Lesson4 Probability of an event [Autosaved].pdf
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About This Presentation
esson4 Probability of an event [Autosaved].pdf
Slide Content
ةعماج فيوس ينب
Probability and Statistics for Engineers
Probability and Statistics for Engineers
Probability
Chapter 2: Lesson 2
Chapter 2: Lesson 2
2.4.ProbabilityofanEvent:Textbookpage48
·Toeverypoint(outcome)inthesample
spaceofanexperimentS,weassigna
weight(orprobability),rangingfrom0to1,
suchthatthesumofallweights
(probabilities)equals1.
·Theweight(orprobability)ofanoutcome
measuresitslikelihood(chance)of
occurrence.
·TofindtheprobabilityofaneventA,we
sumallprobabilitiesofthesamplepoints
inA.Thissumiscalledtheprobabilityof
theeventAandisdenotedbyP(A).
·Toeverypoint(outcome)inthesample
spaceofanexperimentS,weassigna
weight(orprobability),rangingfrom0to1,
suchthatthesumofallweights
(probabilities)equals1.
·Theweight(orprobability)ofanoutcome
measuresitslikelihood(chance)of
occurrence.
·TofindtheprobabilityofaneventA,we
sumallprobabilitiesofthesamplepoints
inA.Thissumiscalledtheprobabilityof
theeventAandisdenotedbyP(A).
Definition2.8:
TheprobabilityofaneventAisthesumof
theweights(probabilities)ofallsample
pointsinA.Therefore,()10 AP
1)()1=SP
2)()0=P
3)
TheprobabilityofaneventAisthesumof
theweights(probabilities)ofallsample
pointsinA.Therefore,()10 AP
1)()1=SP
2)()0=P
3)
Example2.22:
Abalancedcoinistossedtwice.Whatisthe
probabilitythatatleastoneheadoccurs?
Solution:
S={HH,HT,TH,TT}
A={atleastoneheadoccurs}={HH,HT,TH}
Sincethecoinisbalanced,theoutcomesare
equallylikely;i.e.,alloutcomeshavethe
sameweightorprobability.
Abalancedcoinistossedtwice.Whatisthe
probabilitythatatleastoneheadoccurs?
Solution:
S={HH,HT,TH,TT}
A={atleastoneheadoccurs}={HH,HT,TH}
Sincethecoinisbalanced,theoutcomesare
equallylikely;i.e.,alloutcomeshavethe
sameweightorprobability.
Outcome Weight
(Probability)
4w=1w=1/4=0.25
P(HH)=P(HT)=P(TH)=P(TT)=0.25HH
HT
TH
TT
P(HH) = w
P(HT) = w
P(TH) = w
P(TT) = w
sum 4w=1
(Probability)
4w=1w=1/4=0.25
P(HH)=P(HT)=P(TH)=P(TT)=0.25HH
HT
TH
TT
P(HH) = w
P(HT) = w
P(TH) = w
P(TT) = w
sum 4w=1
Theprobabilitythatatleastonehead
occursis:
P(A)=P({atleastoneheadoccurs})=P({HH,
HT,TH})
=P(HH)+P(HT)+P(TH)
=0.25+0.25+0.25
=0.75
occursis:
P(A)=P({atleastoneheadoccurs})=P({HH,
HT,TH})
=P(HH)+P(HT)+P(TH)
=0.25+0.25+0.25
=0.75
Sinoutcomesofno
Ainoutcomesofno
N
An
Sn
An
AP
.
.)(
)(
)(
)( === Theorem2.9:
Ifanexperimenthasn(S)=Nequallylikely
differentoutcomes,thentheprobabilityof
theeventAis:
Ainoutcomesofno
N
An
Sn
An
AP
.
.)(
)(
)(
)( === Theorem2.9:
Ifanexperimenthasn(S)=Nequallylikely
differentoutcomes,thentheprobabilityof
theeventAis:
Example2.25:
Amixtureofcandiesconsistsof6mints,4
toffees,and3chocolates.Ifapersonmakes
arandomselectionofoneofthesecandies,
findtheprobabilityofgetting:
(a)amint
(b)atoffeeorchocolate.
Solution:
Definethefollowingevents:
M={gettingamint}
T={gettingatoffee}
C={gettingachocolate}
Amixtureofcandiesconsistsof6mints,4
toffees,and3chocolates.Ifapersonmakes
arandomselectionofoneofthesecandies,
findtheprobabilityofgetting:
(a)amint
(b)atoffeeorchocolate.
Solution:
Definethefollowingevents:
M={gettingamint}
T={gettingatoffee}
C={gettingachocolate}
Experiment:selectingacandyatrandom
from13candies
n(S)=no.ofoutcomesoftheexperimentof
selectingacandy.
