CS145: Probability & Computing
Lecture 2: Axioms of Probability,
Conditioning & Bayes’ Rule
Instructor: Erik Sudderth
Brown University Computer Science
January 29, 2015
Course Enrollment
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Overall course grades will be determined as follows: 50% homeworks, 20% midterm exam, 30%
final exam. The midterm exam will be given during the normal lecture time on Thursday, March
12. The final exam will be given on Wednesday, May 6 from 9:00am-12:00pm.
Homework Policies
Homework 1: Out 1/29, due 2/5, NO MATLAB.
Homework Assignments
There will be ten homework assignments, each due one week after it is handed out. Homework
problems will emphasize probabilistic derivations, calculations, and reasoning. Most homeworks
will also have one problem requiring Matlab implementation of simple methods from probability
or statistics. The scores of all ten assignments will be averaged equally to determine an overall
homework score (we will not “drop” any homeworks). Homeworks will be submitted electronically.
Collaboration Policy Students may discuss and work on homework problems in groups. HowCollaboration Policy Students may discuss and work on homework problems in groups. However, each student must write up their solutions independently, and do any required programming
ever, each student must write up their solutions independently, and do any required programming
independently. List the names of any collaborators on the front page of your solutions. You may
independently. List the names of any collaborators on the front page of your solutions. You may
not directly copy solutions from other students, or from materials distributed in previous versions
not directly copy solutions from other students, or from materials distributed in previous versions
of this or other courses. You may not make your solutions available to other students: files in your
of this or other courses. You may not make your solutions available to other students: files in your
home directory may not be world-readable, and you may not post your solutions to public websites.
home directory may not be world-readable, and you may not post your solutions to public websites.
Late Submission Policy Homework assignments are due by 11:59pm on Thursday evenings.
Late Submission Policy Homework assignments are due by 11:59pm on Thursday evenings.
Your answers may be submitted up to 4 days late (by Monday evening); after this point, solutions
Your answers may be submitted up to 4 days late (by Monday evening); after this point, solutions
will be distributed and handins will no longer be accepted. You may submit up to two late
will be distributed and handins will no longer be accepted. You may submit up to two late
assignments without penalty. For each subsequent late assignment, 20 points (out of a maximum
assignments without penalty. For each subsequent late assignment, 20 points (out of a maximum
of 100) will be deducted from the overall score. Exceptions to this policy are only given in very
of 100) will be deducted from the overall score. Exceptions to this policy are only given in very
unusual circumstances, and any extensions must be requested in advance by e-mail to the instructor.
unusual circumstances, and any extensions must be requested in advance by e-mail to the instructor.
Syllabus: Summary of Course Topics
Reasoning Under Uncertainty
• 1/1000 of tourists who visit tropical country X return with a
dangerous virus Y .
• There is a test to check for the virus. The test has 5% false
positive rate and no false negative error.
• You returned from country X , took the test, and it was
positive. Should you take the painful treatment for the virus?
Elements of a Probabilistic Model
• The sample space Ω, which is the set of all possible outcomes of an
experiment.
• The probability law, which assigns to a set A of possible outcomes
(also called an event) a nonnegative number P(A) (called the probability of A) that encodes our knowledge or belief about the collective
“likelihood” of the elements of A. The probability law must satisfy
certain properties to be introduced shortly.
Sample Spaces and
Probability Laws
Probability
Law
Event B
Experiment
Event A
Sample Space
(Set of Outcomes)
P(B)
P(A)
A
B
Events
1.2: The main
ingredients
of a probabilistic
model. Introduction to Probability, 2008
Some figures Figure
and materials
courtesy
Bertsekas
& Tsitsiklis,
“likelihood” of the elements ofSample
A. The
probability
law must satisfy
Space
and Probability
Chap. 1
certain properties to be introduced shortly.
of every Sn , and xn ∈ ∩n Snc . This shows that (∪n Sn )c ⊂ ∩n Snc . The converse
inclusion is established by reversing the above argument, and the first law follows.
The argument for the second law is similar.
6
Defining a Probabilistic Model
Probability
Law
Event B
P(B)
1.2 PROBABILISTIC MODELS
Experiment
Event A
P(A)
A probabilistic model is a mathematical description of an uncertain situation.
Sample Space
It must be in accordance with a(Set
fundamental
framework that we discuss in this
of Outcomes)
A
B
section. Its two main ingredients are listed below and are visualized in Events
Fig. 1.2.
