25083436 Chapter 4 Probability

8/12/2019 25083436 Chapter 4 Probability

1/24

Chapter 4: Probability 1

Chapter 4

Probability

LEARNING OBJECTIVES

The main objective of Chapter 4 is to help you understand the basic principles ofprobability, specifically enabling you to

1. Comprehend the different ways of assigning probability.

. !nderstand and apply marginal, union, joint, and conditional probabilities.

". #elect the appropriate law of probability to use in solving problems.

4. #olve problems using the laws of probability, including the law of addition, thelaw of multiplication , and the law of conditional probability.

$. %evise probabilities using &ayes' rule.

CHAPTER TEACHING STRATEGY

#tudents can be motivated to study probability by reali(ing that the field of

probability has some stand)alone application in their lives in such applied areas human

resource analysis, actuarial science, and gaming. *n addition, students should understandthat much of the rest of the course is based on probabilities even though they will not be

directly applying many of these formulas in other chapters.

This chapter is frustrating for the learner because probability problems can beapproached by using several different techni+ues. hereas, in many chapters of this te-t,

students will approach problems by using one standard techni+ue, in chapter 4, different

students will often use different approaches to the same problem. The te-t attempts toemphasi(e this point and underscore it by presenting several different ways to solve

probability problems. The probability rules and laws presented in the chapter can

virtually always be used in solving probability problems. owever, it is sometimes easierto construct a probability matri- or a tree diagram or use the sample space to solve the

problem. *f the student is aware that what they have at their hands is an array of tools or

techni+ues, they will be less overwhelmed in approaching a probability problem. /n


2/24

Chapter 4: Probability

attempt has been made to differentiate the several types of probabilities so that students

can sort out the various types of problems.

*n teaching students how to construct a probability matri-, emphasi(e that it is

usually best to place only one variable along each of the two dimensions of the matri-.

0That is place astercard with yes2no on one a-is and 3isa with yes2no on the otherinstead of trying to place astercard and 3isa along the same a-is.

This particular chapter is very amenable to the use of visual aids. #tudents enjoyrolling dice, tossing coins, and drawing cards as a part of the class e-perience.

5f all the chapters in the boo6, it is most imperative that students wor6 a lot of

problems in this chapter. Probability problems are so varied and individuali(ed that asignificant portion of the learning comes in the doing. 7-perience is an important factor

in wor6ing probability problems.

#ection 4.8 on &ayes9 theorem can be s6ipped in a one)semester course withoutlosing any continuity. This section is a prere+uisite to the chapter 18 presentation of

revising probabilities in light of sample information 0section 18.4.

CHAPTER OUTLINE

4.1 *ntroduction to Probability

4. ethods of /ssigning ProbabilitiesClassical ethod of /ssigning Probabilities

%elative ;re+uency of 5ccurrence

#ubjective Probability

4." #tructure of Probability

7-periment7vent

7lementary 7vents

#ample #pace

!nions and *ntersectionsutually 7-clusive 7vents

*ndependent 7vents

Collectively 7-haustive 7ventsComplimentary 7vents

Counting the Possibilities

The mn Counting %ule #ampling from a Population with %eplacement

Combinations: #ampling from a Population ithout %eplacement

4.4 arginal, !nion,


3/24

Chapter 4: Probability "

4.$ /ddition =aws

Probability atrices

Complement of a !nion#pecial =aw of /ddition

4.> ultiplication =aws?eneral =aw of ultiplication

#pecial =aw of ultiplication

4.@ Conditional Probability

*ndependent 7vents

4.8 %evision of Probabilities: &ayes' %ule

EY TER!S

/ Priori *ntersection

&ayes' %ule


4/24


B1/>, B"/4, /$/>

There are 1$ members of the sample space

The probability of selecting e-actly one defect out of

two is:

D21$ "#$

4. E F1, ", $, @, 8, DG, H F, 4, @, DG

and I F1, , ", 4, @,G

a EI %&' (' )' 4' *' +' ,' -.

b E H %+' -.

c E I %&' )' +.

d EHI %&' (' )' 4' *' +' ,' -.

e E H I %+.

f 0EHI F1, , ", 4, $, @, 8, DG F1, , ", 4, @G %&' (' )' 4' +.g 0H I0E H F, 4, @GF@, DG %(' 4' +' -.

h E or H E H %&' (' )' 4' *' +' ,' -.

i H and I H I %(' 4' +.