=no.ofdifferentwaysofselectinga
candyfrom13candies.13
1
13
=
=
from13candies
n(S)=no.ofoutcomesoftheexperimentof
selectingacandy.
=no.ofdifferentwaysofselectinga
candyfrom13candies.13
1
13
=
=
Theoutcomesoftheexperimentare
equallylikelybecausetheselectionismade
atrandom.
(a)M={gettingamint}
n(M)=no.ofdifferentwaysofselecting
amintcandyfrom6mintcandies6
1
6
=
=
P(M)=P({gettingamint})=()
()13
6
=
Sn
Mn
equallylikelybecausetheselectionismade
atrandom.
(a)M={gettingamint}
n(M)=no.ofdifferentwaysofselecting
amintcandyfrom6mintcandies6
1
6
=
=
P(M)=P({gettingamint})=()
()13
6
=
Sn
Mn
+
=
1
4 734
1
3
=+=
(b)TC={gettingatoffeeorchocolate}
n(TC)=no.ofdifferentwaysof
selectingatoffeeorachocolate
candy
=no.ofdifferentwaysofselecting
atoffeecandy+no.ofdifferent
waysofselectingchocolate
candy
=
1
4 734
1
3
=+=
(b)TC={gettingatoffeeorchocolate}
n(TC)=no.ofdifferentwaysof
selectingatoffeeorachocolate
candy
=no.ofdifferentwaysofselecting
atoffeecandy+no.ofdifferent
waysofselectingchocolate
candy
=no.ofdifferentwaysof
selectingacandyfrom7candies7
1
7
=
=
P(TC)=P({gettingatoffeeorchocolate})
=( )
()13
7
=
Sn
CTn
selectingacandyfrom7candies7
1
7
=
=
P(TC)=P({gettingatoffeeorchocolate})
=( )
()13
7
=
Sn
CTn
Example2.26:
In a poker hand consisting of 5cards, find
the probability of holding 2aces and 3jacks.
In a poker hand consisting of 5cards, find
the probability of holding 2aces and 3jacks.
Example2.26:
In a poker hand consisting of 5cards, find
the probability of holding 2aces and 3jacks.
Solution:
Experiment:selecting5cardsfrom52cards.
n(S)=no.ofoutcomesoftheexperimentof
selecting5cardsfrom52cards.2598960
!47!5
!52
5
52
=
=
=
In a poker hand consisting of 5cards, find
the probability of holding 2aces and 3jacks.
Solution:
Experiment:selecting5cardsfrom52cards.
n(S)=no.ofoutcomesoftheexperimentof
selecting5cardsfrom52cards.2598960
!47!5
!52
5
52
=
=
=
Theoutcomesoftheexperimentareequally
likelybecausetheselectionismadeat
random.
DefinetheeventA={holding2acesand3
jacks}
n(A)=no.ofwaysofselecting2acesand3
jacks
=(no.ofwaysofselecting2aces)(no.
ofwaysofselecting3jacks)
=(no.ofwaysofselecting2acesfrom4
aces)(no.ofwaysofselecting3
jacksfrom4jacks)
likelybecausetheselectionismadeat
random.
DefinetheeventA={holding2acesand3
jacks}
n(A)=no.ofwaysofselecting2acesand3
jacks
=(no.ofwaysofselecting2aces)(no.
ofwaysofselecting3jacks)
=(no.ofwaysofselecting2acesfrom4
aces)(no.ofwaysofselecting3
jacksfrom4jacks)
=
2
4
3
4 !2!2
!4
=
2446
!1!3
!4
==
P(A )= P({holding 2 aces and 3 jacks })()
()
000009.0
2598960
24
===
Sn
An
=
2
4
3
4 !2!2
!4
=
2446
!1!3
!4
==
P(A )= P({holding 2 aces and 3 jacks })()
()
000009.0
2598960
24
===
Sn
An
Additive Rule
2.5AdditiveRules:
Theorem2.10:
IfAandBareanytwoevents,then:
P(AB)= P(A) + P(B) −P(AB)
Corollary1:
IfAandBaremutuallyexclusive(disjoint)
events,then:
P(AB)= P(A) + P(B)
Theorem2.10:
IfAandBareanytwoevents,then:
P(AB)= P(A) + P(B) −P(AB)
Corollary1:
IfAandBaremutuallyexclusive(disjoint)
events,then:
P(AB)= P(A) + P(B)
Corollary2:
IfA
1
,A
2
,…,A
n
arenmutuallyexclusive
(disjoint)events,then:
P(A
1
A
2
… A
n
)= P(A
1
) + P(A
2
) +… + P(A
n
)()
==
=
n
i
i
n
i
i APAP
11
)(
IfA
1
,A
2
,…,A
n
arenmutuallyexclusive
(disjoint)events,then:
P(A
1
A
2
… A
n
)= P(A
1
) + P(A
2
) +… + P(A
n
)()
==
=
n
i
i
n
i
i APAP
11
)(
Corollary 3:
If A
1, A
2, …, A
nis a partition of sample space
S, then
P(A
1A
2….A
n) =
P(A
1) + P(A
2) …+ P(A
n) = P(S) = 1.
