1
Probability with Permutations.
• Five groups agree to randomly select the order of
performance by picking cards out of a hat, one at a time.
What is the probability of the Rolling Stones performing
fourth and the Beatles last?
• Solution:
P (Rolling Stones fourth, Beatles last) =
Number of permutations with Rolling Stones fourth, Beatles last
Total number of possible permutations
• Use the Fundamental Counting Principle to find the total
number of possible permutations.
• There are 5·4·3·2·1 =120 possible permutations.
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Example 1 continued
• Use the Fundamental Counting Principle to find the
number of permutations with the Rolling Stones
performing fourth and the Beatles performing last.
• There are 3·2·1·1·1 = 6 possible permutations.
• Putting both values in the original equation.
P (Rolling Stones fourth, Beatles last) = 6 = 1
120 20
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Section 11.5
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Example 2
Probability and Combinations: Winning the Lottery
• Florida’s lottery game, LOTTO, is set up so that each
player chooses six different numbers from 1 to 53.
With one LOTTO ticket, what is the probability of
winning this prize?
• Solution:
Because the order of the six numbers does not matter,
this situation involves combinations.
P (winning) =
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number of ways of winning
total number of possible combinations
Section 11.5
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Example 2 continued
P (winning) =
number of ways of winning
total number of possible combinations
Using the combinations formula, r = 6 and n = 53.
Cr  53 C6 
n
53!
53!

 22,957,480
(53  6)!6! 47!6!
If a person buys only one ticket, then that person has
selected only one combination, thus
P (winning) = number of ways of winning ___
= __1___
total number of possible combinations 22,957,480
≈ 0.0000000436
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Section 11.5
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Example 3
Probability and Combinations
• A club consists of five men and seven
women. Three members are selected
at random to attend a conference.
Find the probability that the selected
group consists of 3 men.
• Solution:
Order of selection does not matter, so this is a problem
involving combinations.
P(3 men) = number of ways of selecting 3 men
total number of possible combinations
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Example 3 continued
Consider the denominator: We are selecting r = 3 people
from a total group of n = 12.
12!
12!
C3 

 220
12
(12  3)!3! 9!3!
Consider the numerator. We are interested in selecting 3
(r = 3)men from 5 (n = 5) men.
5!
5!
C3 

