Chapter 6 – The Study of Randomness
• The idea of probability is that chance behavior is
unpredictable on the short run, but has a regular
pattern in the long run…
• Behavior is random if, while individual
outcomes are uncertain, for a large number
of repetitions, outcomes are regularly
distributed.
• Example: If I roll a die once, I can’t predict
with any certainty what number it will land
on, but if I roll sixty times, I can expect it to
land on 1 ten times, 2 ten times, 3 ten times,
etc.
• In the short run, we
have no idea what is
going to happen with a
random event…
• BUT, in the long run, a
pattern emerges.
• The probability of an outcome is the
proportion of times the outcome would
occur for a large number of repetitions.
(long term relative frequency)
• Example: The probability of a die landing on
4 is the proportion of times a die lands on 4
for a large number of repetitions.
• The set of all possible outcomes of an event
is the sample space, S , of the event.
• Example: For the event “roll a die and
observe what number it lands on” the
sample space contains all possible numbers
the die could land on.
• S = { 1, 2, 3, 4, 5, 6 }
• An event is an outcome
(or a set of outcomes)
from a sample space.
• An event is usually
denoted by a capital
letter. For example,
call getting two tails
event A.
• The probability of event
A is denoted P(A).
• Example 1:
When flipping three coins,
an event may be getting all
heads (HHH).
– In this case, the event is one
outcome from the sample
space.
• Example 2:
When flipping three coins,
an event may be getting
two tails.
– In this case, the event is a set
of outcomes (HTT, TTH,
THT) from the sample space.
Event A = Three Heads
P(A) = 1 / 8 = 0.125
Event B = Two Tails
P(B) = 3 / 8 = 0.375
• The probability of any event is between 0
and 1, inclusive.
0  P(A)  1
• A probability of 0 indicates the event will
never occur.
• A probability of 1 indicates the event will
always occur.
• If S is the sample space, then P(S) = 1.
Some outcome in the sample space will
occur.
• The probability that event A does not occur
is one minus the probability that A does
c
occur.
P(A )  1  P(A)
That A will not occur is called the
complement of A and is denoted Ac.
• Example: When flipping two coins, the
probability of getting two heads is 0.25.
The probability of not getting two heads
is 1 – 0.25 = 0.75.
Event A = Three Heads
P(A) = 1 / 8 = 0.125
P(Ac)=?
P(Ac) = 1-P(A)
P(Ac) = 1-0.125 = 0.875
• If events A and B are disjoint they have
no outcomes in common.
A
B
DISJOINT
A
B
NOT DISJOINT
Example:
• Let event A be rolling a die and landing on an even number.
• Let event B be rolling a die and landing on an odd number.
• The outcomes for A are {2, 4, 6}
• The outcomes for B are {1, 3, 5}
• Events A and B are disjoint because they have no outcomes
in common.
• So the probability of A or B (landing on either an even or an
odd number) equals the probability of A plus the probability
of B.
P(A or B) = P(A)+P(B)
• Another common term that means the SAME thing
as DISJOINT is Mutually Exclusive.
• DISJOINT = MUTUALLY EXCLUSIVE
• The terms are interchangeable…
• Events A and B are independent if knowing that one
occurs does not change the probability that the
other occurs.
Example:
• Roll a yellow die and a red die.
• Event A is the yellow die landing on an even number
• Event B is the red die landing on an odd number.
• These two events are independent, because the
outcome of A does not change the probability of B.
• If events A and B are independent, then the
probability of A and B equals the probability of A
multiplied by the probability of B.
P(A and B) = P(A)×P(B)
• Example: The probability than the yellow die lands
on an even number and the red die lands on an odd
number is:
1 1 1
P(A and B) = P(A)×P(B)   
2 2 4
• If events A and B are
independent,
–then their complements, Ac and Bc
are also independent,
–and Ac is independent of B
–and Bc is independent of A.
• What does Mutually Exclusive mean?
• What does Independent mean?
• Venn Diagram
Food that students ate.
• How many students are
represented?
– 140
• How many students ate
curly fries?
– 62
• How many students did
not eat a hot dog?
Hamburger
– 92
34
45
• How many students ate a
Hamburger and Fries?
12
– 46
1
Curly
Fries
7
6
10
Hot Dog
25
• Are Hamburgers and hot
dogs disjoint?
– NO!
