Download Conditional Probability and Introduction to Distributions

Combining Probabilities and
Conditional Probability
Stat 700 Lecture 4
9/11/2001-9/13/2001
Overview of Lecture
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Algebra of Events and Probabilities
Conditional Probability and its Importance
Probability Updating: Bayes Rule
Independent Events
Discrete Random Variables
Introduction to Probability Distributions
Parameters of Distributions
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Algebra of Events and Probabilities
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Given an event A, its complement, denoted by Ac, is
the event whose elements are those that are not in A.
Thus, the event Ac is the opposite of event A.
Note that (Ac)c = A. (Rule of double complementation)
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Rule 1 (Complementation Rule): For any event A,
P(A) = 1 - P(Ac).
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From this rule it follows immediately that P() = 0,
since P(S) = 1.
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A Simple Application
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Example: Consider the experiment of rolling two fair dice
simultaneously, and let A be the event that the sum of the
outcomes is at most 11. We seek P(A).
Solution: Since there are (6)(6) = 36 simple events, and we can
assume that these events are equally likely, then each has
probability of 1/36. Out of these 36 events, only one of them, the
simple event {(6,6)}, has a sum that exceeds 11, so Ac = {(6,6)}.
Therefore, by the complementation rule,
P( A)  1  P( Ac )
1
35
 1