 10
5
(5  3)!3! 2!3!
Therefore: P(3 men ) = 10 = 1
220 22
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Section 11.5
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11.6
Events Involving Not and Or; Odds
Objectives
1. Find the probability that an event will not
occur.
2. Find the probability of one event or a
second event occurring.
3. Understand and use odds.
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Probability of an Event Not Occurring
• Complement of E: If we know P(E), the probability of an
event E, we can determine the probability that the event
will not occur, denoted by P(not E).
• The probability that an event E will not occur is equal to
1 minus the probability that it will occur.
P(not E) = 1 – P(E)
• The probability that an event E will occur is equal to 1
minus the probability that it will not occur.
P(E) = 1 – P(not E)
Using set notation, if E’ is the complement of E, then
P(E’) = 1 – P(E) and P(E) = 1 – P(E’)
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Example 1
The Probability of an Event Not Occurring
• If you are dealt one card from a standard 52-card deck,
find the probability that you are not dealt a queen.
• Solution:
Because P(not E) = 1 – P(E) then
P(not a queen) = 1 – P(queen).
• There are 4 queens in a standard deck.
• The probability of being dealt a queen is 4 =1
52 13
• P(not a queen) = 1 – P(queen) = 1 – 1 = 12
13 13
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Or Probabilities with Mutually Exclusive Events
• Mutually Exclusive Events: Events A and B are
mutually exclusive if it is impossible for them to occur
simultaneously.
• Or Probabilities with Mutually Exclusive Events:
– If A and B are mutually exclusive events, then
P(A or B) = P(A) + P(B)
– Using set notation,
P(A U B) = P(A) + P(B)
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Example 2
The Probability of Either of Two Mutually
Exclusive Events Occurring
• If one card is randomly selected from a deck of cards,
what is the probability of selecting a king or a queen?
• Solution:
P ( king or queen) = P(king) + P(queen)
= 4+4= 8=2
52 52 52 13
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Or Probabilities with Events That Are Not
Mutually Exclusive
• If A and B are not mutually exclusive events then the
probability that A or B will occur is determined by adding
their individual probabilities and then subtracting the
probability that A and B occur simultaneously.
– P(A or B) = P(A) + P(B) – P(A and B)
– Using set notation,
P(A U B) = P(A) + P(B) – P(A ∩B)
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Example 3
An OR Probability With Events That Are Not
Mutually Exclusive
• In a group of 25 baboons, 18 enjoy grooming their
neighbors, 16 enjoy screeching wildly, while 10 enjoy
doing both. If one baboon is selected at random, find
the probability that it enjoy grooming its neighbors or
screeching wildly.
• Solution:
Since 10 of the baboons enjoy both grooming their
neighbors and screeching wildly, these events are not
mutually exclusive.
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Example 3 continued
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Probability to Odds
• If we know the probability of an event, we can determine
the odds in favor, or the odds against, the event.
Probability to Odds:
1. The odds in favor of E are found by taking the
probability that E will occur and dividing by the probability
that E will not occur.
Odds in favor of E = P(E) .
P(not E)
2. The odds against E are found by taking the probability
that E will not occur and dividing by the probability that E
will occur.
Odds against E = P(not E)
P(E)
The odds against E can also be found by reversing the ratio
representing the odds in favor of E.
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Example 4
From Probability to Odds
• You roll a single, six-sided die.
Find the odds in favor of rolling a 2
Solution:
Let E represent the event of rolling a 2
• First, find the probability of E occurring and the
probability of E not occurring.
S = {1,2,3,4,5,6}
E = {2}
P(E) = 1 and P(not E) = 1 – P(E) = 1 – 1 = 5
6
6 6
1
• Odds in favor of E( rolling a 2 ) = P(E) . = 6 = 1 .6 =1
P(not E) 5 6 5 5
6
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Example 5
From Odds to Probability
Odds to Probability
If the odds in favor of event E are a to b, then the probability
of the event is given by:
P(E) = a .
a+b
• The odds in favor of a particular horse winning a race are
2 to 5. What is the probability this horse will win the race?
• Solution.
Because odds in favor, a to b, means a probability of a ,
then the odds in favor, 2 to 5,
a+b
means a probability of:
2 =2
2+5 7
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Section 11.7
Events Involving And; Conditional Probability
Objectives
1. Find the probability of one event and a
second event occurring.
2. Compute conditional probabilities.
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And Probabilities with Independent Events
• Independent Events: Two events are independent
events if the occurrence of either of them has no effect
on the probability of the other.
• And Probabilities with Independent Events
If A and B are independent events, then
P(A and B) = P(A) ∙P(B)
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Example 1
Independent Events on a Roulette Wheel
• A U.S. roulette wheel has 38 numbered slots (1 through
36, 0, and 00. 18 are black, 18 are red, and 2 are green.
The ball can land on any slot with equal probability. What
is the probability of red occurring on 2 consecutive plays.
• Solution: the probability of red occurring on a play is
18 or 9 .
32 19
Thus,
P( red and red) = P( red)∙P( red)
= 9 . 9 = 81 ≈ 0.224
19 19 361
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Section 11.7
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Example 2
Independent Events In a Family
• If two or more events are independent, we can find the
probability of them all occurring by multiplying their
probabilities. The probability of a baby girl is ½, so the
probability of nine girls in a row is ½ used as a factor
nine times.
• Solution:
P( nine girls in a row)
= ½∙ ½∙ ½∙ ½∙ ½∙ ½∙ ½∙ ½∙ ½
= (½)² = 1 .
512
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Example 3
The Probability of an Event Happening At
Least Once
• P( event happening at least once)
= 1 – P( event does not happen)
The probability that South Florida will not be hit by a
hurricane in ten years was computed to be 0.047.
What is the probability that South Florida will be hit by a
hurricane at least once in the next 10 years?
Solution: P( at least one hurricane in 10 years)
= 1 – P( no hurricane for ten years)
= 1 – 0.047 = 0.953
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And Probabilities with Dependent Events
• Dependent Events: Two events are dependent events if
the occurrence of one of them has an effect on the
probability of the other.
• And Probabilities with Dependent Events
If A and B are dependent events, then
P(A and B) = P(A) ∙ P(B given that A has occurred).
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Example 4
And Probabilities with Dependent Events
• There are 20 chocolates in a box,
5 of which are chocolate-covered cherries.
What is the probability of choosing
2 chocolate-covered cherries?
• Solution:
P( chocolate-covered cherry and chocolate-covered cherry) =
P( chocolate-covered cherry) ∙ P( chocolate-covered cherry) =
(given that one was selected )
1∙4=1
4 19 19
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Example 5
And Probability With Three Dependent
Events
• If two or more events are dependent, we can find the
probability of them all occurring by multiplying their
probabilities.
• Three people are randomly selected, one person at a
time, from 5 freshmen, 2 sophomores, and 4 juniors.
Find the probability that the first two people selected are
freshmen and the third is a junior.
• Solution:
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Conditional Probability
• The probability of event B, assuming that the event A has
already occurred, is called the conditional probability of B,
given A. This probability is denoted by P(B|A).
P(B|A) = n(B  A) = number of outcomes common to B and A
n(A)
number of outcomes in A
• Conditional probability is the probability that event B occurs
if the sample space is restricted to the outcomes associated
with event A.
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Example 6
Finding Conditional Probability
• A letter is randomly selected from the letters of the
English alphabet. Find the probability of selecting a
vowel, given that the outcome is a letter that precedes h.
• Solution:
We are looking for P( vowel | letter precedes h).
This is the probability of a vowel if the sample space is
restricted to the set of letters that precede h. Thus, the
sample space is given by: S = {a,b,c,d,e,f,g}.
There are 7 possible outcomes in the sample space. We
can select a vowel from this set in one of two ways: a or e.
Therefore, P( vowel | letter precedes h) = 2
7
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Section 11.7
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Example 7
Conditional Probabilities with Real-World Data
Mammography Screening on 100,000 U.S. Women, Ages 40 to 50
Breast Cancer
No Breast Cancer
Positive Mammogram
720
6,944
7,664
Negative Mammogram
80
92,256
92,336
800
99,200
100,000
Total
Total
• Assuming that these numbers are representative of all U.S. women
age 40 to 50, find the probability that a women in this age range has a
positive mammogram, given that she does not have breast cancer.
• Solution:
The data states that there are 6944 women with positive mammograms
and 6,944 + 92,256, or 99,200 women without breast cancer. Thus,
P( positive mammogram | no breast cancer) = 6,944 = 0.07
99,200
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Section 11.8
Expected Value
Objectives
1. Compute expected value.
2. Use expected value to solve applied
problems.
3. Use expected value to determine the
average payoff or loss in a game of
chance.
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Expected Value
• Expected value is a mathematical way to use
probabilities to determine what to expect in various
situations over the long run.
• Expected value is used to
– determine premiums on insurance policies.
– weigh the risks versus the benefits in alternatives in
business ventures.
– indicate to a player of any game of chance what will
happen if the game is played repeatedly.
• Standard way to compute expected value is to multiply
each possible outcome by its probability and then add
these products.
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Example 1
Computing Expected Value
• Find the expected value for the number of girls for a
family with three children.
• Solution:
A family with three children can have 0,1,2, or 3 girls.
There are eight ways these outcomes can occur.
The expected value, E, is computed by multiplying each
outcome by its probability and then adding these
products.
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Example 1 continued
1
3
3
1 0  3  6  3 12 3
E  0   1  2   3  