If A
1, A
2, …, A
nis a partition of sample space
S, then
P(A
1A
2….A
n) =
P(A
1) + P(A
2) …+ P(A
n) = P(S) = 1.
Note: Two event Problems:
Total area= P(S)=1* In Venn diagrams,
considerthe probability of an event A as
theareaoftheregioncorrespondingtothe
eventA.
* Total area= P(S)=1
Total area= P(S)=1
Total area= P(S)=1* In Venn diagrams,
considerthe probability of an event A as
theareaoftheregioncorrespondingtothe
eventA.
* Total area= P(S)=1
Total area= P(S)=1
*Examples:
P(A)= P(AB)+ P(AB
C
)
P(AB)= P(A) + P(A
C
B)
P(AB)= P(A) + P(B) −P(AB)
P(AB
C
)= P(A) −P(AB)
P(A
C
B
C
)=1−P(AB)
etc.,
P(A)= P(AB)+ P(AB
C
)
P(AB)= P(A) + P(A
C
B)
P(AB)= P(A) + P(B) −P(AB)
P(AB
C
)= P(A) −P(AB)
P(A
C
B
C
)=1−P(AB)
etc.,
Example2.27:
TheprobabilitythatPaulapasses
Mathematicsis2/3,andtheprobabilitythat
shepassesEnglishis4/9.Iftheprobability
thatshepassesbothcoursesis1/4,whatis
theprobabilitythatshewill:
(a)passatleastonecourse?
(b)passMathematicsandfailEnglish?
(c)failbothcourses?
TheprobabilitythatPaulapasses
Mathematicsis2/3,andtheprobabilitythat
shepassesEnglishis4/9.Iftheprobability
thatshepassesbothcoursesis1/4,whatis
theprobabilitythatshewill:
(a)passatleastonecourse?
(b)passMathematicsandfailEnglish?
(c)failbothcourses?
Solution:
Define the events:
M={Paula passes Mathematics}
E={Paula passes English}
We know that P(M)=2/3, P(E)=4/9, and
P(ME)=1/4.
(a)Probability of passing at least one
course is:
P(ME)= P(M) + P(E) −P(ME)36
31
4
1
9
4
3
2
=−+=
Define the events:
M={Paula passes Mathematics}
E={Paula passes English}
We know that P(M)=2/3, P(E)=4/9, and
P(ME)=1/4.
(a)Probability of passing at least one
course is:
P(ME)= P(M) + P(E) −P(ME)36
31
4
1
9
4
3
2
=−+=
(b)ProbabilityofpassingMathematicsand
failingEnglishis:
P(ME
C
)=P(M)−P(ME)12
5
4
1
3
2
=−=
(c)Probabilityoffailingbothcoursesis:
P(M
C
E
C
)=1−P(ME)36
5
36
31
1 =−=
failingEnglishis:
P(ME
C
)=P(M)−P(ME)12
5
4
1
3
2
=−=
(c)Probabilityoffailingbothcoursesis:
P(M
C
E
C
)=1−P(ME)36
5
36
31
1 =−=
Theorem2.12:
IfAandA
C
arecomplementaryevents,
then:
P(A) + P(A
C
) = 1 P(A
C
) = 1 −P(A)
Proof: Since A U A
C
= S and the sets A and A
C
are disjoint, then
1 = P(S) = P(A U A
C
) = P(A) + P(A
C
).
IfAandA
C
arecomplementaryevents,
then:
P(A) + P(A
C
) = 1 P(A
C
) = 1 −P(A)
Proof: Since A U A
C
= S and the sets A and A
C
are disjoint, then
1 = P(S) = P(A U A
C
) = P(A) + P(A
C
).
Example2.31Iftheprobabilitiesthatan
automobilemechanicwillservice3,4,5,6,7,or
8ormorecarsonanygivenworkdayare,
respectively,0.12,0.19,0.28,0.24,0.10,and
0.07,whatistheprobabilitythathewillservice
atleast5carsonhisnextdayatwork?
automobilemechanicwillservice3,4,5,6,7,or
8ormorecarsonanygivenworkdayare,
respectively,0.12,0.19,0.28,0.24,0.10,and
0.07,whatistheprobabilitythathewillservice
atleast5carsonhisnextdayatwork?