Figure of
1.2:a The
main ingredients
of a probabilistic model.
Elements
Probabilistic
Model
• The sample space Ω, which is the set of all possible outcomes of an
experiment.
Sample
Spaces
and Events
• The
probability
law, which assigns to a set A of possible outcomes
(also called an event) a nonnegative number P(A) (called the probaEvery bility
probabilistic
model
involves
underlying
process,
called
the experiof A) that
encodes
our an
knowledge
or belief
about
the collective
ment, “likelihood”
that will produce
oneofout
several
possible law
outcomes.
The set
of theexactly
elements
A. ofThe
probability
must satisfy
of all possible
outcomes to
is called
the sample
space of the experiment, and is
certain properties
be introduced
shortly.
denoted by Ω. A subset of the sample space, that is, a collection of possible
Background: Sets
Sets:
• A set is a collection of objects, which are elements of the set.
• A set can be finite, S = {1, 2, . . . , n}.
• A set can be countably infinite:
S
= {x | x = 2k + 1 or x =
2k + 1, k integer}
= {1, 1, 3, 3, 5, 5, . . . }.
• A set can be uncountable, S = {x | x 2 [0, 1]}.
• A set can be empty S = ;.
Sets: Elements & Relationships
• x 2 S - the element x is a member of the set S
• x 2
/ S - the element x is not a member of the set S
• 9x - there exists x...
• 8 - for all elements x ...
• T ✓ S - 8x 2 T , x 2 S
• T ⇢ S - 8x 2 T , x 2 S AND 9x 2 S such that x 62 T .
Sets: Combination & Manipulation
• A base set ⌦, all sets are subsets of ⌦
• Basic operations: for S, T ✓ ⌦,
• S [ T = {x | x 2 S or x 2 T }
• S \ T = {x | x 2 S and x 2 T }
• S̄ = S c = {x | x 62 S}
• De Morgan’s laws:
• (S [ T )c = S̄ \ T̄
• (S \ T )c = S̄ [ T̄
S
T
c
S
•
S
=
S̄
Ti2I i c Si2I i
•
= i2I S̄i
i2I Si
  union
  intersection
  complement
T
Venn Diagram ⌦
Visualizing Sets: Venn Diagrams
• A base set ⌦, all sets are subsets of ⌦
• Basic operations: for S, T ✓ ⌦,
• S [ T = {x | x 2 S or x 2 T }
• S \ T = {x | x 2 S and x 2 T }
• S̄ = S c = {x | x 62 S}
• De Morgan’s laws:
• (S [ T )c = S̄ \ T̄
• (S \ T )c = S̄ [ T̄
S
T
c
•
S =
S̄
Ti2I i c Si2I i
•
= i2I S̄i
i2I Si
cases B 1, ... , Bn. Figure 1 shows a subset B of the square is partitioned in three
different ways. However B is partitioned into subsets, or broken up into pieces, the
area in B is the sum of the areas of the pieces. This is the addition rule for area.
Partitions of a Set
FIGURE 1. Partitions of a set B.
B
The addition rule is satisfied by other measures of sets instead of area, for example,
length, volume, and the number or proportion of elements for finite sets.
A set B is partitionedTheinto
n subsets if:
addition rule now appears as one of the three basic rules of proportion. No
matter how probabilities are interpreted, it is generally agreed they must satisfy the
same three rules:
B1 [ B2 [ · · · [ Bn = B
Bi \ Bj = ; for any i 6= j
mutually disjoint
Pitman’s Probability, 1999
The Sample Space
⌦=
“Omega”
a set, or unordered “list”, of possible outcomes
from some random (not deterministic) experiment
The list defining the sample space must be:
  Mutually exclusive: Each experiment has a unique outcome.
  Collectively exhaustive: No matter what happens in the
experiment, the outcome is an element of the sample space.
  An art: Choosing the “right” granularity, to capture the
phenomenon of interest as simply as possible.
Modeling in science and engineering involves tradeoffs
between accuracy, simplicity, & tractability.
A Finite Sample Space
• Two rolls of a tetrahedral die
–You
Sample
space vs. sequential
description
roll a tetrahedral
(4-sided) die
2 times.
⌦ = (x, y) | x 2 {1, 2, 3, 4}, y 2 {1, 2, 3, 4}
4
1
1,1
1,2
1,3
1,4
2
Y = Second 3
roll
3
2
1
1
2
3
X = First roll
4
4
4,4
  Formally, sample space is a set of 42=16 discrete outcomes
  Can also model outcome via tree-based sequential description
uld not be unique.
