Theoretical Probability
Sample Space: A listing of all the possible
outcomes from a chance experiment being
considered. All individual outcomes in a sample
space are equally likely.
The sample space is {1, 2, 3, 4, 5, 6}
P(4) = 1/6; P(1) = 1/6
Section 4.1, Page 76
1
Empirical Probability
P(pick orange M&M) = 137/692 = 0.198 = 19.8%
Section 4.1, Page 75
2
Theoretical Probability
Sample Space for rolling two die
P(2) = 1/36
P(5) = 4/36 = 1/9
P(7) = 6/36 = 1/6
Section 4.1, Page 77
3
Theoretical Probability
Sample Space for Gender of 3 children.
P(3 boys) = 1/8
P(same Gender) = 2/8
P(at least one girl) = 7/8
P(at most one girl) = 4/8
Section 4.1, Page 78
4
Properties of Probability
Section 4.1, Page 79
5
Subjective Probability
A subjective probability is a personal judgment determined
by an observer with incomplete information.
TV news says chance of rain is 70%.
Students says the chance of 4.0 is statistics is
80%.
Section 4.1, Page 78
6
Law of Large Numbers
Roll die 6 times in each trial, record # of 1s.
Results of individual
trials
(1+2)/12=.25
As the number of trials
increases the cumulative
long term frequency
approaches the theoretical
probability: 1/6.
Section 4.1, page 80
7
Probability as Odds
Las Vegas Gamblers say the odds that the
Seahawks win the Super Bowl are 1 to 50.
What is the probability they win?
1/(1+50) = .0196
What is probability that they will not win?
50/(1+50) = .9804.
Section 4.1, Page 81
8
Probability vs. Statistics
If a chip is drawn at random from a bag
containing these chips, the probability that
it will be green is 20/60 =1/3.
A sample of ten 10 is drawn from the bag.
There were 3 green chips. We are 95%
sure that the true proportion of green chips
is between .25 and .35.
Section 4.3, Page 82
9
Problems
Problems, Page 94
10
Problems
Problems, Page 95
11
Problems
a.
a. What percentage of the class is not watching
television on school nights?
b.
If a child is picked at random from the class, what is
the probability the child is not watching television on
school nights?
c.
What percentage of the class is watching at least 4
hours of television on school nights?
d.
If a child is picked at random from the class, what is
the probability the child is watching at least 4 hours of
television on school nights?
Problems, Page 95
12
Problems
Problems, Page 95
13
Conditional Probability of Events
The probability of surviving sinking of the Titanic was .44.
The probability of surviving sinking, given a crew member,
was .09.
P (survived) = .44
P (survived | crew member) = .09
Section 4.2, Page 83
14
Conditional Probability Problem
Section 4.2, Page 82
15
Conditional Probability Problem
Section 4.2, Page 83
16
Problem
Problems, Page 95
17
Standard Deck of Cards
52 Total Cards
4 Suites – Clubs, Diamonds, Hearts, Spades
3 Face Cards in each suite – Jack, Queen, King
Problems
4.91 You draw a card at random from a standard deck of
52 cards. Find each of the conditional probabilities.
a.The card is a heart, given that it is red.
b.The card is a jack, given that it is a heart
c.The card is an ace, given that it is red.
d.The card is a queen, given that it is a face card.
Problems, Section 4.2
19
Problems
4.92 In its monthly report, the local animal shelter states
that it currently has 24 dogs and 18 cats available for
adoption. Eight of the dogs and 6 of the cats are male.
Find each of the conditional probabilities if an animal is
selected at random.
a. The pet is male, given that it is a cat.
b. The pet is a cat, given that it is a female
c. The pet is a female, given that it is a dog.
Problems, Secion 4.2
20
Probability of Not A
P(A) 1 P(A)