4." *f / F, >, 1, 4G and the population is the positive even numbers through "J,

/9 %4' ,' &$' &4' ' &,' ($' ((' (#' (,' )$.

4.4 >040"0" (

4.$ 7numeration of the si- parts: B1, B, /1, /, /", /4B Befective part

/ /cceptable part

#ample #pace:

B1B/1, B1B/, B1B/",

B1B/4, B1/1/, B1/1/",

B1/1/4, B1//", B1//4,

B1/"/4, B/1/, B/1/",

B/1/4, B//", B//4,B/"/4, /1//", /1//4,

/1/"/4, //"/4

Combinations are used to counting the sample space because sampling is done

without replacement.


5/24

Chapter 4: Probability $

>C"K"K"

K> J

Probability that one of three is defective is:

12J "2$ "#$

There are J members of the sample space and 1 of them have 1 defective part.

4.> 1J@ 1J,JJJ,JJJ different numbers

4.@ JC>K14K>

KJ ),'+#$

*t is assumed here that > different 0without replacement employees are to be

selected.

4.8 P0/ .1J, P0& .1, P0C .1

P0/ C .J$ P0& C .J"

a P0/C P0/ L P0C ) P0/ C .1J L .1 ) .J$ "(#

b P0&C P0& L P0C ) P0&C .1 L .1 ) .J" ")$

c *f /, & mutually e-clusive, P0/& P0/ L P0& .1J L .1 "((

4.D

B 7 ;

/ $ 8 1 $

& 1J > 4 J

C 8 $ 1$

" 1> 1 >J

a P0/ B P0/ L P0B ) P0/ B $2>J L "2>J ) $2>J 4"2>J "++

b P07 & P07 L P0& ) P07& 1>2>J L J2>J ) >2>J "J2>J "*$$$

c P0B 7 P0B L P07 "2>J L 1>2>J "D2>J "#*$$


6/24

Chapter 4: Probability >

d P0C ; P0C L P0; ) P0C ; 1$2>J L 12>J ) $2>J "12>J "*+

4.1J

7 ;

/ .1J .J" .1"

& .J4 .1 .1>

C .@ .J> .""

B ."1 .J@ ."8

.@ .8 1.JJ

a P0/ ; P0/ L P0; ) P0/ ; .1" L .8 ) .J" "),

b P07& P07 L P0& ) P07& .@ L .1> ) .J4 ",4

c P0&

C P0& L P0C .1> L ."" "4-d P07; P07 L P0; .@ L .8 &"$$$

4.11 / event ) flown in an airplane at least onceT event ) ridden in a train at least once

P0/ .4@ P0T .8

P 0ridden either a train or an airplane P0/ T P0/ L P0T ) P0/ T .4@ L .8 ) P0/ T

Cannot solve this problem without 6nowing the probability of the intersection.

e need to 6now the probability of the intersection of / and T, the proportionwho have ridden both.

4.1 P0= .@$ P0 .@8 P0 = .>1

a P0 = P0 L P0= ) P0 = .@8 L .@$ ) .>1 "-(

b P0

= but not both P0

= ) P0

= .D ) .>1 ")&c P0A A= 1 ) P0= 1 ) .D "$,

4.1" =et C have cable T3

=et T have or more T3 sets

P0C .>@, P0T .@4, P0C T .$$

a P0CT P0C L P0T ) P0C T .>@ L .@4 ) .$$ ",#


7/24

Chapter 4: Probability @

b P0CT but not both P0CT ) P0CT .8> ) .$$ ")&

c P0AC AT 1 ) P0C T 1 ) .8> "&4

d The special law of addition does not apply because P0C T is not .JJJJ.

Possession of cable T3 and or more T3 sets are not mutually e-clusive.

4.14 =et T review transcript; consider faculty references

P0T .$4 P0; .44 P0T ; ."$

a P0;T P0; L P0T ) P0; T .44 L .$4 ) ."$ "#)

b P0;T ) P0; T .>" ) ."$ "(,

c 1 ) P0;T 1 ) .>" ")+

d

H A H ."$ .1D .$4

A .JD ."@ .4>

.44 .$> 1.JJ

4.1$

C B 7 ;

/ $ 11 1> 8 4J

& " $ @ 1@

@ 14 1 1$ $@

a P0/ 7 1>2$@ "(,$+

b P0B & "2$@ "$*(#

c P0B 7 "$$$$

d P0/ & "$$$$

4.1>

B 7 ;

/ .1 .1" .J8 .""