Solution: Let Ebe the event that at least 5 cars
are serviced. Now, P(E) = 1 —P(E
c
),
where E
c
is the event that fewer than 5 cars are
serviced. Since
P(E
c
) = 0.12+ 0.19 = 0.31,
it follows from Theorem 2.12 that
P(E) = 1 -0.31 = 0.69.
are serviced. Now, P(E) = 1 —P(E
c
),
where E
c
is the event that fewer than 5 cars are
serviced. Since
P(E
c
) = 0.12+ 0.19 = 0.31,
it follows from Theorem 2.12 that
P(E) = 1 -0.31 = 0.69.
Example2.32Supposethemanufacturer
specificationsofthelengthofacertaintypeof
computercableare2000±10millimeters.In
thisindustry,itisknownthatsmallcableisjust
aslikelytobedefective(notmeeting
specifications)aslargecable.Thatis,the
probabilityofrandomlyproducingacablewith
lengthexceeding2010millimeters
specificationsofthelengthofacertaintypeof
computercableare2000±10millimeters.In
thisindustry,itisknownthatsmallcableisjust
aslikelytobedefective(notmeeting
specifications)aslargecable.Thatis,the
probabilityofrandomlyproducingacablewith
lengthexceeding2010millimeters
isequaltotheprobabilityofproducingacable
withlengthsmallerthan1990millimeters.The
probabilitythattheproductionproceduremeets
specificationsisknowntobe0.99.
(a)Whatistheprobabilitythatacableselected
randomlyistoolarge?
(b)Whatistheprobabilitythatarandomly
selectedcableislargerthan1990
millimeters?
withlengthsmallerthan1990millimeters.The
probabilitythattheproductionproceduremeets
specificationsisknowntobe0.99.
(a)Whatistheprobabilitythatacableselected
randomlyistoolarge?
(b)Whatistheprobabilitythatarandomly
selectedcableislargerthan1990
millimeters?
Solution:LetMbetheeventthatacablemeets
specifications.LetSandLbetheevents
thatthecableistoosmallandtoolarge,
respectively.Then
(a)P(M)=0.99andP(S)=P(L)= =0.005.
(b)DenotingbyXthelengthofarandomly
selectedcable,wehave2
99.01−
specifications.LetSandLbetheevents
thatthecableistoosmallandtoolarge,
respectively.Then
(a)P(M)=0.99andP(S)=P(L)= =0.005.
(b)DenotingbyXthelengthofarandomly
selectedcable,wehave2
99.01−
P(1990<X<2010)=P(M)=0.99.
SinceP(X>2010)=P(L)=0.005
then
P(X>1990)=P(M)+P(L)=0.995.
ThisalsocanbesolvedbyusingTheorem2.12:
P(X>1990)+P(X<1990)=1.
Thus,P(X>1990)=1–P(S)
=1-0.005=0.995.
SinceP(X>2010)=P(L)=0.005
then
P(X>1990)=P(M)+P(L)=0.995.
ThisalsocanbesolvedbyusingTheorem2.12:
P(X>1990)+P(X<1990)=1.
Thus,P(X>1990)=1–P(S)
=1-0.005=0.995.
Exercise
Exercise
Exercise
Exercise
Conditional
Probability
Probability
The probability of an event Boccurring
when it is known that some event Ahas
occurred is called a conditional
probabilityand is denoted by P(B|A). The
symbol P (B | A) is usually read "the
probability that Boccurs given that A
occurs"
or simply "the probability of B, given A."
when it is known that some event Ahas
occurred is called a conditional
probabilityand is denoted by P(B|A). The
symbol P (B | A) is usually read "the
probability that Boccurs given that A
occurs"
or simply "the probability of B, given A."
Definition 2.9: The conditional probability
of B, given A, denoted by P(B|A)is defined
by
??????(�|�)=
)??????(�∩�
)??????(�
provided P(A) > 0.
of B, given A, denoted by P(B|A)is defined
by
??????(�|�)=
)??????(�∩�
)??????(�
provided P(A) > 0.
Example:
Suppose that our sample space Sis
the population of adults in a small
town who have completed the
requirements for a college degree.
We shall categorize them according
to gender and employment status.
The data are given in the following
table.
Suppose that our sample space Sis
the population of adults in a small
town who have completed the
requirements for a college degree.
We shall categorize them according
to gender and employment status.
The data are given in the following
table.
Employed Unemployed Total
Male 460 40 500
Female 140 260 400
Total 600 300 900
Male 460 40 500
Female 140 260 400
Total 600 300 900
One of these individuals is to be selected
at random for a tour throughout the
country to publicize the advantages of
establishing new industries in the town.
We shall be concerned with the
following events:
M: a man is chosen,
E: the one chosen is employed.
What is ????????????|�?
at random for a tour throughout the
country to publicize the advantages of
establishing new industries in the town.
We shall be concerned with the
following events:
M: a man is chosen,
E: the one chosen is employed.
What is ????????????|�?
Solution:
Let )??????(�denote the number of elements in any
set A, then we write
??????(??????|�)=
)??????(??????∩�
)??????(�
=
)??????(??????∩�
)??????(??????