.
36 36
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Algebra … continued
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Given two events, A and B, the union of A and B,
denoted by (A  B), is the event whose elements are
those that belong to either A or B or both. It
represents the occurrence of at least one of the
events A or B.
We also write (A or B) for (A  B).
We could generalize this to the union of several
events, e.g., (A  B  C  D) which would then
represent the occurrence of at least one of these 4
events.
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Algebra … continued
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Given two events, A and B, the intersection of A and
B, denoted by (A  B), is the event whose elements
are those that belong to both A and B. It represents
the simultaneous occurrence of A and B.
We also write (A and B) and (AB) for (A  B).
We could generalize this to the intersection of several
events, e.g., (A  B  C  D) which would then
represent the simultaneous occurrence of all 4
events.
If A  B =  = {empty event}, we say that A and B are
disjoint or mutually exclusive.
Generalization: notion of pairwise disjoint events.
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Finite Additivity Properties
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Rule 2: Given disjoint events A and B, then
P(A  B) = P(A) + P(B).
Extended Rule 2: Given pairwise disjoint
events A1, A2, …, Ak, then
P( A1  A2  ...  Ak )  P( A1 )  P( A2 )  ...  P( Ak )
k
  P( Ai ).
i 1
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Addition and Generalized Addition
Rules
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Rule 3 (Addition Rule): Given events A and B (which
are not necessarily disjoint), then
 P(A  B) = P(A) + P(B) - P(A  B).
Extended Rule 3 (Inclusion-Exclusion Principle):
Given three events A, B, and C (which are not
necessarily pairwise disjoint), then
P( A  B  C )  {P ( A)  P( B )  P (C )}
 {P ( AB)  P ( AC )  P ( BC )}
 P ( ABC ).
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Some Concrete Applications of the
Probability Rules
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Example 1: A study is to be performed to examine the
association between the occurrence of lung cancer and
smoking. Suppose that one person is to be randomly chosen
and classified into either a smoker or a nonsmoker, and whether
he/she has lung cancer or not. For this experiment, the sample
space is:
S ={(Nonsmoker, No lung cancer), (Nonsmoker, With lung
cancer), (Smoker, No lung cancer), (Smoker, With lung cancer)}.
Assume that the proportion in the population who are smokers is
0.15, while the proportion in the population who have lung
cancer is 0.05. Furthermore, assume that the proportion in the
population who are smokers and with lung cancer is 0.009.
We seek the probability that the person chosen will either be a
smoker or has lung cancer.
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Information for Example
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The table below shows the information provided for in
the description of the problem.
Such a table could also have been used to compute
the desired probabilities … illustrated in class.
Smokers
With lung cancer
Nonsmokers Marginal Proportions
0.009
0.05
Without lung cancer
Marginal Proportions
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0.15
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Applications … continued
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Solution of Example 1: Let A be the event that the
person is a smoker, and B be the event that the
person has lung cancer. From the given information
we have that P(A) = 0.15, P(B) = 0.05, and P(AB) =
0.009. Therefore, by the addition rule, the desired
probability is
P( A or B)  P( A)  P( B)  P( AB)
 0.15  0.05  0.009
 0.191.
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DeMorgan’s Rules
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DeMorgan’s rules state that:
– the complement of the union of events equals the
intersection of their complements;
– the complement of the intersections of events
equals the union of their complements.
Formally, for two events A and B, we have:
A  B
c
 ( Ac  B c );
( A  B)  ( A  B ).
c
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Example … continued
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Suppose that in the preceding example we were instead
interested in the probability that the person is neither a smoker
nor has lung cancer. Then, in formal notation, we want:
P(Ac  Bc) = P{(not a smoker) and (is free of lung cancer)}.
By virtue of deMorgan’s rule, we have that Ac  Bc = (A  B)c.
Applying the complementation rule, we therefore obtain
P( Ac  B c )  P(( A  B) c )
 1  P( A  B) since ( A  B) cc  A  B
 1  0.191  0.809.
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Another Approach: By Completing the
Table of Probabilities
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The answers obtained in the preceding slides
could also be derived by simply completing
the table of probabilities:
Smokers
Nonsmokers Marginal Proportions
With lung cancer
0.009
0.041
0.05
Without lung cancer
0.141
0.809
0.95
Marginal Proportions
0.15
0.850
1.00
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A Matching Problem
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Example 2: Four people, Peter, Paul, Mary, and
Magdalene, write their names on identically-sized
chips of papers. These chips are then placed on a
box and thoroughly shuffled. Each of them (in the
order given above) then randomly draws a chip from
the box, with the drawing being without replacement.
We seek the probabilities of the following events:
– a) probability that Paul draws his name;
– b) probability that either Paul or Magdalene (or both) draw
their respective names; and
– c) probability that at least one of them draws his/her name.
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Matching … continued
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Solution: We shall let A = {Peter draws his name}; B =
{Paul draws his name}; C = {Mary draws her name};
and D = {Magdalene draws her name}. Observe that
the number of outcomes {e.g., (Paul, Mary,
Magdalene, Peter)} in the sample space S is
(4)(3)(2)(1) = 4! = 24, and these outcomes are