Given two die, find the probability that the sum of a
random throw is at least 3.
P(sum at least 3) = P(sum ≥ 3) =
1 – P(sum < 3) = 1 –1/36 = 35/36.
Section 4.3, Page 84
21
Probability of A or B
General Addition Rule
Position on Budget Proposal
P(voter in favor or a republican) =
(136+314+14+88)/800=552/800=0.69
P(voter in favor) = 464/800 = 0.58
P(voter a republican) = 224/800 = 0.28
P(voter in favor and a republican) =
136/800 = 0.17.
P(in favor or a republican= P(in favor) +
P(republican) – P(in favor and republican)
= 0.58 +0.28 – 0.17 = 0.69
Section 4.3, Page 84
22
Problems
4.93 Real estate ads suggest that 64%of homes for
sale have garages, 21 % have swimming pools, and
17% have both features. What is the probability that a
home for sale has a pool or a garage?
4.94 Employment data at a large company revealed
that 72% of the workers are married, 44% are college
graduates, and 25% are both married and college
graduates. What the probability that a person is
married or a college graduate?
Problems, Section 4.3
23
Problems
4.95 You draw one card at random from a deck of
cards. What is the probability that the card is a
a.An ace or a heart?
b.A king or a red card?
Problems, Section 4.3
24
Probability of A and B
Probability a voter is in favor of budget and a
Republican = 136/800 = 0.17
Using Formula:
P(in favor of budget) = 464/800 = 0.58
P(Republican | in favor of budget) = 136/464 = 0.2931
P(in favor of budget and a republican) = P(in favor of
budget) × P(Republican | in favor of budget) =
0.58 × .2931 = .17
Section 4.3, Page 85
25
Problems
4.13 Seventy percent of kids who visit a doctor
have a fever, and 30% of the kids with a fever
have sore throats. What is the probability that a
kid who goes to the doctor has a fever and a sore
throat?
Problems, Section 4.3
26
Problems
Problems, Page 97
27
Mutually Exclusive Events
Section 4.4, Page 87
28
Mutually Exclusive Events
A and B = ϕ, the empty set. P(A and B) = 0
A and B are disjoint or mutually exclusive events.
B and C = {(5,5)}, P(B and C) = 1/36. B and C are
not disjoint events
Section 4.4, Page 89
29
Special Addition Rule
The general rule for addition:
P(A or B) = P(A) + P(B) – P(A and B)
If A and B are disjoint or mutually exclusive:
P(A and B) = 0
P(A or B) = P(A) + P(B)
P(A or B or C or …E) = P(A)+P(B) +P(C) …+P(E)
A and B are disjoint
P(A or B) = P(A) + P(B) = 6/36 + 3/36 = ¼
A and C are disjoint,
P(A or C) = P(A) + P(C) =6/36 +6/36 = ⅓
Section 4.4, Page 90
30
Problems
Problems, Page 98
31
Problems
4.31 The Masterfoods company says that for a large
bag of candies, yellow candies made up 20%, red
another 20%, orange and blue and green each 10%,
and the rest are brown. If you pick a candy at random
from the bag, what is the probability that
a. It is yellow or brown?
b.It is red or orange?
c.It is not green?
Problems, Section 4.4
32
Problems
4.32 The American Red Cross says that about
45% of the U. S. Population has Type O Blood,
40% type A, 11% type B, and the rest Type AB.
If a volunteer is selected at random, what is the
probability that her blood type is
a. Type O or type B?
b. Type A or type AB?
c. Not Type A?
Problems, Section 4.4
33
Independent Events
Section 4.5, Page 90
34
Independent Events
A = Ace on draw.
B = Ace on 2nd draw.
Draw two cards
replacing the first card
in the deck before
drawing the second
card.
P(Ace on 1st draw) = 4/52
P(Ace on 2nd draw | Ace on 1st draw) = 4/52
P(Ace on 2nd draw | Not Ace on 1st draw) = 4/52
Since P(B) = P(B | A) = P(B | Not A) drawing two
cards with replacement are independent events.
Draw two cards without replacing the first card
before the second draw.
P(Ace on 1st draw) = 4/52
P(Ace on 2nd draw | Ace on 1st draw) = 3/51
P(Ace on 2nd draw | Not Ace on 1st draw) = 4/51
Since P(B) ≠ P(B | A) ≠ P(B | Not A) drawing two
cards without replacement are not independent
events, but dependent events.
Section 4.5, Page 91
35
Special Multiplication Rule
The General Rule for multiplication:
P(A and B) = P(A)*P(B|A)
If A and B are independent then:
P(B|A) = P(B):
P(A and B) = P(A)*P(B)
P(A and B and C and… E) = P(A)*P(B)*P(C)*…*P(E)
Example 1: What is the probability of rolling three 6 in
three rolls of a die.
1 
Since each roll is independent, the probability is 6 
3
Example 2: For a certain bowler, the probability of

getting a strike in one roll of the ball is 0.45.
What is
probability that this bowler will roll a perfect game,
twelve strikes in a row?
Since each roll in independent of the other rolls, the
probability of a perfect game is (0.45)12

Section 4.5, Page 93
36
Problems
Problems, Page 98
37
Problems
Problems, Page 99
38
Problems
4.52 The American Red Cross says that about
45% of the U. S. Population has Type O Blood,
40% type A, 11% type B, and the rest Type AB.
If three volunteers are selected at random, what
is the probability that
a. All three are type A?
b. None are Type AB?
c. The first one is type A, the second one is not
Type AB, and the third one is Type O?
Problems, Section 4.5
39
Problems
A large airplane manufacturing company has designed
an new plane. Among the thousands of parts that make
up the plane, there are 30 that are mission critical.
Failure of a mission critical part means that the airplane
will crash. The company has designed each of these
parts to be 99.9% reliable.
a.What is the probability that the plane will crash on a
given flight, that is, at least one of the mission critical
parts fails?
b.Would you fly on this airplane?
Problems, Section 4.5
40
Summary of Probability Formulas
Equally Likely Outcomes: P(A) = n(A)/n
Complement: P(A) = 1- P(not A); P(not A) =1- P(A)
General Addition Rule:
P(A or B) = P(A) + P(B) – P(A and B)
If A and B are disjoint, P(A and B) = 0
Then the Special Addition Rule:
Then P(A or B) = P(A) + P(B)
General Multiplication Rule:
P(A and B) = P(A)×P(B|A)
If A and B are independent, P(B|A) = P(B)
Then the Special Multiplication Rule:
P(A and B) = P(A)×P(B)
Odds
If the odds for A are a:b, then the odds against A are
b:a. The probability of A is a/(a+b). The probability of
not A is b/(b+a)
Chapter 4
41