#### Transcript 12 Probability

```Copyright © 2005 Pearson Education, Inc.
SEVENTH EDITION and EXPANDED SEVENTH EDITION
Slide 12-1
Chapter 12
Probability
12.1
The Nature of Probability
Definitions



An experiment is a controlled operation that
yields a set of results.
The possible results of an experiment are called
its outcomes.
An event is a subcollection of the outcomes of
an experiment.
Slide 12-4
Definitions continued


Empirical probability is the relative frequency
of occurrence of an event and is determined by
actual observations of an experiment.
Theoretical probability is determined through
a study of the possible outcome that can occur
for the given experiment.
Slide 12-5
Empirical Probability
P (E ) 


number of times event E has occurred
total number of times the experiment has been performed
Example: In 100 tosses of a fair die, 19 landed
showing a 3. Find the empirical probability of the
die landing showing a 3.
Let E be the event of the die landing showing a 3.
P (E ) 
19
 0.19
100
Slide 12-6
The Law of Large Numbers

The law of large numbers states that
probability statements apply in practice to a
large number of trials, not to a single trial. It is
the relative frequency over the long run that is
accurately predictable, not individual events or
precise totals.
Slide 12-7
12.2
Theoretical Probability
Equally likely outcomes

If each outcome of an experiment has the same
chance of occurring as any other outcome, they
are said to be equally likely outcomes.
P (E ) 
number of outcomes favorable to E
total number of outcomes
Slide 12-9
Example






A die is rolled. Find the probability of rolling
a) a 3.
b) an odd number
c) a number less than 4
d) a 8.
e) a number less than 9.
Slide 12-10
Solutions: There are six equally likely
outcomes: 1, 2, 3, 4, 5, and 6.

a) P (2)  number of outcomes that will result in a 2  1
total number of possible outcomes

b) There are three ways an odd number can
occur 1, 3 or 5.
3 1
P (odd) 

6
6

2
c) Three numbers are less than 4.
3 1
P (number less than 4)  
6 2
Slide 12-11
Solutions: There are six equally likely
outcomes: 1, 2, 3, 4, 5, and 6. continued

d) There are no outcomes that will result in
an 8.
0
P (number greater than 8) 

6
0
e) All outcomes are less than 10. The event
must occur and the probability is 1.
Slide 12-12
Important Facts




The probability of an event that cannot occur is 0.
The probability of an event that must occur is 1.
Every probability is a number between 0 and 1
inclusive; that is, 0  P(E) 1.
The sum of the probabilities of all possible outcomes
of an experiment is 1.
Slide 12-13
Example







A standard deck of cards is well shuffled. Find the
probability that the card is selected.
a) a 10.
b) not a 10.
c) a heart.
d) a ace, one or 2.
f) a card greater than 4 and less than 7.
Slide 12-14
Example continued

a) a 10

There are four 10’s in a
deck of 52 cards.
4
1
P (10) 

52 13

b) not a 10
P (not a 10)  1  P (10)
1
 1
13
12

13
Slide 12-15
Example continued

c) a heart

d) an ace, 1 or 2

There are 13 hearts in
the deck.

There are 4 aces, 4 ones
and 4 twos, or a total of
12 cards.
13 1
P (heart) 

52 4
12 3
P (A, 1 or 2) 

52 13
Slide 12-16
Example continued


e) a card greater than 4
and less than 7

The word and means
both events must occur.
This is not possible.

The cards greater than 4
and less than 7 are 5’s,
and 6’s.
0
0
52
8
2
P (E ) 

52 13
Slide 12-17
12.3
Odds
Odds Against
 Odds against event =


P (event fails to occur)
P (failure)

P (event occurs)
P (success )
Example: Find the odds against rolling a 5 on
one roll of a die.
1
P (5) 
6
P(odds) 
5
P (fails to roll a 5) 
6
5
6
1
6
5 6 5
  
6 1 1
The odds against
rolling a 5 are 5:1.
Slide 12-19
Odds in Favor
P (event occurs)
Odds in favor of event =
P (event fails to occur)
P (success )

P (failure)
Slide 12-20
Example

Find the odds in favor of landing on blue in one
spin of the spinner.
3
P (blue)=
8
5
P (not blue)=
8
P(odds in favor) 
3
8
5
8
3 8 3
  
8 5 5
The odds in favor of spinning blue are 3:5.
