Transcript Probability

Probability
The Study of Randomness
The language of probability
Random in statistics does not mean
“haphazard”.
Random is a description of a kind of
order that emerges only in the long run
even though individual outcomes are
uncertain.
The probability of any outcome of a
random phenomenon is the proportion
of times the outcome would occur in a
very long series of repetitions.
Probability Models
The sample space of a random event is
the set of all possible outcomes.
What is the sample space for rolling a sixsided die?
S = {1, 2, 3, 4, 5, 6}
What is the sample space for flipping a
coin and then choosing a vowel at
random?
Tree diagram
H
a
e
i
o
u
a
T
e
i
o
u
S={Ha, He, Hi, Ho, Hu, Ta, Te, Ti, To, Tu}
• What is the sample space for answering
one true/false question?
• S = {T, F}
• What is the sample space for answering
two true/false questions?
• S = {TT, TF, FT, FF}
• What is the sample space for three?
Tree diagram
True
True
False
True
True
False
False
True
True
False
False
True
False
False
S = {TTT, TTF, TFT, FTT, FFT, FTF,
TFF, FFF}
Intuitive Probability
An event is an outcome or set of
outcomes of a random phenomenon.
An event is a subset of the sample
space.
For probability to be a mathematical
model, we must assign proportions for
all events and groups of events.
Basic Probability Rules
The probability P(A) of any event A
satisfies 0 < P(A) < 1.
Any probability is a number
between 0 and 1, inclusive.
If S is the sample space in a probability
model, then P(S) = 1.
All possible outcomes together
must have probability of 1.
Complement Rule
The complement of any event A is the
event that A does not occur, written as Ac.
The complement rule states that
P(Ac) = 1 – P(A)
The probability that an event does
not occur is 1 minus the probability
that the event does occur.
Venn diagram: complement
S
A
Ac
General Addition Rule for
Unions of Two Events
For any two events A and B,
P(A or B) = P(A) + P(B) – P(A and B)
P(AB) = P(A) + P(B) – P(AB)
The simultaneous occurrence of two
events is called a joint event.
The union of any collections of event
that at least one of the collection
occurs.
Venn diagram: {A and B}
S
A
B
Venn diagram: disjoint events
(Mutually Exclusive)
S
A
B
Addition Rule
Two events A and B are disjoint (also called
Mutually Exclusive) if they have no
outcomes in common and so can never occur
simultaneously. If A and B are disjoint,
P(A or B) = P(A) + P(B)
If two events have no outcomes in
common, the probability that one or
the other occurs is the sum of their
individual probabilities.
General Multiplication Rule
The joint probability that both of two
events A and B happen together can be
found by
P(A and B) = P(A) P(B|A)
P(B|A) is the conditional probability that
B occurs given the information that A
occurs.
Definition of Conditional Probability
When P(A)>0, the conditional
probability of B given A is
P(A and B)
P(B|A) =
P(A)
OR
P(AB) = P(AB)
P(B)
Multiplication Rule
If one event does not affect the
probability of another event, the
probability that both events occurs
is the product of their individual
probabilities.
Two events A and B are independent if
knowing that one occurs does not change the
probability that the other occurs. If A and B
are independent,
P(A and B) = P(A)P(B)
Question #3
Suppose that 60% of all customers of a large insurance
agency have automobile policies with the agency, 40%
have homeowner’s policies, and 25% have both types of
policies. If a customer is randomly selected, what is the
probability that he or she has at least one of these two
types of policies with the agency? (Hint: Venn diagram)
P(A or B) = P(A) + P(B) – P(A and B)
P(auto or home) = .60 + .40 - .25 = .75
Question #6
Drawing two aces with replacement.
 4  4 
P(2 aces)=      .0059
 52   52 
Drawing three face cards with replacement.
 12   12   12 
P(3 face)=        .0123
 52   52   52 
Multiplication Rule Practice
Draw 5 reds cards without replacement.
 26   25   24   23   22 
P(5 red)=            .0253
 52   51   50   49   48 
Draw two even numbered cards without
replacement.
 20   19 
P(2 even)=      .1433
 52   51 
Multiplication Rule Practice
Draw three odd numbered red cards
with replacement.
3
 8 
P(3 red, odd)=    .0036
 52 
Back to Flipchart
Question #7
Nondefective Defective
Company 1
10
5
Company 2
8
2
What is the probability of a GFI switch from a selected spa
is from company 1?
15
P  company 1 
 .6
25
What is the probability of a GFI switch from a selected spa
is defective?
7
P  defective  
 .28
25
Question #7
Nondefective Defective
Company 1
10
5
Company 2
8
2
What is the probability of a GFI switch from a selected spa
is defective and from company 1?
5
P  company 1  defective  
 .2
25
What is the probability of a GFI switch from a selected spa
is from company 1 given that it is defective?
5
P  company 1|defective    .7143
7
Question #7
Nondefective Defective
Company 1
10
5
Company 2
8
2
P(A and B) = P(A) P(B|A)
5
P  company 1  defective  
25
7
P  defective  
25
5
P  company 1|defective  
7
5
7 5
 