& .18 .JD .J4 ."1

C .J> .4 .J> .">


8/24


."> .4> .18 1.JJ

a P07 & "$-

b P0C; "$#

c P07B "$$

4.1@ =et B Befective part

a 0without replacement

P0B1 B P0B1 P0BB1 4$J

"J

4D

$

$J

>= "$&((

b 0with replacement

P0B1 B P0B1 P0B $JJ

">

$J

>

$J

>=

"$&44

4.18 =et ! !rban* care for *ll relatives

a P0! * P0! P0* !

P0! .@8 P0* .1$ P0*! .11P0! * [email protected] "$,*,

b P0!

A* P0! P0A*! but P0*! .11 #o, P0A*! 1 ) .11 .8D and P0!A*

P0! P0A*! [email protected] "#-4(

c

!

Hes Ao

* Hes .1$

Ao .8$

.@8 .

The answer to a is found in the H7#)H7# cell. To compute this cell, ta6e 11M


9/24

Chapter 4: Probability D

or .11 of the total 0.@8 people in urban areas. 0.110.@8 .J8$8 which belongs in

the H7#)H7#N cell. The answer to b is found in the Hes for ! and no for * cell.

*t can be determined by ta6ing the marginal, .@8, less the answer for a, .J8$8.

d. P0A! * is found in the no for ! column and the yes for * row 01strow and

nd

column. Ta6e the marginal, .1$, minus the yes)yes cell, .J8$8, to get"$#4(.

4.1D =et # stoc6holder=et C college

P0# .4" P0C ."@ P0C# .@$

a P0A# 1 ) .4" "*+

b P0#C P0# P0C# 0.4"0.@$ ")((* c P0#C P0# L P0C ) P0#C .4" L ."@ ) ."$ "4++*

d P0A#AC 1 ) P0#C 1 ) .4@@$ "*((* e P0A#AC P0A# L P0AC ) P0A#AC .$@ L .>" ) .$$ "#++*

f P0C A# P0C ) P0C # ."@ ) ."$ "$4+*

4.J =et ; fa- machine=et P personal computer

?iven: P0; .1J P0P .$ P0P; .D1

a P0; P P0; P0P; 0.1J0.D1 "$-&

b P0; P P0; L P0P ) P0; P .1J L .$ ) .JD1 "*(-

c P0; AP P0; P0AP;

#ince P0P; .D1, P0AP; 1 ) P0P; 1 ) .D1 .JD P0; AP 0.1J0.JD "$$-

d P0A; AP 1 ) P0; P 1 ) .$D "4+&

e P0A; P P0P ) P0; P .$ ) .JD1 "4(-

P AP ; .JD1 .JJD .1J

A; .4D .4@1 .DJ

.$J .48J 1.JJ

4.1


10/24

Chapter 4: Probability 1J

=et # safety

=et / age

P0# ."J P0/ ."D P0/# .8@

a P0#

A/ P0# P0A/#

but P0A/# 1 ) P0/# 1 ) .8@ .1" P0# A/ 0."J0.1" "$)-

b P0A# A/ 1 ) P0#/ 1 ) OP0# L P0/ ) P0#/

but P0#/ P0# P0/# 0."J0.8@ .>1 P0A# A/ 1 ) 0."J L ."D ) .>1 "*+&

c P0A#/ P0A# ) P0A#A/

but P0A# 1 ) P0# 1 ) ."J .@J

P0A#/ .@J ) $@1 "&(-

4. =et C ceiling fans

=et 5 outdoor grill

P0C .>J P05 .D P0C 5 .1"

a P0C 5 P0C L P05 ) P0C 5 .>J L .D ) .1" "+#

b P0AC A5 1 ) P0C 5 1 ) .@> "(4

c P0AC 5 P05 ) P0C 5 .D ) .1" "

d P0C A5 P0C ) P0C 5 .>J ) .1" "4+

4."

7 ; ?

/ 1$ 1 8 "$

& 11 1@ 1D 4@

C 1 " @ 8J

B 18 1" 1 4"

>$ @4 >> J$

a P0?/ 82"$ "((,#


11/24


b P0&; 1@2@4 "((-+

c P0C7 12>$ ")()&

d P07? "$$$$

4.4

C B

/ ."> .44 .8J

& .11 .JD .J

.4@ .$" 1.JJ

a P0C/ .">2.8J "4*$$

b P0&B .JD2.$" "-,

c P0/& "$$$$

4.$Calculator

Hes Ao

Computer

Hes 4> " 4D

Ao 11 1$ >

$@ 18 @$

#elect a category from each variable and test

P0313 P031.