Slide 12-21
Probability from Odds



Example: The odds against spinning a blue on
a certain spinner are 4:3. Find the probability
that
a) a blue is spun.
b) a blue is not spun.
Slide 12-22
Solution


Since the odds are 4:3
the denominators must
be 4 + 3 = 7.
The probabilities ratios
are:

4
P (blue)=
7
3
P (not blue)=
7
Slide 12-23
12.4
Expected Value (Expectation)
Expected Value

E = x1P(x1) + x2P(x2) + x3P(x3) +…+ xnP(xn)
or E = P(x1)x1 + P(x2)x2 + P(x3)x3 +…+ P(xn)xn
(order doesn’t matter)

The symbol P1 represents the probability that
the first event will occur, and A1 represents the
net amount won or lost if the first event occurs.
Slide 12-25
Example

Teresa is taking a multiple-choice test in which there are
four possible answers for each question. The instructor
indicated that she will be awarded 3 points for each
correct answer and she will lose 1 point for each
incorrect answer and no points will be awarded or
 If Teresa does not know the correct answer to a
guess?
 If she can eliminate one of the possible choices, is it
Slide 12-26
Solution

Expected value if Teresa guesses.
1
P (guesses correctly) 
4
3
P (guesses incorrectly) 
4
1
3
Teresa's expectation = (3)  ( 1)
4
4
3 3
  0
4 4
Slide 12-27
Solution continued—eliminate a choice

1
P (guesses correctly) 
3
2
P (guesses incorrectly) 
3
1
2
Teresa's expectation = (3)  ( 1)
3
3
2 1
 1 
3 3
Slide 12-28
Example: Winning a Prize


When Calvin Winters attends a tree farm event,
he is given a free ticket for the \$75 door prize. A
total of 150 tickets will be given out. Determine
his expectation of winning the door prize.
1
149
Expectation =
(75) 
(0)
150
150
1

2
Slide 12-29
Example

When Calvin Winters attends a tree farm event,
he is given the opportunity to purchase a ticket
for the \$75 door prize. The cost of the ticket is
\$3, and 150 tickets will be sold. Determine
Calvin’s expectation if he purchases one ticket.
Slide 12-30
Solution
 1 
 149
E ( x)  72
  3

 150
 150
72 447


150 150
375

150
 2.5

Calvin’s expectation is \$2.50 when he
purchases one ticket.
Slide 12-31
Fair Price
Fair price = expected value + cost to play
Slide 12-32
Example



Suppose you are playing a game in which you spin the
pointer shown in the figure, and you are awarded the
amount shown under the pointer. If is costs \$10 to play
the game, determine
a) the expectation of the
person who plays the
\$10
\$2
game.
\$15
\$20
b) the fair price to play the
game.
\$10
\$2
Slide 12-33
Solution

Amt. Shown
on Wheel
\$2
\$10
\$15
\$20
Probability
3/8
3/8
1/8
1/8
Amt.
Won/Lost
\$8
\$0
\$5
\$10
3
3
1
1
Expectation = ( \$8)  (\$0)  (\$5)  (\$10)
8
8
8
8
24
5 10

0 
8
8 8
9

 1.125  1.13
8
Slide 12-34
Solution

Fair price = expectation + cost to play
= \$1.13 + \$10
= \$8.87
Thus, the fair price is about \$8.87.
Slide 12-35
12.5
Tree Diagrams
Counting Principle

If a first experiment can be performed in M
distinct ways and a second experiment can be
performed in N distinct way, then the two
experiments in that specific order can be
performed in M • N ways. (Note: Order is
important.)
Slide 12-37
Definitions

Sample space: A list of all possible outcomes of
an experiment.

Sample point (simple event): Each individual
outcome in the sample space.
Slide 12-38
Example





Two balls are to be selected without replacement from a bag that
contains one purple, one blue, and one green ball.
a) Use the counting principle to determine the
number of points in the sample space.
b) Construct a tree diagram and list the sample
space.
c) Find the probability that one blue ball is selected.
d) Find the probability that a purple ball followed by a green ball
is selected.
Slide 12-39
Solutions


a) 3 • 2 = 6 ways
b)
P
B
G
B
G
P
G

PB
PG

4 2
c) P (blue)= 
6 3
d) P(purple then green =
1/6
BP
BG
P
GP
B
GB
Note: The probability of
each simple event is
1 1 1
  .