  1.5
8
8
8
8
8
8 2
• The expected value is 1.5. This means that if we record
the number of girls in many different three-child families,
the average number of girls for all these families will be
1.5 or half the children.
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Example 2
Determining an Insurance Premium
• An automobile insurance company has Amount Probability
of Claim
determined the probabilities for various
$0
0.70
claim amounts (to the nearest $2000) for
$2000
0.15
drivers ages 16 through 21 as shown in
$4000
0.08
the table. Calculate the expected value
0.05
and describe what this means in practical $6000
$8000
0.01
terms.
$10,000
0.01
E = $0(0.70) + $2000(0.15) +
$4000(0.08) + $6000(0.05) + $8000(0.01)+$10,000(0.01)
= $0+$300+ $320+$300+ $80+$100=$1100.
• This means that in the long run, the average cost of a claim
is $1100 which is the very least the insurance company
should charge to break even.
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Example 3
Expect Value and Games of Chance
• To find the expected value of a game, multiply the gain
or loss for each possible outcome by its probability.
Then add the products.
• Find the expected value of betting $1 on the number 20
in roulette.
– If the ball lands on that number , you are awarded
$35 and get to keep the $1 that you paid to play.
– If the ball lands on any of the other 37 slots, you are
awarded nothing and the $1 that you bet is collected.
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Example 3 continued
Playing one Number with a 35 to 1 payoff in roulette
Outcome
Gain or Loss
Ball lands on 20
$35
Ball doesn’t land on 20
-$1
Probability
1
38
37
38
 1
 37  $35  $37  $2
E  35   ( $1)  

 $0.05
38
38
 38 
 38 
• The expected value is approximately -$0.05.
This means that in the long run, a player can
expect to lose about 5¢ for each game played.
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