A given physical situation may be modeled in several different ways, dending on the kind of questions that we are interested in. Generally, the sample
ace chosen for a probabilistic
modela must
be collectively
exhaustive, in the
You toss
(2-sided)
coin 10 times.
se that no matter what happens in the experiment, we always obtain an outTwo possible sample spaces for this experiment:
me that has been included in the sample space. In addition, the sample space
You
record
thetonumber
ofbetween
times the
coin comes
up heads:
ouldA.
have
enough
detail
distinguish
all outcomes
of interest
to the
⌦ = irrelevant
{0, 1, 2,details.
. . . , 9, 10}
deler, while
avoiding
A Finite Sample Space
B.  You record the full sequence of head-tail outcomes:
10
all 210 possible H-T sequences
⌦
=
{H,
T
}
Example 1.1. Consider two alternative games, both involving ten successive coin
tosses:
Which
is better? It depends on what you want to model:
Game 1: We receive $1 each time a head comes up.
Game 2: We receive $1 for every coin toss, up to and including the first time
a head comes up. Then, we receive $2 for every coin toss, up to the second
time a head comes up. More generally, the dollar amount per toss is doubled
each time a head comes up.
TABLE 1. Translations between events and sets. To interpret the Venn diagrams in terms
Events and Sets
of events,
imagine that a point is picked at random from the square. Each point in the square then represents
an outcome, and each region of the diagram represents the event that the point is picked from that
region.
Event language
et language
outcome space
univer al s t
event
ub et of n
et notation
Venn diagram
n
A , B, C, etc.
impo ible event
empty set
0
not A, oppo ire of A
complement of A
AC
either A or B or both
union of A and B
AUB
both A and B
inter ction of A and B
AB, AnB
A and B are mutually exclusive
A and B are disjoint
AB=0
if A then B
A is a ub et of B
It.JI
Ibbl
I cgysl
Pitman’s Probability, 1999
impo ible event
The
Axioms
of Probability
Probability
axioms
empty set
complement
of A
A, oppo airesubset
of A of the sample
•notEvent:
space
• Probability is assigned to events
union of A and B
either A or B or both
0
Probability
AC
Y=S
AUB
r
It.JI
Ibbl
  These axioms are appropriate for any finite sample
space.
I
cgysl
Infinite spaces require generalization of additivity axiom.
Axioms:
both
A and B
1. Nonnegativity:
P(A) ⌅ 0
inter ction of A and B
AB, AnB
2. Normalization: P( ) = 1
A
B are mutually
A and
B ⇧
areB)
disjoint
P(A
= P(A) + PAB=0
(B)
3.and
Additivity:
If Aexclusive
⌃ B = Ø, then
A is a ub et of B
if A then B
• P({s1, s2, . . . , sk }) = P({s1}) + · · · + P({sk })
= P(s1) + · · · + P(sk )
• Let every po
– P((X, Y ) is
– P({X = 1})
– P
(X + Y is
  The nonnegativity and additivity axioms are fundamental
to
– P(min(X, Y
our intuitive understanding of probability and uncertainty
  Unit normalization is just a convention, and other options
could also be used (e.g., probability between 0% and 100%)
• Axiom 3 needs strengthening
• Do weird sets have probabilities?
The Discrete Uniform Law
Discrete
uniform
law sampling. Formalizes the idea of
“completely
random”
• Let all outcomes be equally likely
• Then,
number of elements of A
P(A) =
total number of sample points
• Computing probabilities ⇥ counting
• Defines fair coins, fair dice, well-shu⇤ed decks
Sample space vs. sequential description
Uniform Law– for
a Finite Sample Space
You roll a tetrahedral (4-sided) die 2 times.
Probability law: Example with finite sample
space
4
1
1,1
1,2
1,3
1,4
2
Y = Second 3
4
roll
Y = Second 3
3
2
roll
2
1
1
1
2
3
4
X = First roll
• Let every possible outcome have probability 1/16
– P((X, Y ) is (1,1) or (1,2)) =
– P({X = 1}) =
– P(X + Y is odd) =
– P(min(X, Y ) = 2) =
1
2
3
X = First roll
4
4
4,4
Probability axioms
Probability
Example 1.5. Romeo and Juliet have a date at a given time, and each will arrive
the meeting
place with
delaysample
between space
0 and 1 hour, with all pairs of delays
• at
Event:
a subset
of athe
being equally likely. The first to arrive will wait for 15 minutes and will leave if the
Sec. 1.2
Probabilistic Models
• other
Probability
assigned
events
has not yetisarrived.