3 2 6
Slide 12-40
12.6
Or and And Problems
Or Problems


P(A or B) = P(A) + P(B)  P(A and B)
Example: Each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9,
and 10 is written on a separate piece of paper. The 10
pieces of paper are then placed in a bowl and one is
randomly selected. Find the probability that the piece of
paper selected contains an even number or a number
greater than 5.
Slide 12-42
Solution

P(A or B) = P(A) + P(B)  P(A and B)
 even or

 even and

P
  P (even)  P (greater 5)  

greater
than
5
greater
than
5




5
5
3



10 10 10
7

10

Thus, the probability of selecting an even number or a
number greater than 5 is 7/10.
Slide 12-43
Example

Each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and
10 is written on a separate piece of paper. The
10 pieces of paper are then placed in a bowl
and one is randomly selected. Find the
probability that the piece of paper selected
contains a number less than 3 or a number
greater than 7.
Slide 12-44
Solution

2
P (less than 3) 
10
3
P (greater than 7) 
10

There are no numbers that are both less than 3
and greater than 7. Therefore,
 2
10

3
5 1
0 

10
10 2
Slide 12-45
Mutually Exclusive

Two events A and B are mutually exclusive if it
is impossible for both events to occur
simultaneously.
Slide 12-46
Example

One card is selected from a standard deck of
playing cards. Determine the probability of the
following events.




a) selecting a 3 or a jack
b) selecting a jack or a heart
c) selecting a picture card or a red card
d) selecting a red card or a black card
Slide 12-47
Solutions

a) 3 or a jack

b) jack or a heart
4
4

52 52
8
2


52 13
P (3)  P ( jack) 
 jack and  4 13 1
P (jack)  P (heart)  P 



 heart
 52 52 52
16 4


52 13
Slide 12-48
Solutions continued

c) picture card or red card
 picture &  12 26 6
P (picture)  P (red)  P 



 red card  52 52 52
32 8


52 13

d) red card or black card
26 26
P (red)  P (black) 

52 52
52

1
52
Slide 12-49
And Problems


P(A and B) = P(A) • P(B)
Example: Two cards are to be selected with
replacement from a deck of cards. Find the
probability that two red cards will be selected.
P ( A)  P (B )  P (red )  P (red )
26 26


52 52
1 1 1
  
2 2 4
Slide 12-50
Example

Two cards are to be selected without
replacement from a deck of cards. Find the
probability that two red cards will be selected.
P (red and red )  P(red ) P(red )
26 25 650
25




52 51 2652 102
Slide 12-51
Independent Events

Event A and Event B are independent events if the
occurrence of either event in no way affects the
probability of the occurrence of the other event.

Experiments done with replacement will result in
independent events, and those done without
replacement will result in dependent events.
Slide 12-52
Example

A package of 30 tulip bulbs contains 14 bulbs for red
flowers, 10 for yellow flowers, and 6 for pink flowers.
Three bulbs are randomly selected and planted. Find
the probability of each of the following.




All three bulbs will produce pink flowers.
The first bulb selected will produce a red flower, the
second will produce a yellow flower and the third will
produce a red flower.
None of the bulbs will produce a yellow flower.
At least one will produce yellow flowers.
Slide 12-53
Solution

30 tulip bulbs, 14 bulbs for red flowers,
10 for yellow flowers, and 6 for pink flowers.
All three bulbs will produce pink flowers.
P  3 pink   P (pink 1)  P (pink 2)  P (pink 3)
6 5 4
= 

30 29 28
1

203
Slide 12-54
Solution
30 tulip bulbs, 14 bulbs for red flowers,
0010 for yellow flowers, and 6 for pink flowers.
The first bulb selected will produce a red flower,
the second will produce a yellow flower and the
third will produce a red flower.

P  red,yellow,red   P (red)  P (yellow)  P (red)
14 10 13


30 29 28
13

174
=
Slide 12-55
Solution
30 tulip bulbs, 14 bulbs for red flowers,
0010 for yellow flowers, and 6 for pink flowers.

None of the bulbs will produce a yellow flower.
 first not   second not   third not 
P  none yellow   P 
P 
P 

yellow
yellow
yellow

 
 

20 19 18
=


30 29 28
57

203
Slide 12-56
Solution
30 tulip bulbs, 14 bulbs for red flowers,
0010 for yellow flowers, and 6 for pink flowers.