)??????(�
)??????(??????
=
)??????(??????∩�
)??????(�
where )??????(??????∩�and )??????(�are found from the
original sample space S. To verify this result,
note that ??????(�)=
���
���
=
�
�
,??????(??????∩�)=
���
���
=
��
��
Hence ??????(??????|�)=
ൗ
��
��
ൗ
�
�
=
��
��
Let )??????(�denote the number of elements in any
set A, then we write
??????(??????|�)=
)??????(??????∩�
)??????(�
=
)??????(??????∩�
)??????(??????
)??????(�
)??????(??????
=
)??????(??????∩�
)??????(�
where )??????(??????∩�and )??????(�are found from the
original sample space S. To verify this result,
note that ??????(�)=
���
���
=
�
�
,??????(??????∩�)=
���
���
=
��
��
Hence ??????(??????|�)=
ൗ
��
��
ൗ
�
�
=
��
��
The probability that a regularly scheduled
flight departs on time is ??????�=�.��;
the probability that it arrives on time is ??????�
=�.�2; and the probability that it
departs and arrives on time is ??????�∩�
=�.��. Find the probability that a plane
(a)arrives on time given that it departed on
time,
(b)departed on time given that it has arrived
on time.
flight departs on time is ??????�=�.��;
the probability that it arrives on time is ??????�
=�.�2; and the probability that it
departs and arrives on time is ??????�∩�
=�.��. Find the probability that a plane
(a)arrives on time given that it departed on
time,
(b)departed on time given that it has arrived
on time.
(a)The probability that a plane arrives
on time given that it departed on
time is
??????�|�=
??????�∩�
??????�
=
�.��
�.��
=�.��
(b) The probability that a plane
departed on time given that it has
arrived on time is
??????�|�=
??????�∩�
??????�
=
�.��
�.��
=�.��
on time given that it departed on
time is
??????�|�=
??????�∩�
??????�
=
�.��
�.��
=�.��
(b) The probability that a plane
departed on time given that it has
arrived on time is
??????�|�=
??????�∩�
??????�
=
�.��
�.��
=�.��
Independent Events :
Although conditional probability allows for an
alteration of the probability of an event in the light
of additional material, it also enables us to
understand better the very important concept of
independence or, in the present context,
independent events. Consider the situation where
we have events �and Band ??????�|�=??????�
That is, the occurrence of Bhad no impact on the
odds of occurrence of A.
Although conditional probability allows for an
alteration of the probability of an event in the light
of additional material, it also enables us to
understand better the very important concept of
independence or, in the present context,
independent events. Consider the situation where
we have events �and Band ??????�|�=??????�
That is, the occurrence of Bhad no impact on the
odds of occurrence of A.
Definition 2.10:
Two events Aand Bare independent if
and only if
??????�|�=??????�or ??????�|�=??????�,
provided the existences of the
conditional probabilities. Otherwise A
and B are dependent.
Two events Aand Bare independent if
and only if
??????�|�=??????�or ??????�|�=??????�,
provided the existences of the
conditional probabilities. Otherwise A
and B are dependent.
Example:
Consider an experiment in which 2cards are
drawn in succession from an ordinary deck,
with replacement. The events are defined as
A: the first card is an ace,
B: the second card is a spade.
Find ??????�|�,??????�|�.
Consider an experiment in which 2cards are
drawn in succession from an ordinary deck,
with replacement. The events are defined as
A: the first card is an ace,
B: the second card is a spade.
Find ??????�|�,??????�|�.
Solution:
Since the first card is replaced, our
sample space for both the first and
second draws consists of 52cards,
containing 4 aces and 13 spades.
Hence
??????�|�=
��
��
=
�
�
, ??????�=
��
��
then ??????�|�=??????�=
�
�
and ??????�|�=
�
��
=
�
��
,??????�=
�
��
then ??????�|�=??????�=
�
��
Since the first card is replaced, our
sample space for both the first and
second draws consists of 52cards,
containing 4 aces and 13 spades.
Hence
??????�|�=
��
��
=
�
�
, ??????�=
��
��
then ??????�|�=??????�=
�
�
and ??????�|�=
�
��
=
�
��
,??????�=
�
��
then ??????�|�=??????�=
�
��
Multiplicative Rules:
Multiplying the formula of
Definition 2.9 by P(A), we obtain
the following important
multiplicative rule, which enables
us to calculate the probability that
two events will both occur.
Multiplying the formula of
Definition 2.9 by P(A), we obtain
the following important
multiplicative rule, which enables
us to calculate the probability that
two events will both occur.
Theorem 2.13:
If in an experiment the events A
and Bcan both occur, then
??????�∩�=??????�??????�|�
provided P(A) > 0.