;or e-ample, P0Hes ComputerHes Calculator P0Hes ComputerQ

@$4D

$@4>

= Q

.8J@J .>$""

Variable o/ Co0p1ter 2ot i23epe23e2t o/ Variable o/ Cal1lator"


12/24


4.>

=et C construction

=et # #outh /tlantic

8","84 total failures

1J,8>@ failures in construction 8,J1J failures in #outh /tlantic

1,$8 failures in construction and #outh /tlantic

a P0# 8,J1J28","84 "$-#$#

b P0C # P0C L P0# ) P0C #

1J,8>@28","84 L 8,J1J28","84 ) 1,$828","84 1@,>1D28","84 "(&&)

c P0C #

"84,8"

8J1J"84,8"

1$8

00 =

SP

SCP "&*+$*

d P0# C

"84,8"

8>@,1J

"84,8"

1$8

0

0=

CP

SCP "&&*+#

e P0A#AC 0

01

0

0

NCP

SCP

NCP

NCNSP =

but AC 8","84 ) 1J,8>@ @,$1@

and P0AC @,$1@28","84 .8>D>@$

Therefore, P0A#AC 01 ) .11"20.8>D>@$ "-$#-

f P0A# C 0

00

0

0

CP

SCPCP

CP

CNSP =

but P0C 1J,8>@28","84 .1"J"

P0C # 1,$828","84 .J1$1

Therefore, P0A#C 0.1"J" ) .J1$12.1"J" ",,4(

4.@ =et 7 7conomy=et R Rualified


13/24

Chapter 4: Probability 1"

P07 .4> P0R ."@ P07R .1$

a P07R P07R2P0R .1$2."@ "4$*4

b P0R7 P07R2P07 .1$2.4> ")(#&

c P0RA7 P0RA72P0A7

but P0R A7 P0R ) P0R7 ."@ ) .1$ .

P0A7 1 ) P07 1 ) .4> .$4

P0RA7 .2.$4 "4$+4

d P0A7AR 1 ) P07R 1 ) OP07 L P0R ) P07R

1 ) O.4> L ."@ L .1$ 1 ) 0.>8 ")(

4.8

=et / airline tic6ets

=et T transacting loans

P0/ .4@ P0T/ .81

a P0/T P0/ P0T/ [email protected] "),$+b P0AT/ 1 ) P0T/ 1 ) .81 "&-

c P0AT / P0/ ) P0/ T .4@ ) ."8J@ "$,-)

4.D =et hardware

=et # software

P0 ."@ P0# .$4 P0# .D@

a P0A# 1 ) P0# 1 ) .D@ "$)

b P0#A P0#A2P0A but P0 # P0 P0# 0."@0.D@ ."$8D

so P0A

# P0# ) P0

# .$4 ) ."$8D .1811

P0A 1 ) P0 1 ) ."@ .>"

P0#A 0.181120.>" "(,+*

c P0A# P0A#2P0# .181122$4 "))*4


14/24


d P0AA# P0AA#2P0A# but P0AA# P0A ) P0A # .>" ) .1811 .448D

and P0A# 1 ) P0# 1 ) .$4 .4>

P0AA# .448D2.4> "-+*-

4."J =et / product produced on achine /

& product produces on achine &C product produced on achine C

B defective product

P0/ .1J P0& .4J P0C .$J

P0B/ .J$ P0B& .1 P0BC .J8

7vent Prior Conditional


15/24

Chapter 4: Probability 1$

7vent Prior Conditional $"4(#- C .$ .J$ .J1$ .J1$2.1>$"$-,,

P0*.1>$

%evised: P0/* .J>JJ2.1>$ "4+4)

P0&* .J$4J2.1>$ "4(#-

P0C* .J1$2.1>$ "$-,,

d

7vent Prior Conditional J ."D>J2.8@"$"4*))