At least one will produce yellow flowers.
P(at least one yellow) = 1  P(no yellow)
57
 1
203
146

203
Slide 12-57
12.7
Conditional Probability
Conditional Probability
In general, the probability of event E2
occurring, given that an event E1 has
happened is called conditional probability
and is written
P(E2|E1).

Slide 12-59
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.
Slide 12-60
Solutions


a) two girls
There are four possible outcomes BB, BG, GB,
and GG.
1
P (2 girls) =

4
b) two girls if you know that at least one of the
children is a girl
1
P (both girls|at least one is a girl) =
3
Slide 12-61
Solutions continued

two girls given that the older child is a girl
S = {BB, BG, GB, GG}
Older is a Girl BG, GG
Two girls given older is a girl GG
1
P (both girls|older child is a girl) =
2
Slide 12-62
Conditional Probability

For any two events, E1, and E2,
P( E2 | E1 ) 
n( E2 AND E1 ) n( E1 AND E2 )

n( E1 )
n( E1 )
P( E1 AND E2 ) P( Both Events)


P( E1 )
P(Given Event)
Exam ple:
P (2 girlsE2 | olderis a girlE1 ) 
n(2 girlsE2 AND olderis a girlE1 )
n(olderis a girlE1 )
1

2

S = {BB, BG, GB, GG}
Older is a Girl BG, GG
Two girls given older is a girl GG
Slide 12-63
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
Prefers
Peppermint
Prefers
Wintergreen
Total
Men
70
55
125
Women
60
72
132
Total
130
127
257
Slide 12-64
Example continued

a) prefers peppermint
130
P (peppermint) 
257

b) is a woman
132
P (woman) 
257
Slide 12-65
Example continued

c) prefers peppermint, given that a
woman is selected
60 15
P (peppermint|woman) 

132 33

d) is a man, given that the person
prefers wintergreen
55
P (man|wintergreen) 
127
Slide 12-66
12.8
The Counting Principle and
Permutations
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 ways.
Slide 12-68
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) and
the first digit cannot be 0, and repetition of
letters and digits is not permitted.
Slide 12-69
Solutions

repetition of letters and digits is permitted.

There are 26 letters and 10 digits. We have 6
positions to fill.
L
L
L
D D D
26 26 26 10 10 10
L
L
L
D D D
 17,576,000
Slide 12-70
Solution

repetition of letters and digits is not permitted.
L
L
L
D D D
26 25 24 10 9
L
L
L
8
D D D
 11,232,000
Slide 12-71
Solution

the first letter must be a vowel (a, e, i, o, u)
and the first digit cannot be 0, and repetition
of letters and digits is not permitted.
L
L
L
D D D
5 25 24
9
L
D D D
L
L
9
8
 1,944,000
Slide 12-72
Permutations

A permutation is any ordered arrangement
of a given set of objects. (Order is important)
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)
Slide 12-73
Example

How many different ways can 6 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.
Slide 12-74
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.
Slide 12-75
Permutation Formula

The number of permutations possible when r
objects are selected from n objects is found
by the permutation formula
n!
n Pr 
 n  r !
Slide 12-76
Example

The swimming coach has 8 swimmers who can
compete in the 100 m butterfly, he can only enter 3
swimmers in the event. In how many ways could he
select the 3 swimmers?
8!
8!

8 P3 
 8  3 ! 5!
8  7  6  5  4  3  2 1
5  4  3  2 1
 336

Slide 12-77
Permutations of Duplicate Objects

The number of distinct permutations of n
objects where n1 of the objects are identical,
n2 of the objects are identical, …, nr of the
objects are identical is found by the formula
n!
n1 ! n2 !...n3 !
Slide 12-78
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.
10!
10  9  8  7  6  5  4  3  2  1

3!3!2!
3  2  1  3  2  1 2  1

100,800
 50,400
2
Slide 12-79
12.9
Combinations
Combination

A combination is a distinct group (or set) of
objects without regard to their arrangement.

The number of combinations possible when r
objects are selected from n objects is found by
n!
n Cr 
 n  r ! r !
Slide 12-81
Example

A student must select 4 of 7 essay questions to be answered on a
test. In how many ways can this selection be made?
7!
7!

7 C4 
(7  4)!4! 3!4!