equally likely since the draws are done at random.
Now, for problem (a), we want: P(B) = N(B)/N(S). But
N(B) = (3)(1)(2)(1) = 6 since we need “Paul” to be the
second chip drawn. Therefore, P(B) = 6/24 = 1/4 =
0.25.
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Continued ...
• For
problem (b) we want P(B or D). By the
addition rule, we have P(B or D) = P(B) + P(D) P(B and D). Analogously to problem (a), we
have P(D) = (3)(2)(1)(1)/24 = 1/4.
• On the other hand, P(B and D) = (2)(1)(1)(1)/24
= 1/12 since both Paul and Magdalene must get
their names, respectively.
•Therefore, P(B or D) = 1/4 + 1/4 - 1/12 = 5/12
= 0.4167.
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Solution … continued
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For problem (c) where we are seeking the probability
of having at least one of them draw their own name,
we want:
P{A  B  C  D} = P{at least one of A, B, C, or D
occurs}.
But by the generalized addition rule (inclusionexclusion principle) for 4 events, this is:
P{A  B  C  D} = {P(A)+P(B)+P(C)+P(D)} - {P(AB)
+ P(AC) + P(AD) + P(BC) + P(BD) + P(CD)} +
{P(ABC) + P(ABD) + P(ACD) + P(BCD)} - P(ABCD}.
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Continued ...
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Similar calculations as in the previous two
problems now yield that (coloreds means the
number of summands):
P{A  B  C  D} = (4)(1/4) - (6)(1/12) +
(4)(1/24) - (1)(1/24) = 1 - 1/2 + 1/6 - 1/24 = 1 1/2! + 1/3! - 1/4! = 0.625.
Food for Thought!! What do you think will
happen to this probability if there were 1000
people, instead of 4 people??
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Another Solution: The “Brute Force”
Approach
Sample Space: Possible Outcomes of the Draws
(Legend: 1=Peter, 2=Paul, 3=Mary, 4=Magdalene)
(1,2,3,4)
(1,2,4,3)
(1,3,2,4)
(1,3,4,2)
(1,4,2,3)
(1,4,3,2)
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(2,1,3,4)
(2,1,4,3)
(2,3,1,4)
(2,3,4,1)
(2,4,1,3)
(2,4,3,1)
(3,1,2,4)
(3,1,4,2)
(3,2,1,4)
(3,2,4,1)
(3,4,1,2)
(3,4,2,1)
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(4,1,2,3)
(4,1,3,2)
(4,2,1,3)
(4,2,3,1)
(4,3,1,2)
(4,3,2,1)
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Event B: “Paul” is Matched
Sample Space: Possible Outcomes of the Draws
(Legend: 1=Peter, 2=Paul, 3=Mary, 4=Magdalene)
(1,2,3,4)
(1,2,4,3)
(1,3,2,4)
(1,3,4,2)
(1,4,2,3)
(1,4,3,2)
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(2,1,3,4)
(2,1,4,3)
(2,3,1,4)
(2,3,4,1)
(2,4,1,3)
(2,4,3,1)
(3,1,2,4)
(3,1,4,2)
(3,2,1,4)
(3,2,4,1)
(3,4,1,2)
(3,4,2,1)
(4,1,2,3)
(4,1,3,2)
(4,2,1,3)
(4,2,3,1)
(4,3,1,2)
(4,3,2,1)
Therefore, P(B) = 6/24 = 1/4 = 0.25.
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Event (B or D): Either “Paul” or
“Magdalene” or Both Are Matched
Sample Space: Possible Outcomes of the Draws
(Legend: 1=Peter, 2=Paul, 3=Mary, 4=Magdalene)
(1,2,3,4)
(1,2,4,3)
(1,3,2,4)
(1,3,4,2)
(1,4,2,3)
(1,4,3,2)
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(2,1,3,4)
(2,1,4,3)
(2,3,1,4)
(2,3,4,1)
(2,4,1,3)
(2,4,3,1)
(3,1,2,4)
(3,1,4,2)
(3,2,1,4)
(3,2,4,1)
(3,4,1,2)
(3,4,2,1)
(4,1,2,3)
(4,1,3,2)
(4,2,1,3)
(4,2,3,1)
(4,3,1,2)
(4,3,2,1)
Therefore, P(B or D) = 10/24 = 5/12 = 0.4167.
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Event (A or B or C or D): At Least One
of Them Gets a Match
Sample Space: Possible Outcomes of the Draws
(Legend: 1=Peter, 2=Paul, 3=Mary, 4=Magdalene)
(1,2,3,4)
(1,2,4,3)
(1,3,2,4)
(1,3,4,2)
(1,4,2,3)
(1,4,3,2)
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(2,1,3,4)
(2,1,4,3)
(2,3,1,4)
(2,3,4,1)
(2,4,1,3)
(2,4,3,1)
(3,1,2,4)
(3,1,4,2)
(3,2,1,4)
(3,2,4,1)
(3,4,1,2)
(3,4,2,1)
(4,1,2,3)
(4,1,3,2)
(4,2,1,3)
(4,2,3,1)
(4,3,1,2)
(4,3,2,1)
Thus, P(A or B or C or D) = 15/24 = 5/8 = 0.625.
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Conditional Probability: Motivation
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In many situations occurring in the sciences, both
natural and social, one is interested in the probability
of an event given other information.
For example, one may not be interested in the
probability of a person getting lung cancer, but rather
might be interested in the probability that the person
will get lung cancer given the information that this
person is a cigarette smoker.
Or, one maybe interested in knowing the probability
that someone is HIV-infected given a positive result
from an ELISA test (a test for HIV-infection).
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Definition of Conditional Probability
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In such situations, we are interested in the
conditional probability of an event.
Given events A and B (for some experiment),
the conditional probability of B given A is:
 P( A  B)
whenever P( A)  0
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P( B | A)   P( A)
.
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not defined whenever P( A)  0
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Justification of the Definition
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If we are given the information that event A has
occurred, then we know that the outcome of the
experiment is in event A.
Therefore, given this information, event B could have
occurred only if the outcome is in the intersection of A
and B. Dividing P(A and B) by P(A) serves to
standardize P(A and B) since, given that A has
occurred, the effective sample space becomes A.
If P(A) = 0, then conditioning on an event that never
occurs is clearly not of any interest!
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A Simple Example
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Consider the experiment of tossing three fair
coins simultaneously.
Thus, S = {HHH, HHT, HTH, THH, HTT, THT,
TTH, TTT}, and each outcome has probability
of 1/8.
Define A = identical outcomes = {HHH, TTT},
and B = at least two heads = {HHT, HTH,
THH, HHH}. Thus, P(A) = 2/8 and P(B) = 4/8.
Note that (A  B) = {HHH} so P(A  B) = 1/8.
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Continued ...
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P(B|A) = P(A  B)/P(A) = (1/8)/(2/8) = 1/2.
This conditional probability is clearly intuitive
since if we know that A has occurred [so the
outcome was either (HHH) or (TTT)], then the
only way that B could have occurred is if the
outcome was (HHH)