• What is the probability of
randomly selecting a
student that had a hot
dog and curly fries?
– 11/140 = 0.07857
• Tree Diagram
Flipping three coins.
Probabilities
Sample Space
0.5 Head
Head
Flip 2
0.5
0.5
Tail
Flip 1
0.5
0.5 Head
Tail
Flip 2
0.5
Tail
0.5 Head
Flip 3
0.5 Tail
0.5 Head
Flip 3
0.5 Tail
0.5 Head
Flip 3
0.5 Tail
0.5 Head
Flip 3
0.5 Tail
HHH
0.125
HHT
0.125
HTH
0.125
HTT
0.125
THH
0.125
THT
0.125
TTH
0.125
TTT
0.125
• To determine the number of outcomes in a
sample space that has multiple actions
– for example, roll 3 dice or flip four coins,
• you can find the total number of outcomes by
multiplying the individuals.
• roll three dice:
6 outcomes X 6 outcomes X 6 outcomes
= 216 total
• flip four coins:
2 outcomes X 2 outcomes X 2 outcomes X 2 outcomes
= 16 total
• Replacement
– when drawing cards, or numbers from a hat, etc. and
return the number used.
• Without Replacement
– when drawing cards, or numbers from a hat, etc. and
you do not return the previously drawn item. This
changes the number of total outcomes.
• Example: Draw three cards from a standard deck of
cards without replacement. How many outcomes
are there?
52 X 51 X 50 = 132,600 total outcomes
Class
Freshman
Sophomore
Junior
Senior
Probability
0.36
0.25
0.20
0.19
• This table is an example of a probability model
This model shows the probability of a student being in a
certain class if selected at random from a high school.
• What do you notice about the sum of the
probabilities?
• Find each of the following…
P(Freshman)
P(Seniorc)
• Equally Likely Outcomes
– If a random phenomenon has k possible outcomes,
all equally likely, then each individual outcome has
probability 1/k.
• Event Probabilities
– To find the probability of an event, take the number of
outcomes in event A and divide it by the total number
of outcomes in the sample space S.
count of outcomes in A
P(A)=
count of outcomes in S
• OR

–Union of two or more events
• AND

–Intersection of two or more
events
P(A or B)
P(A  B)
P(A and B)
P(A  B)
P(A or B)
P(A  B)
P(A and B)
P(A  B)
• What does Mutually Exclusive mean?
P(AB) = P(A) + P(B) – P(AB)
A
B
Special Case: Disjoint
A
B
P(AB) = P(A) + P(B)
No overlap means P(AB) = 0
• P(A) = 0.3
• P(B) = 0.4
• P( A or B ) = 0.65
• Are events A and B
mutually exclusive?
• How do you know?
• If events A and B are DISJOINT, then the fact that A
occurs tells us that B can not occur…
• Disjoint events are NOT Independent
• The first transatlantic telegraph cable was laid in
1866. The first telephone cable across the Atlantic
did not appear until 1956 – the barrier was designing
“repeaters,” amplifiers needed to boost the signal,
that could operate for years on the sea bottom. The
first cable had 52 repeaters. The copper cable, laid
in 1963 and retired in 1994, had 662 repeaters. The
first fiber optic cable, laid in 1988 and has 109
repeaters. There are now more than 400,000 miles
of undersea cable, with more being laid every year
to handle the flood of internet traffic.
• Repeaters in undersea cables must be very reliable.
To see why, suppose that each repeater has
probability 0.999 of functioning without failure for
25 years. Repeaters fail independently of each
other.
• What is the probability that two repeaters will work
for 25 years?
• 10 repeaters?
• 662 repeaters?
• P354 (6.15)
• The probability that event A occurs if we
know for certain that event B will occur is
called conditional probability.
• The conditional probability of A given B is denoted:
P( A | B )
• If events A and B are independent, then knowing
that event B will occur does not change the
probability of A so for independent events:
P(A | B)=P(A)
P(B| A)=P(B)
• Example: When flipping a coin twice, what is the
probability of getting heads on the second flip if the
first flip was a head?
– Event A: getting head on first flip
– Event B: getting head on second flip
• Events A and B are independent since the outcome
of the first flip does not change the probability of the
second flip, so…
1
P(B | A) = P(B) =
2
Venn Diagram: Flip of 2 Fair Coins
0.25
0.25
P(A) = 0.5
0.25
P(B) = 0.5
0.25
P(AB) = P(A)  P(B|A)
A
B
Special Case: Independent
P(AB) = P(A)P(B)
Knowing A occurred does not
effect B so…P(B|A) = P(B)
• Choose a new car or light truck at random and note
its color. Here are the probabilities of the most
popular colors for vehicles in North America in 2000.