C .$ .D$ ."@$ ."@$2.8@"$"(+&-

P0S.8@"$

4." =et T lawn treated by Tri)state? lawn treated by ?reen Chem

3 very healthy lawn

A not very healthy lawn

P0T .@ P0? .8 P03T ."J P03? .J

7vent Prior Conditional .J$>2.@"($*-P03.@

%evised: P0T3 .1>2.@ "+-4&

P0?3 .J$>2.@ "($*-

4."" =et T training


16/24

Chapter 4: Probability 1>

=et # small

P0T .>$

P0#T .18 P0A#T .8P0#AT .@$ P0A#AT .$

7vent Prior Conditional $ .8 .$""J .$""J2.>J$",*-$

AT ."$ .$ .J8@$ .J8@$2.>J$.141J

P0A#"#($*

4."4

3ariable 1

B 7

/ 1J J

3ariable & 1$ $

C "J 1$

$$ 4J D$

a P07 4J2D$ "4(&$*

b P0&B P0& L P0B ) P0&B

J2D$ L $$2D$ ) 1$2D$ >J2D$ "#)&*,

c P0/7 J2D$ "(&$*)

d P0&7 $24J "&(*$e P0/& P0/ L P0& "J2D$ L J2D$

$J2D$ "*(#)(

f P0&C .JJJJ 0mutually e-clusive

g P0BC "J24$ "####+

h P0/& P0/ & .JJJJ "$$$$ 01t1ally e6l17i8e P0& J2D$


17/24

Chapter 4: Probability 1@

i P0/ P0/BQQ

"J2D$ 1J2D$ QQ

."1$@D .1818Ao, 3ariables 1 and are not independent.

4."$

B 7 ; ?

/ " D @ 1 "1

& 8 4 > 4

C 1J $ " @ $

1 18 1> " @8

a P0;/ @2@8 "$,-+4

b P0/& P0/ & .JJJJ .JJJJP0& 2@8

c P0& 2@8 "(,($*

d P07; "$$$$ !1t1ally E6l17i8e

e P0B& 82 ")#)#4

f P0&B 821 "),$-*g P0BC 12@8 L $2@8 1J2@8 ">2@8 "4#&*

h P0; 1>2@8 "($*&)

4.">

/ge0years

U"$ "$)44 4$)$4 $$)>4 V>$

?ender ale .11 .J .1D .1 .1> .@8

;emale .J@ .J8 .J4 .J .J1 ..18 .8 ." .14 .1@ 1.JJ

a P0"$)44 "(,

b P0oman4$)$4 "$4


18/24


c P0an"$)44 P0an L P0"$)44 ) P0an"$)44 .@8 L .8 ) .J "

,#

d P0U"$$$)>4 P0U"$ L P0$$)>4 .18 L .14 ")(

e P0oman4$)$4 P0oman4$)$42P04$)$4 .J42." "&+)-f P0AA $$)>4 .11 L .J L .1D L .1> "##

4."@ =et T thoroughness

=et S 6nowledge

P0T .@8 P0S .4J P0TS .@

a P0TS P0T L P0S ) P0TS

.@8 L .4J ) .@ "-&

b P0ATAS 1 ) P0TS 1 ) .D1 "$-

c P0ST P0ST2P0T .@2.@8 ")4#(d P0ATS P0AT ) P0ATAS

but P0AT 1 ) P0T .

P0ATS . ) .JD "&)

4."8 =et % retirement=et = life insurance

P0% .4 P0= .>1 P0%= .""

a P0%= P0%=2P0= .""2.>1 "*4&$b P0=% P0%=2P0% .""2.4 "+,*+c P0=% P0= L P0% ) P0=%

.>1 L .4 ) ."" "+$

d P0%A= P0% ) P0%= .4 ) ."" .$-

e P0A=% P0A=%2P0% .JD2.4 .(&4)

4."D P0T .1> P0T .J P0TA7 .1@

P0 .1 P0A7 .J

a P0T P0 P0T 0.10.J "$4(b P0A7T P0A7 P0TA7 0.J0.1@ "$)4c P0T P0T2P0T 0.J420.1> "(#(*d P0A7AT P0A7AT2P0AT FP0A7 P0ATA7G2P0AT