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Example … continued
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On the other hand, if we are given the information
that B has occurred, then
P(A|B) = P(A  B)/P(B) = (1/8)/(4/8) = 1/4.
Again, this is intuitive because the information that B
occurred tells us that the outcome is either (HHT),
(HTH), (THH), or (HHH). In order for A to have
occurred, then (HHH) must be the outcome, hence
the conditional probability of A given B is 1/4.
From these two examples, note that:
P(B|A) and P(A|B) need not be identical.
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Another Example
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Consider the hypothetical population proportions
regarding smoking and the presence of lung cancer
which were used in previous examples. The
information is reproduced in the table below. Recall
that the experiment is to randomly choose one person
from this population.
Smokers
Nonsmokers Marginal Proportions
With lung cancer
0.009
0.041
0.05
Without lung cancer
0.141
0.809
0.95
Marginal Proportions
0.15
0.850
1.00
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Example … continued
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Let us denote by A the event that the person chosen
is a smoker, and by B the event that the person
chosen has lung cancer.
Suppose that we are given the information that the
person chosen is a smoker. Then
P(has lung cancer | smoker) = P(B|A) = P(AB)/P(A)
= 0.009/0.15 = 0.06.
On the other hand, if we know that the person is a
nonsmoker, then
P(has lung cancer | nonsmoker) = P(B|Ac) =
P(AcB)/P(Ac) = 0.041/0.85 = 0.048.
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Example … continued
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By comparing the conditional probabilities P(B|A) =
0.060 and P(B|Ac) = 0.048 [and we could do such a
comparison since the process of conditioning has
standardized the probabilities for the two groups],
one can conclude (in this hypothetical situation) that
the prevalence of lung cancer among smokers is
slightly higher than the prevalence of lung cancer
among nonsmokers.
One may also look at the “inverse” probabilities:
P(A|B) = P(A and B)/P(B) = 0.009/0.05 = 0.18; and
P(A|Bc) = P(A and Bc)/P(Bc) = 0.141/0.95 = 0.148.
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Multiplication Rule
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Given events A and B, if we are given P(A) > 0
and the conditional probability P(B|A), then the
probability P(A  B) could be obtained by
inverting the conditional probability formula to
get the multiplication rule:
P( A  B)  P( A) P( B | A).
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Utility of Multiplication Rule
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The multiplication rule is what enables us to multiply
the (conditional) probabilities in the branches of a
tree diagram to get the (joint) probabilities of the
outcomes of the experiment.
On the other hand, the multiplication rule is only
usable if we are able to determine the conditional
probability of B given A without recourse to the
conditional probability rule [since this latter rule
requires P(AB)].
Obtaining P(B|A) this way is usually done by
examining the situation at hand.
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An Illustration
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Example: Suppose you have two boxes with
Box I containing 4 red and 7 blue balls, and
Box II containing 6 red and 2 blue balls.
The two-step experiment is to pick a ball at
random from Box I which is then transferred
to Box II. A ball is then drawn from Box II.
Let A = event that a red ball is transferred
from I to II; and
let B = event that a red ball is drawn from II.
We seek: P(A and B) = P({both balls are red})
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Example … continued
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Clearly, P(A) = 4/11, while without using the
conditional probability formula, we obtain:
P(B|A) = 7/9 since, after the transfer of a red
ball, there will be 7 reds and 2 blues in Box II.
Consequently, by the multiplication rule:
P(A and B) = P(A)P(B|A) = (4/11)(7/9) = 28/99
= 0.2828.
In class we illustrate this computation via a
tree diagram.
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Example … continued
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Similarly, we obtain the joint probability of
getting a blue ball from Box I and a red ball
from Box II, which is symbolically represented
by (Ac  B) as follows:
P(Ac) = P(blue from Box I) = 7/11;
P(B|Ac) = P(red from Box II | blue ball
transferred) = 6/9, so
P(Ac  B) = P(Ac)P(B|Ac) = (7/11)(6/9) =
42/99 = 0.4242.
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Example … continued
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Let us now consider the problem of determining the
(marginal) probability of getting a red ball from Box II,
that is, we want P(B).
Notice that we are not anymore directly interested in
what happens in step 1 (getting a ball from Box I), but
at the same time we see that the occurrence of event
B depends on what will happen in step 1.
We are therefore faced with the problem of
combining the probabilities arising from whether we
transfer a red ball to Box II or we transfer a blue ball
to Box II.
The question is: how do we combine?
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Analyzing the Situation
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First we note that the event B could arise in
two ways:
– (draw a red from I, then draw a red from II); or
– (draw a blue from I, then draw a red from II).
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Symbolically, this is could be represented via:
B = (A  B)  (Ac  B)
Furthermore, note that since A and Ac are
disjoint (could not occur simultaneously), then
(A  B) and (Ac  B) are also disjoint events.
We could therefore apply the addition rule.
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The Combined Probability
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By the addition rule (or the finite additivity property),
we obtain:
P(B) = P{(A  B)  (Ac  B)} = P(A  B) +
P(Ac  B) = P(A)P(B|A) + P(Ac)P(B|Ac).
From our earlier calculations where we used the
multiplication rule, we obtained:

P(A  B) = 28/99 and P(Ac  B) = 42/99.

Therefore, P(B) = 28/99 + 42/99 = 70/99 =
.7070. This is the probability of getting a red ball
from Box II (taking into proper account what could
happen with the draw from Box I).
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Theorem of Total Probabilities
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The calculation of P(B), where B is an event
pertaining to the result of the second step of the
experiment, is a special case of what is referred to as
the “Theorem of Total Probabilities,” a method for
combining probabilities from the different possibilities
arising from the first step of the experiment.
In its simplest form, when there are only two
possibilities, A and Ac, from the first step of the
experiment, the theorem states that:
P( B)  P( A) P( B | A)  P( A ) P( B | A ).
c
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c
41
Updating of Probabilities






Let us go back to the example we were considering.
Evidently, it is immediate that
P(A) = P(red from Box I) = 4/11 = .3636; and
P(Ac) = P(blue from Box I) = 7/11 = .6363.
These are our prior probabilities for A and Ac.
Suppose that when we performed the experiment, we
did not look at the color of the ball that we transferred
from Box I to Box II. Furthermore, suppose that when
we looked at the ball drawn from Box II it is red.
Given this information, how do we update our
knowledge of what we transferred from Box I to II?
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Updating … continued

Since we are interested in determining the probability
of event A (and also of Ac), given event B, then the
desired probabilities are just the conditional
probabilities. Therefore, applying the conditional
probability rule, we have:
P( A  B) P( A) P( B | A)
P( A | B) 

; and
P( B)
P( B)
c
c
c
P
(
A

B
)
P
(
A
)
P
(
B
|
A
)
c
P( A | B) 

P( B)
P( B)
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The Updating … continued




From our earlier calculations, we have obtained:
P(A  B) = 28/99 and P(Ac  B) = 42/99
P(B) = 70/99
Substituting these values in the formulas of the
preceding slide, we obtain our updated probabilities,
also called the posterior probabilities, to be:
(28 / 99) 28
P( A | B) 

 0.40;
(70 / 99) 70
(42 / 99) 42
c
P( A | B) 

 0.60.
(70 / 99) 70
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Comparison: Priors and Posteriors



We may compare the posterior probabilities of P(A|B)
= 0.40 and P(Ac|B) = 0.60 with the prior probabilities
of P(A) = 0.3636 and P(Ac) = 0.6363.
These values indicate that the information that we got
a red ball from Box II had increased the probability
that we transferred a red ball from Box I to Box II
from 0.3636 to 0.40, and decreased the probability
that we transferred a blue ball from 0.6363 to 0.60.
The directions of change in the values are clearly
intuitive, but the exact magnitudes of the changes
cannot be obtained without recourse to the formulas
and reasoning we had employed.
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(Reverend) Bayes Theorem


The procedure we have just employed to update our
prior probabilities of A and Ac, given the occurrence of
event B, is a special case of Bayes Theorem.
For the situation where in the first step of the
experiment we only have two possibilities: A and Ac,
and B is an event pertaining to the outcome of the
second step of the experiment, Bayes Theorem states
that:
P( A) P( B | A)
P( B)
P( A) P( B | A)

.
c
c
P( A) P( B | A)  P( A ) P( B | A )
P( A | B) 
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Another Example