7  6  5  4  3  2 1

 35
3  2  1 4  3  2  1

There are 35 different ways that 4 of 7 questions can be selected.
Slide 12-82
Example

Toastline Bakery is testing 5 new wheat breads,
market 2 of the wheat breads, 2 of the bran
different combinations are possible?
Bread choices  5C2  4C2  3 C1
 10  6  3
 180
Slide 12-83
12.10
Solving Probability Problems
by Using Combinations
Example

A club consists of 5 men and 6 women. Four members are to be
selected at random to form a committee. What is the probability that
the committee will consist of two women?

P (two women) =
# of possible committees with two women
total number of 4-member committees
6 C2

11C4
15
1


330 22
Slide 12-85
Example

The Honey Bear is testing 10 new flavors of ice cream.
They are testing 5 vanilla based, 3 chocolate based and
2 strawberry based ice creams. If we assume that each
of the 10 flavors has the same chance of being selected
and that 4 new flavors will be produced, find the
probability that



a) no chocolate flavors are selected.
b) at least 1 chocolate is selected.
c) 2 vanilla and 2 chocolate are selected.
Slide 12-86
Solution


5 vanilla, 3 chocolate, 2 strawberry
selecting 4 flavors
a) P(no chocolate)  7C4  35  1
10

C4
210
6
b) P (at least 1 chocolate)  1  P (no chocolate)
 1
1 5

6 6
Slide 12-87
Solution continued

C2  3 C2
P (2 vanilla and 2 chocolate) 
10 C4
5

10  3 30
1


210 210 7
Slide 12-88
12.11
Binomial Probability Formula
To Use the Binomial Probability Formula

There are n repeated independent trials.

Each trial has two possible outcomes, success
and failure.

For each trial, the probability of success (and
failure) remains the same.
Slide 12-90
Binomial Probability Formula

The probability of obtaining exactly x
successes, P(x), in n independent trials is given
by
x nx
P( x )   nCx  p q
where p is the probability of success on a single
trial and q is the probability of failure on a single
trial.
Slide 12-91
Example

A basket contains 5 pens: one of each color red, blue,
black, green and purple. Five pens are going to be
selected with replacement from the basket. Find the
probability that




no blue pens are selected
exactly 1 blue pen is selected
exactly 2 blue pens are selected
exactly 3 blue pens are selected
Slide 12-92
Solution

no blue pens

P ( x )   n Cx  p x q n  x
0
 1  4
P (0)   5 C0     
5 5
4
 (1)(1)  
5
5
5
1024
4
  
3125
5
exactly 1 blue pen
P ( x )   n Cx  p x q n  x
5 0
1
 1  4
P (0)   5 C1     
5 5
5 1
4
 1  4 
 (5)    
 5  5 
 1   256  256
 5 


 5   625  625
Slide 12-93
Solution continued

exactly 2 blue pens

P ( x )   n Cx  p x q n  x
P ( x )   n Cx  p x q n  x
2
 1  4
P (0)   5 C2     
5 5
2
3
52
 1  4
 (10)    
5 5
 1   64  128
 10 




 25   125  625
exactly 3 blue pens
3
 1  4
P (0)   5 C3     
5 5
3
5 3
2
 1  4
 (5)    
5 5
 1   16  32
 10 




 125   25  625
Slide 12-94
Example

A manufacturer of lunch boxes knows that 0.7% of the
lunch boxes are defective.
 Write the binomial probability formula that would be use
to determine the probability that exactly x our of n lunch
boxes produced are defective.
 Write the binomial probability formula that would be
used to find the probability that exactly 4 lunch boxes
out of 60 produced will be defective.
Slide 12-95
Solution

P ( x )   n Cx  p x q n  x
  n Cx  (0.007)x ((0.993)n  x

P ( x )   n Cx  p x q n  x
  60 C4  (0.007) ((0.993)
4
56
Slide 12-96
Example

The probability that an egg selected at random
has a weak shell and will crack is 0.2. Find the
probability that


none of 8 eggs selected at random has a weak
shell.
at least one of 8 eggs selected has a weak shell.
Slide 12-97
Solution


p = 0.2
q = 1  0.2 = 0.8
We want 0 successes in 8 trials.
P ( x )   n Cx  p x q n  x
  8 C0  (0.2)0 (0.8)8 0
 1(1)(0.167772)
 0.167772

at least 1egg = 1  0.167772 = 0.832228