Whatto
is the
probability that they will meet?
Let us use as sample space the square Ω = [0, 1] × [0, 1], whose elements are
the possible pairs of delays for the two of them. Our interpretation of “equally
Axioms:
likely” pairs of delays is to let the probability of a subset of Ω be equal to its area.
probability law satisfies
the ⌅
three
1. This
Nonnegativity:
P(A)
0 probability axioms. The event that Romeo
B
A
and Juliet will meet is the shaded region in Fig. 1.5, and its probability is calculated
2. toNormalization:
P( ) = 1
be 7/16.
Sec. 1.2
Probabilistic Models
Sec.
1.2
Probabilistic
Models
3. Additivity: If A ⌃ B = Ø, then P(A ⇧ B) = P(A) + P(B)
Properties of Probability Laws
U
B
A
U
Probability laws have a number of properties, which can be deduced from the
P(s1) + below.
· · · + P(sk )
axioms. Some of them are=
summarized
B
A
A)
– P({X = 1}
A
A
B
– P(X + Y is
C
Y
– cP(min(X,
c
(a) If A ⊂ B, then P(A) ≤ P(B).
A
U
(b)
(a)
B
U
(a)
Consider a probability law, and let A, B, and C be events.
1
A
c
A B– P
A((X,
B Y ) is
• P({s1, s2, . . . , sk }) = P({s1}) + · · · + P({sk })
• Axiom 3 needs strengthening
Properties
of Probability
Laws
• Some
Do weird
sets have
probabilities?
r
Let every po
•
(a)
Properties of Probability Laws
Y=S
C
(c
A
A
c
U
B
C
C
U
U
!
c
A
laws using Venn diagrams. If A ⊂ B,
c
c C
c
c
events
A A
C A ∩ B; seeAdiagram
B and
B (a).
have
c
A B
(c) P(B) = P(A) + P
U
(d) P(A ∪ B ∪ C) = P(A) + P(Ac ∩ B) + P(Ac ∩ B c ∩ C).
BFigure 1.6: Visualization and verificat
U
(c) P(A ∪ B) ≤ P(A) + P(B).
B
U
(b) P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
certain properties to be introduced shortly.
Axioms of Probability for Infinite Spaces
Probability
Law
Event B
P(B)
Probability law: Ex. w/countably infinite sample space
Experiment
• Sample space: {1, 2, . . .}
Event A
n
– We are given P(n) = 2 , n = 1, 2, . . Sample
.
Space
(Set
of
Outcomes)
– Find P(outcome Probability
is even)
axioms
P(A)
A
B
Probability
Events
p
• Event: a subset of the sample space
1/2
• Probability is assigned
to The
events
Figure 1.2:
main ingredients of a probabilistic model.
1/4
…..
Axioms:
1
3
4
SamplePSpaces
and
Events
1. Nonnegativity:
(A)
⌅2 0
1
1
1
1
P
({2,
4,
6,
.
.
.})
=
P
(2)
+
P
(4)
+
·
·
·
=
+
+
+
·
·
·
=
2. Normalization: P( ) = 1
2
4
6
1/8
• Turn in recitation/tu
Y = Sw
• Tutorials start next
r
1/16
2
3
2an underlying
2
Every probabilistic model involves
process,
called the experiment,
produce
several
possible
(A ⇧one
B)out
= of
P(A)
+P
(B) outcomes. The set
3.
If A that
⌃ B will
=axiom:
Ø,
thenexactly
3.• Additivity:
Countable
additivity
Pfor
Countable additivity
axiom
(needed
this
calculation):
of all possible outcomes is called the sample space of the experiment, and is
If A1, A2, . . .denoted
are disjoint
then:
by Ω. events,
A subset
of the sample space, that is, a collection of possible
⇧A
· · · )1=
+P
P({s
(A2k)})
+ ···
, sk1})
=2 ⇧
P({s
}) P
+(A
· ·1·)+
• P({s1, s2, . .P. (A
= P(s1) + · · · + P(sk )
• Let every po
– P((X, Y ) is
– P({X = 1})
uniform law
y likely
Continuous uniform law
• Two
“random” numbers in
[0, 1].