Color
Silver
White
Black
D Green
D Blue
M Red
Prob
0.176
0.172
0.113
0.089
0.088
0.067
Color
Silver
White
Black
D Green
D Blue
M Red
Prob
0.176
0.172
0.113
0.089
0.088
0.067
• P(color other than listed) =
= 1-(0.176+0.172+0.113+0.089+0.088+0.067)
= 0.295
• P(Silver or White) =
= P(Silver) + P(White) *if disjoint
= 0.176 + 0.172 = 0.348
Color
Silver
White
Black
D Green
D Blue
M Red
Prob
0.176
0.172
0.113
0.089
0.088
0.067
• Let S = Silver and W = White
• P(SS)
= P(S)P(S) *if independent
= (0.176)(0.176) = 0.030976
• P(W  W)
= P(W)P(W) *if independent
= (0.172)(0.172) = 0.029584
• P((S  S)(W W)) = P(S S)+P(W W)
= 0.0605
• The 2000 census allowed each
person to choose one or more
from a long list of races. That
is, in the eyes of the Census
Bureau, you belong to
whatever race or races you say
you belong to.
“Hispanic/Latino” is a separate
category; Hispanics may be of
any race. If we choose a
resident of the United States
at random, the 2000 census
gives these probabilities
Hispanic
NonHispanic
Asian
0.000
0.036
Black
0.003
0.121
White
0.060
0.691
Other
0.062
0.027
• Let A be the event that a randomly
chosen American is Hispanic and let B
be the event that the person chosen is
white.
• P(S) =
1
• P(A) =
0.000+0.003+0.060+0.062 = 0.125
• P(BC)=
1 – (0.060+0.691) = 0.249
• P(Non-Hispanic White)
P(ACB)
=0.691
Hispanic
NonHispanic
Asian
0.000
0.036
Black
0.003
0.121
White
0.060
0.691
Other
0.062
0.027
• The type of medical care a
patient receives may vary with
the age of the patient. A large
study of women who had a
breast lump investigated
whether or not each woman
received a mammogram and a
biopsy when the lump was
discovered.
YES
NO
Under 65
0.321
0.124
65 +
0.365
0.190
• P(Under65)
= 0.321+0.124 = 0.445
• P(65+)
YES
NO
Under 65
0.321
0.124
65 +
0.365
0.190
= 0.365+0.190 = 0.555
• P(Yes)
= 0.321+0.365 = 0.686
• P(No)
= 0.124+0.190 = 0.314
A = 65+
B = YES
• Are A & B Independent?
• P(AB) ?=? P(A)P(B)
From Table
P(A B) = 0.365
Mult Rule
P(A) = 0.555 P(B) = 0.686
P(A)P(B) = (0.555)(0.686) = 0.3807
YES
NO
Under 65
0.321
0.124
65 +
0.365
0.190
• The Values are not the same so they are
NOT Independent!
• Deborah and Matthew are anxiously awaiting word
on whether they have been made partners of their
law firm. Deborah guesses that her probability of
making partner is 0.7 and that Matthew's is 0.5.
Deborah also guesses that the probability that both
she and Matthew are made partners is 0.3.
• P(at least one is promoted) = ?
P(D or M)=
P(D) + P(M) – P(D and M)
0.7 + 0.5 - 0.3
= 0.9
D
C
A
B
• Event A = The woman chosen is young (18 to 29)
• Event B = The woman chosen is married
Event A = The woman chosen is young (18 to 29)
Event B = The woman chosen is married
P(A) =
22,512
 0.217
103,870
7,842
P(A and B) =
 0.075
103,870
7,842
 0.348
P(B|A) =
22,512
The Multiplication Rule
P(AB) = P(A)P(B|A)
Only 5% of male high
school basketball,
baseball, and
football players go
on to play at the
college level. Of
these only 1.7%
enter major
league
professional
sports.
A = competes in
college
B = competes
professionally
P(A and B)=
0.05  0.017  0.00085
P(Ac and Bc)=
0.95  0.9999  0.9499
P(Bc|A)=
0.983