Situation: A medical test for HIV-infection has the
following characteristics:
– if the person is HIV-infected, the (conditional)
probability that the test will be positive is 0.98, so
the (conditional) probability that it will turn up
negative is 0.02; while
– if the person is not HIV-infected, the (conditional)
probability that the test will be negative is 0.99, so
the conditional probability that it will turn up
positive is 0.01.
Assume that the prevalence of HIV-infection in the
population is 0.005.
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Characteristics of the Medical Test for
Infection

Below we present in tabular form the
characteristics of the test for HIV-infection.
Note that the probabilities are conditional
probabilities.
THEN the
test for
HIVinfection
will be
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WITH conditional
probability of
Positive (+)
Negative (-)
IF the person being tested is
HIV-infected
Not HIVinfected
0.98
0.01
0.02
Lecture 4: More Probabilities and
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0.99
48
Example … continued

The Experiment: Suppose that a person is randomly
chosen from the population, and this person is
subjected to the test for HIV-infection. We are
interested in:
– a) The probability that the person will be HIVinfected (prior to the test).
– b) The probability that the test for HIV-infection
will show a positive result.
– C) Given that the test showed a positive result,
the (updated) probability that the person is HIVinfected. [QUESTION: Without computing the
probability, what is your best estimate??]
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Example … continued







Solution: We let A be the event that the person
chosen is HIV-infected, and by B the event that the
test for HIV-infection will show a positive result. From
the given information, we have:
P(A) = 0.005 so P(Ac) = 1 - P(A) = 1-0.005 = 0.995.
P(B|A) = 0.98 and P(B|Ac) = 0.01.
By Theorem of Total Probabilities:
P(B) = P(A)P(B|A) + P(Ac) P(B|Ac) = (.005)(.98) +
(.995)(.01) = .0049 + .00995 = .01485
By Bayes Rule:
P(A|B) = P(A)P(B|A)/P(B) = (.005)(.98)/.01485 = 0.33.
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Generalizations of the Combining and
Updating Rules


For completeness of our discussion, we
present generalizations of the following:
– Multiplication Rule
– Theorem of Total Probabilities
– Bayes Rule
The application of these rules are analogous
to the preceding examples except for the
possibility that there might be more than 2
outcomes in the “first step” of the experiment.
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Independent Events




Let A and B be events of an experiment. We say that
A and B are independent if and only if:
 P(B|A) = P(B) or P(A|B) = P(A)
That is, knowledge of the occurrence of one event
does not provide information about whether the other
event has also occurred.
By the multiplication rule and the above definition, if A
and B are independent, then
 P(A  B) = P(A)P(B)
That is, joint probability = product of marginal
probabilities for independent events.
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Examples






Suppose that two fair coins are tossed so S = {HH,
HT, TH, TT} and each sample outcome has
probability of 1/4.
Let A = {outcome of first toss is H} = {HH, HT} so P(A)
= 2/4 = 1/2.
Let B = {outcome of second toss is T} = {HT, TT} so
P(B) = 2/4 = 1/2.
Note that since A  B = {HT} then P(A  B) = 1/4.
Therefore, P(B|A) = (1/4)/(1/2) = 1/2 = P(B), so A and
B are independent events.
Note also that P(A  B) = (1/4) = (1/2)(1/2) =
P(A)P(B), another way to check independence.
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Examples … continued

This preceding example is the reason
why we say that the outcomes of the
tosses of a coin (or dice) are
independent. In a practical sense, the
outcome of the first toss does not have
any bearing on the outcome of the
second toss.
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Other Comments About Independence




The main utility of the property of independence
among events is it enables us to calculate joint
probabilities by simply multiplying the marginal
probabilities.
Thus, if A, B, C, D are independent, then P(ABCD) =
P(A)P(B)P(C)P(D).
Many events however are not independent! When
independence is assumed, you should have a good
reason (by examining the situation) why the events
are independent.
Saying: The weather in Kenya tomorrow will be
affected by the flutter of the wings of a monarch
butterfly in Mexico today!
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A Final Example





Consider the experiment of observing whether two
people, Paul and Paula, who do not know each other
and are in different places, get the flu in the next 6
months.
Let A be the event that Paul gets the flu, and B be the
event that Paula gets the flu. Assume that A and B
are independent, and furthermore, P(A) = 0.20 and
P(B) = 0.05.
Want: P(at least one of them gets the flu) = P(A or B)
By the addition rule and independence:
P(A or B) = P(A) + P(B) - P(AB) = P(A) + P(B) P(A)P(B) = 0.20 + 0.05 - (0.20)(0.05) = 0.24.
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