An (Infinite) Continuous
Probability
Model
y
You throw
a dart
at a Continuous
1-meter square
region.
Sample
space:
example
1
er of elements of A
mber of sample points
= {(x, y) | 0 ⇤ x, y ⇤ 1}
counting
1
y
• Uniform law: Probability = Area
well-shu⇤ed decks
1
– P(X + Y ⇤ 1/2) = ?
– P( (X, Y ) = (0.5, 0.3) )
1
x
x
Conditional
Probabilities
and
• Divide and conquer
Bayes’ Rule
• Partition of sample space into A , A , A
Total probability theorem
1
• Have P(B | Ai), for every i
A1
2
• “Prior”
– initial “
3
• We know
• Wish to
– revise “
B
A2
A3
• One way of computing P(B):
Some figures and materials courtesy
Tsitsiklis,
P(B) =Bertsekas
P(A &)P
(B | A )Introduction to Probability, 2008
1
1
P(A ∪ C) ≤ P(A) + P(C) translates to the new fact
roll
P(A ∪ C | B) ≤ P(A | B) + P(C | B).
Conditional Probability
Let us summarize the conclusions reached so far.
• P(A |ofB)
= probability
Properties
Conditional
Probabilityof
A,
given that
B occurred
probability
of an event
A, given an
• The conditional
P(B) > 0, is defined by
event B with
– B is our new universe
P(A | B) =
P(A ∩ B)
,
P(B)
• Definition: Assuming P(B) =
⇥ 0,
and specifies a new (conditional) probability law on the same sample
P(A ⌅ofB)
space Ω. In particular, all known properties
probability laws remain
P
(A
|
B)
=
valid for conditional probability laws.
P(B)
A
• Let B b
B•
Let M =
• Conditional probabilities can also be viewed as a probability law on a
P(A
| B)B,undefined
if conditional
P(B) =probability
0
new
universe
because all of the
is concentrated on B.
• P⌦
(M =
• In the case where the possible outcomes are finitely many and equally
likely, we have
• P(M =
P(A | B) =
number of elements of A ∩ B
.
number of elements of B
– Sample space vs. sequential description
Conditional
Probability Example
You roll a tetrahedral (4-sided) die 2 times.
Die roll example
Probability
law: Example with finite sample space
4
4
4
Y = Second 3
roll
roll
roll
1
1
3
2
2
1
2
Y = Second 3
Y = Second 3
2
1
1,1
1,2
1,3
1,4
1
2
1
3
2
3
4
4
X = First roll
X = First roll
• Let every possible outcome have probability 1/16
1
2
3
X = First roll
4
4
4,4
P((X,
Y )the
is (1,1)
(1,2)) Y=) = 2
• – Let
B be
event:ormin(X,
P({M
X=
=max(X,
1}) = Y )
• – Let
– P(X + Y is odd) =
• P(M = 1 | B) =
– P(min(X, Y ) = 2) =
• P(M = 2 | B) =
Note that conditional probabilities of
M given B satisfy axioms, i.e. they
are non-negative and sum to one.
P(A)=0.05
– B is our new universe
Models Based on Conditional
Probabilities
Multiplication
rule
• Definition: Assuming
P(B)
=
⇥ 0,
Models based on conditional
probabilities
Multiplication rule
P(Ac)=0.95
P(A | B) =
P(A
P(C | A B)
A ⌅BB) =
U
P(B) =
U
Bc
A Bc C
U
false alarm
U
P(B) =
A Bc C
U
A
P(Ac)=0.95
P(A ⌅ B) =
c | A)
P(B
P(A
| B) =
U
P(A)
P(Bc | Ac)=0.90
A B C
P(B | A)
missed detection
A
P(B | Ac)=0.10
U
P(A)=0.05
P(Bc |
P(A | B) undefined if P(B) = 0
P(B | A)=0.99
P(Bc | A)=0.01
P(B)
U U
• Event A: Airplane is flying above
Event B: Something registers on radar
screen
P
B)
(A∩⌅B
B∩
⌅ C)
·P
P(A
· P(C
(C ||P(B
A∩
⌅| B
A
PP
(A
C) =
= PP(A)
(A)
(B |⌅A)
A)
P
A
P(Ac)
Ac
Total probability th
• Divide and conquer
(B c
|Ac
Multiplication Rule
)=
0.9
0
Multiplication
rule
Multiplication rule
(A∩⌅B
B∩
⌅ C)
P(B | A)
· P(C
(C ||A
A∩
⌅ B)
PP
(A
C) =
= PP(A)
(A)·P
A)P
B)
U
P(C | A B)
U U
U
A B
A B C
Figure 1.8: Sequential description of the sample space for the radar detection
problem
in Example 1.9
.