If in an experiment the events A
and Bcan both occur, then
??????�∩�=??????�??????�|�
provided P(A) > 0.
Example:
Suppose that we have a fuse box
containing 20 fuses, of which 5 are
defective. If 2 fusesare selected at
random and removed from the box
in succession without replacing the
first, what is the probability that
both fuses are defective?
Suppose that we have a fuse box
containing 20 fuses, of which 5 are
defective. If 2 fusesare selected at
random and removed from the box
in succession without replacing the
first, what is the probability that
both fuses are defective?
Solution:
Let Abe the event that the first fuse is
defective and Bthe event that the second
fuse is defective; then we interpret ??????(
)
�
∩�as the event that Aoccurs, and then
Boccurs after Ahas occurred. The
probability of first removing a defective
fuse is
�
�
; then the probability of
removing a second defective fuse from
the remaining 4is
�
��
. Hence
??????�∩�=
�
�
�
��
=
�
��
Let Abe the event that the first fuse is
defective and Bthe event that the second
fuse is defective; then we interpret ??????(
)
�
∩�as the event that Aoccurs, and then
Boccurs after Ahas occurred. The
probability of first removing a defective
fuse is
�
�
; then the probability of
removing a second defective fuse from
the remaining 4is
�
��
. Hence
??????�∩�=
�
�
�
��
=
�
��
Theorem 2.14:
Two events Aand Bare independent if and only
if
??????�∩�=??????�??????�
Therefore, to obtain the probability that two
independent events will both occur, we simply
find the product of their individual probabilities.
Two events Aand Bare independent if and only
if
??????�∩�=??????�??????�
Therefore, to obtain the probability that two
independent events will both occur, we simply
find the product of their individual probabilities.
Example:
An electrical system consists of four
components as illustrated in Figure 1 below.
The system works if components A and Bwork
and either of the components Cor Dwork. The
reliability (probability of working) of each
component is also shown in Figure 1. Find the
probability that
(a)the entire system works, and
(b)the component Cdoes not work, given that
the entire system works. Assume that four
components work independently.
An electrical system consists of four
components as illustrated in Figure 1 below.
The system works if components A and Bwork
and either of the components Cor Dwork. The
reliability (probability of working) of each
component is also shown in Figure 1. Find the
probability that
(a)the entire system works, and
(b)the component Cdoes not work, given that
the entire system works. Assume that four
components work independently.
Figure 1: An electrical system
0.9
0.8
0.9
0.8
0.9
0.8
0.9
0.8
Solution:
In this configuration of the system, A, B, and
the subsystem Cand Dconstitute
a serial circuit system, whereas the subsystem
Cand Ditself is a parallel circuit system.
(a) Clearly the probability that the entire
system works can be calculated as the
following:
In this configuration of the system, A, B, and
the subsystem Cand Dconstitute
a serial circuit system, whereas the subsystem
Cand Ditself is a parallel circuit system.
(a) Clearly the probability that the entire
system works can be calculated as the
following:
??????�∩�∩�∪�=??????�??????�??????�∪�
=??????�??????��−??????�
′
∩�
′
=??????�??????��−??????�
′
??????�
′
=(�.�)(�.�) )�−(�−(�.�))(�−(�.�)=�.����
The equalities above hold because of the
independence among the four components.
=??????�??????��−??????�
′
∩�
′
=??????�??????��−??????�
′
??????�
′
=(�.�)(�.�) )�−(�−(�.�))(�−(�.�)=�.����
The equalities above hold because of the
independence among the four components.
Conditional
Probability -Bayes’
Rule
Probability -Bayes’
Rule
Theorem 2.15:
If, in an experiment, the events �
�,�
�,…,�
??????
can occur, then
If the events �
�,�
�,…,�
??????are independent, then
??????�
�∩⋯∩�
??????=??????�
�??????�
�…??????�
??????
If, in an experiment, the events �
�,�
�,…,�
??????
can occur, then
If the events �
�,�
�,…,�
??????are independent, then
??????�
�∩⋯∩�
??????=??????�
�??????�
�…??????�
??????
Example:
Three cards are drawn in succession,
without replacement, from an ordinary
deck of playing cards. Find the
probability that the event ??????(
)
�
�∩�
�
∩�
�occurs, where
�
�is the event that the first card is a red
ace, �
�is the event that the second card
is a 10 or a jack, and �
�is the event that
the third card is greater than 3 but less
than 7.
Three cards are drawn in succession,
without replacement, from an ordinary
deck of playing cards. Find the
probability that the event ??????(
)
�
�∩�
�
∩�
�occurs, where
�
�is the event that the first card is a red
ace, �
�is the event that the second card
is a 10 or a jack, and �
�is the event that
the third card is greater than 3 but less
than 7.