25 25 7
Remember
Two events A and B are independent if
knowing that one occurs does not change the
probability that the other occurs. If A and B
are independent,
P(A and B) = P(A)P(B)
The joint probability that both of two events
A and B happen together can be found by
P(A and B) = P(A) P(B|A)
How can we use the formulas to test for
independence?
Independence and Mutually
Exclusivity
Independence means knowing
something about one tells you nothing
about the other.
Mutually exclusive events cannot
happen at the same time.
Are independent events mutually
exclusive?
Independent Events
Two events A and B that both have
positive probability are independent if
P(B|A) = P(B)
Back to flipchart
13. Jack and Jill have finished conducting taste tests with 100 adult
from their neighborhood. They found that 60 of them correctly
identified the tap water. The data is displayed below.
Yes
No
Total
Male
21
14
35
Female
39
26
65
Total
60
40
100
Is the event that a participant is male and the
event that he correctly identified tap water
independent?
In order for a participant being male and the event that
he correctly identified tap water to be independent, we
know that
Yes
No
Total
Male
21
14
35
Female
39
26
65
Total
60
40
100
P(male|yes) = P(male)
or
P(yes|male) = P(yes)
In order for a participant being male and the event that
he correctly identified tap water to be independent, we
know that P(yes|male) = P(yes)
Yes
No
Total
Male
21
14
35
Female
39
26
65
Total
60
40
100
21
60
21 60
We know P(yes|male) =
and P(yes) =
. Since

35
100
35 100
we can conclude the participant being male and their ability to
correctly identify tap water are independent.
In order for a participant being male and the event that
he correctly identified tap water to be independent, we
know that P(male|yes) = P(male)
Yes
No
Total
Male
21
14
35
Female
39
26
65
Total
60
40
100
21
35
21 35
We know P(male|yes) =
and P(male) =
. Since

60
100
60 100
we can conclude the participant being male and their ability to
correctly identify tap water are independent.
Question #16
DNA 
DNA -
Has TB
14
12
26
Does Not
0
14
181 193
181 207
What is the probability of an individual having tuberculosis
given the DNA test is negative? 12
.0622
193
12
P(DNA - TB)
207 .0622

P(TB|DNA - ) =
193
P(DNA - )
207
Conditional Probability with
Tree Diagrams
17. Dr. Carey has two bottles of sample pills
on his desk for the treatment of arthritic pain.
He often grabs a bottle without looking and
takes the medicine. Since the first bottle is
closer to him, the chances of grabbing it are
0.60. He knows the medicine from this bottle
relieves the pain 70% of the time while the
medicine in the second bottle relieves the pain
90% of the time. What is the probability that
Dr. Carey grabbed the first bottle given his
pain was not relieved?
st
P(1
bottle  not relieved)
st
P(1 bottle|pain not relieved) 
P(pain not relieved)
.6 .3

 .8182
.6 .3 .4 .1
.6
.4
.7
relieved
.3
not
.9
relieved
1st
2nd
.1
not