In mathematical terms, we are dealing with an event A which occurs if and
only if each one of several events A1 , . . . , An has occurred, i.e., A = A1 ∩ A2 ∩
· · · ∩ An . The occurrence of A is viewed as an occurrence of A1 , followed by
the occurrence of A2 , then of A3 , etc, and it is visualized as a path on the tree
with n branches, corresponding to the events A1 , . . . , An . The probability of A
is given by the following rule (see also Fig. 1.9).
P(B | A)
Multiplication Rule
P(Bc | A)
Assuming that all of the conditioning events have positive probability, we
have
A Bc C
U
P(A)
U
A
U
A
Bc
A Bc Cc
!
"
"
!
P ∩ni=1 Ai = P(A1 )P(A2 | A1 )P(A3 | A1 ∩ A2 ) · · · P An | ∩n−1
i=1 Ai .
U
U
The multiplication rule can be verified by writing
P(Ac)
P
Ac
!
∩ni=1
Ai
"
!
"
P ∩ni=1 Ai
P(A1 ∩ A2 ) P(A1 ∩ A2 ∩ A3 )
= P(A1 )
· · · ! n−1 " ,
P(A1 )
P(A1 ∩ A2 )
P ∩i=1 Ai
and by using the definition of conditional probability to rewrite the right-hand
side above as
"
!
P(A1 )P(A2 | A1 )P(A3 | A1 ∩ A2 ) · · · P An | ∩n−1
i=1 Ai .
Total probability theorem
Total Probability
Theorem
• Divide and conquer
• Partition of sample space into A1, A2, A3
• Have P(B | Ai), for every i
A1
A2
• “Prior”
– initial
• We kno
• Wish to
– revise
B
A3
• One way of computing P(B):
P(B) =
P(A1)P(B | A1)
+ P(A2)P(B | A2)
+ P(A3)P(B | A3)
P(Ai |
Bayes’ Rule
mputing P(B):
P(A1)P(B | A1)Bayes’ rule
+ P(A2)P(B | A2)
• “Prior” probabilities P(Ai)
+ P(A
)P(B | A3)
– 3 initial “beliefs”
• Wish to compute P(Ai | B)
– revise “beliefs”, given that B occurred
=
=
B
A2
P(Ai | B) =
A3
P(Ai | B) =
• We know P(B | Ai) for each i
A1
A2
A3
P(Ai ⌅ B)
2
P(Ai ⌅ B)
P(B)
P(Ai)P(B | Ai)
P(B)
P(Ai)P(B | Ai)
j P(Aj )P(B | Aj )
Counterintuitive?
Reasoning
Under Uncertainty
Counterintuitive?
1/1000 of
of tourists
touristswho
whovisit
visittropical
tropicalcountry
country
return
with
•• 1/1000
XX
return
with
a a
dangerous
virus
. visit tropical country X return with a
dangerous
YY.who
• 1/1000 ofvirus
tourists
dangerous
virusto
Y .check
• There
There
test
to
checkfor
forthe
thevirus.
virus.The
The
test
false
isis aa test
test
hashas
5%5%
false
• There israte
a test
to no
check
fornegative
the
virus. error.
The
test has 5% false
positive
and
positive
rate
and
nofalse
false
negative
error.
positive rate and no false negative error.
• You
and
it was
You returned
returnedfrom
fromcountry
countryXX, took
, tookthe
thetest,
test,
and
it was
• You returned from country X , took the test, and it was
positive.
Should you
take
the painful
treatment
for thethe
virus?
positive.
youtake
take
painful
treatment
positive. Should
Should you
thethe
painful
treatment
for the for
virus? virus?
• A
virus.BB- positive
- positive
in test.
the test.
• A- -has
has the
the virus.
in the
1 1
10001000
1 1
999 999
5
1000 + 1000 100
Pr(A
(A || B)
Pr
B)==
1000
+
20 20
= ⇡ 2% ⇡ 2%
5 1019 1019
=
1000 100
Explanation: Out of 1000 tourist, 1 will have the virus and
Explanation:
of 1000
tourist,
1 will have the virus and
another 50 willOut
be false
positive
in the test.
another 50 will be false positive in the test.