Solution: First we define the events
�
�:the first card is a red ace,
�
�:the. second card is a 10 or a jack,
�
�:the third card is greater than 3 but less than
7. Now
??????�
�=
�
��
,??????�
�|�
�=
�
��
,??????�
�|�
�∩�
�
=
��
��
and hence, by Theorem 2.15,
??????�
�∩�
�∩�
�=??????�
�??????�
�|�
�??????�
�|�
�∩�
�
=
�
��
�
��
��
��
=
�
����
�
�:the first card is a red ace,
�
�:the. second card is a 10 or a jack,
�
�:the third card is greater than 3 but less than
7. Now
??????�
�=
�
��
,??????�
�|�
�=
�
��
,??????�
�|�
�∩�
�
=
��
��
and hence, by Theorem 2.15,
??????�
�∩�
�∩�
�=??????�
�??????�
�|�
�??????�
�|�
�∩�
�
=
�
��
�
��
��
��
=
�
����
Bayes’Rule:
Consider the following figure
Figure 2:Venn diagram for the events A, E,
and E’
E E’
A
A
Consider the following figure
Figure 2:Venn diagram for the events A, E,
and E’
E E’
A
A
Referring to Figure 2, we can write Aas
the union of the two mutually exclusive
events�∩�and �
′
∩�.Hence
�=�∩�∪�
′
∩�
and by additive rule and multiplicative
rule, we can write
??????(�)=??????�∩�∪�
′
∩�
=??????�∩�+??????�
′
∩�
=??????�??????�|�+??????�
′
??????�|�
′
the union of the two mutually exclusive
events�∩�and �
′
∩�.Hence
�=�∩�∪�
′
∩�
and by additive rule and multiplicative
rule, we can write
??????(�)=??????�∩�∪�
′
∩�
=??????�∩�+??????�
′
∩�
=??????�??????�|�+??????�
′
??????�|�
′
A generalization of the foregoing illustration
to the case where the sample space is
partitioned into ksubsets is covered by the
following theorem, sometimes called the
theorem of total probabilityor the rule of
elimination.
to the case where the sample space is
partitioned into ksubsets is covered by the
following theorem, sometimes called the
theorem of total probabilityor the rule of
elimination.
Theorem 2.16:
If, the events �
�,�
�,…,�
??????constitutea
partition of the sample space S such that
??????�
??????≠�for??????=�,�,…,??????
then for any event Aof S,
??????�=σ
??????=�
??????
??????�
??????∩�=σ
??????=�
??????
??????�
????????????�|�
??????.
If, the events �
�,�
�,…,�
??????constitutea
partition of the sample space S such that
??????�
??????≠�for??????=�,�,…,??????
then for any event Aof S,
??????�=σ
??????=�
??????
??????�
??????∩�=σ
??????=�
??????
??????�
????????????�|�
??????.
B2
B1
Bk
Figure 3: Partitioning the sample space S
A
B1
Bk
Figure 3: Partitioning the sample space S
A
Proof:
Consider the Venn diagram of the figure 3.
The events A is seen to be the union of the
mutually exclusive events
�
�∩�,�
�∩�,…,�
??????∩�
that is
�=�
�∩�∪�
�∩�∪⋯∪�
??????∩�
Using theorem 2.10 (additive rule) and
theorem 2.13 (multiplicative rule) we have
Consider the Venn diagram of the figure 3.
The events A is seen to be the union of the
mutually exclusive events
�
�∩�,�
�∩�,…,�
??????∩�
that is
�=�
�∩�∪�
�∩�∪⋯∪�
??????∩�
Using theorem 2.10 (additive rule) and
theorem 2.13 (multiplicative rule) we have
??????�=??????�
�∩�∪�
�∩�∪⋯∪�
??????∩�
=??????�
�∩�+??????�
�∩�+⋯+??????�
??????∩�
=
??????=�
??????
??????�
??????∩�=
??????=�
�
??????�
????????????�|�
??????
�∩�∪�
�∩�∪⋯∪�
??????∩�
=??????�
�∩�+??????�
�∩�+⋯+??????�
??????∩�
=
??????=�
??????
??????�
??????∩�=
??????=�
�
??????�
????????????�|�
??????
Example:
In certain assembly plant In a certain assembly
plant, three machines,�
�,�
�,�
�, make 30%,
45%, and 25%, respectively, of the products. It
is known from past experience that 2%, 3%,
and 2% of the products made by each machine,
respectively, are defective. Now, suppose that a
finished product is randomly selected. What is
the probability that it is defective?
In certain assembly plant In a certain assembly
plant, three machines,�
�,�
�,�
�, make 30%,
45%, and 25%, respectively, of the products. It
is known from past experience that 2%, 3%,
and 2% of the products made by each machine,
respectively, are defective. Now, suppose that a
finished product is randomly selected. What is
the probability that it is defective?
Solution:
Consider the following events:
A: the product, is defective,
�
�: the product is made by machine �
�,
�
�: the product is made by machine �
�,
�
�: product is made by machine �
�.
Applying the rule of elimination, we can
write
??????�
=??????�
�??????�|�
�+??????�
�??????�|�
�
+??????�
�??????�|�
�
Consider the following events:
A: the product, is defective,
�
�: the product is made by machine �
�,
�
�: the product is made by machine �
�,
�
�: product is made by machine �
�.
Applying the rule of elimination, we can
write
??????�
=??????�
�??????�|�
�+??????�
�??????�|�
�
+??????�
�??????�|�
�
B1 P(A|B1)=0.02
P(B1)=0.3 A
P(B2)=0.45 P(A|B2)=0.03
B2 A
P(B3)=0.25
P(A|B3)=0.02
B3 A
Figure 4:Tree diagram
P(B1)=0.3 A
P(B2)=0.45 P(A|B2)=0.03
B2 A
P(B3)=0.25
P(A|B3)=0.02
B3 A
Figure 4:Tree diagram
Referring to the tree diagram of Figure 4, we
find that the three branches give
the probabilities
??????�
�??????�|�
�= (0.3) (0.02) = 0.006.
??????�
�??????�|�
�= (0.45)(0.03) = 0.0135,
??????�
�??????�|�
�= (0.25)(0.02) = 0.005,
and hence
??????�= 0.006 + 0.0135 + 0.005 = 0.0245.
find that the three branches give
the probabilities
??????�
�??????�|�
�= (0.3) (0.02) = 0.006.
??????�
�??????�|�
�= (0.45)(0.03) = 0.0135,
??????�
�??????�|�
�= (0.25)(0.02) = 0.005,
and hence
??????�= 0.006 + 0.0135 + 0.005 = 0.0245.
Theorem 2.17: (Bayes' Rule)
If the events �
�,�
�,…,�
??????constitute a
partition of the sample space Ssuch that
??????�
??????≠�for??????=�,�,…,??????,then for any
event Ain Ssuch that ??????�≠�,
??????�
??????|�=
??????�
??????∩�
σ
??????=�
??????
??????�
??????∩�
=
??????�
????????????�|�
??????
σ
??????=�
??????
??????�
????????????�|�
??????
for??????=�,�,…,??????
If the events �
�,�
�,…,�
??????constitute a
partition of the sample space Ssuch that
??????�
??????≠�for??????=�,�,…,??????,then for any
event Ain Ssuch that ??????�≠�,
??????�
??????|�=
??????�
??????∩�
σ
??????=�
??????
??????�
??????∩�
=
??????�
????????????�|�
??????
σ
??????=�
??????
??????�
????????????�|�
??????
for??????=�,�,…,??????
Proof:
By the: definition of conditional
probability,
??????�
??????|�=
??????�
??????∩�
??????�
and then using Theorem 2.16 in the
denominator, we have
??????�
??????|�=
??????�
??????∩�
σ
??????=�
??????
??????�
??????∩�
By the: definition of conditional
probability,
??????�
??????|�=
??????�
??????∩�
??????�
and then using Theorem 2.16 in the
denominator, we have
??????�
??????|�=
??????�
??????∩�
σ
??????=�
??????
??????�
??????∩�
Applying theorem 2.13 to both numerator and
denominator, we obtain
??????�
??????|�=
??????�
????????????�|�
??????
σ
??????=�
??????
??????�
????????????�|�
??????
which completes the proof.
denominator, we obtain
??????�
??????|�=
??????�
????????????�|�
??????
σ
??????=�
??????
??????�
????????????�|�
??????
which completes the proof.
Example:
With reference to the last example above, if a
product was chosen randomly and found to be
defective, what is the probability that it was
made by machine �
�?
Solution :Using Bayes' rule to write
??????�
�|�
=
??????�
�??????�|�
�
??????�
�??????�|�
�+??????�
�??????�|�
�+??????�
�??????�|�
�
With reference to the last example above, if a
product was chosen randomly and found to be
defective, what is the probability that it was
made by machine �
�?
Solution :Using Bayes' rule to write
??????�
�|�
=
??????�
�??????�|�
�
??????�
�??????�|�
�+??????�
�??????�|�
�+??????�
�??????�|�
�
and then substituting the: probabilities calculated
in last example, we have
??????�
�|�=
�.���
�.���+�.����+�.���
=
�.���
�.����
=
��
��
In view of the fact that a defective product was
selected, this result suggests that it probably was
not made by machine �
�.
in last example, we have
??????�
�|�=
�.���
�.���+�.����+�.���
=
�.���
�.����
=
��
��
In view of the fact that a defective product was
selected, this result suggests that it probably was
not made by machine �
�.
Exercise
Exercise
Exercise
Exercise
Exercise
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