but P0ATA7 1 ) P0TA7 1 ) .1@ .8" and


19/24

Chapter 4: Probability 1D

P0AT 1 ) P0T 1 ) .1> .84

Therefore, P0A7AT FP0A7 P0ATA7G2P0AT

F0.J0.8"G20.84 "&-+#e P0not not A7T P0not not A7T2 P0T

but P0not not A7T

.1> ) P0T ) P0A7T .1> ) .J4 ) .J"4 .J84

P0not not A7T2 P0T 0.J8420.1> "*(*

4.4J =et astercard / /merican 7-press 3 3isa

P0 ."J P0/ .J P03 .$

P0/ .J8 P03 .1 P0/3 .J>

a P03/ P03 L P0/ ) P03/

.$ L .J ) .J> ")-

b P03 P032P0 .12."J "4$c P03 P032P03 .12.$ "4,

d P03 P03QQ

.$ .4J

Possession of 3isa is not independent of

possession of astercard

e /merican 7-press is not mutually e-clusive of 3isa

because P0/3 .JJJJ

4.41

=et # believe ## secure A don't believe ## will be secure

U4$ under 4$ years old

V4$ 4$ or more years old

P0A .$1

Therefore, P0# 1 ) .$1 .4D

P0#V4$ .@JTherefore, P0AV4$ 1 ) P0#V4$ 1 ) .@J ."J


20/24

Chapter 4: Probability J

P0U4$ .$@

a P0V4$ 1 ) P0U4$ 1 ) .$@ "4)

b P0V4$# P0V4$ P0#V4$ 0.$@0.@J ."J1

P0U4$

# P0# ) P0V4$

# .4D ) ."J1 "&,- c P0V4$# P0V4$#2P0# P0V4$ P0#V4$2P0#

0.4"[email protected] "#&4)

d 0U4$A P0U4$ L P0A ) P0U4$A

but P0U4$A P0U4$ ) P0U4$#

.$@ ) .18D ."81

so P0U4$A .$@ L .$1 ) ."81 "#--

Probability atri- #olution for Problem 4.41:

# A

U4$ .18D ."81 .$@

V4$ ."J1 .1D .4"

.4DJ .$1J 1.JJ

4.4 =et e-pect to save more% e-pect to reduce debt

A don't e-pect to save more

A% don't e-pect to reduce debt

P0 .4" P0% .4$ P0% .81 P0A% 1)P0% 1 ) .81 .1D P0A 1 ) P0 1 ) .4" .$@

P0A% 1 ) P0% 1 ) .4$ .$$

a P0% P0 P0% 0.4"0.81 ")4,)b P0% P0 L P0% ) P0%

.4" L .4$ ) ."48" "*)&+

c P0neither save nor reduce debt

1 ) P0% 1 ) .$"1@ "4#,)

d P0A% P0 P0A% 0.4"0.1D "$,&+

Probability matri- for problem 4.4:


21/24


%educe

Hes Ao

#ave

Hes ."48" .J81@ .4"

Ao .1J1@ .4>8" [email protected]$JJ .$$JJ 1.JJ

4.4" =et % read

=et & chec6ed in the with boss

P0% .4J P0& ."4 P0&% .@8

a P0&

% P0% P0&% 0.4J0.@8 ")&(b P0A%A& 1 ) P0%& but P0%& P0% L P0& ) P0%&

.4J L."4 ) ."1 .48

P0A%A& 1 ) .48 "*+(

c P0%& P0%&2P0& 0."120."4 "-&+#

d P0A&% 1 ) P0&% 1 ) .@8 "((e P0A&A% P0A&A%2P0A%

but P0A% 1 ) P0% 1 ) .4J .>J

P0A&A% .$@2.>J "-*))

f Probability matri- for problem 4.4":

& A&

% ."1 .J88 .4J

A% .J8 .$@ .>J

."4J .>>J 1.JJ

4.44

=et: B denial

* inappropriateC customer


22/24

Chapter 4: Probability

P payment dispute

# specialty

? delays getting care% prescription drugs

P0B .1@ P0* .14 P0C .14 P0P .11P0# .1J P0? .J8 P0% .J@

a P0P# P0P L P0# .11 L .1J "(&

b P0%C .$$$$ 0mutually e-clusive

c P0*# P0*#2P0# .JJJJ2.1J .$$$$d P0A?AP 1 ) P0?P 1 ) OP0? L P0P

1 O.J8 L .11 1 ) .1D ",&

4.4$

=et % retention P process

P0% .$> P0P% ."> P0% P .DJ

a P0%AP P0% ) P0P% .$> ) ."> .($

b P0P% P0P%2P0% .">2.$> .#4(-

c P0P QQ

P0%P P0%P2P0P so P0P P0%P2P0%P .">2.DJ .4$

d P0%P P0% L P0P ) P0%P

.$> L .4J ) ."> .#$

e P0A%AP 1 ) P0%P 1 ) .>J .4$

f P0%AP P0%AP2P0AP

but P0AP 1 ) P0P 1 ) .4J .>J

P0%AP .J2.>J .))

4.4> =et mail

# salesP0 ."8 P0# .JJJJ P0AA# .41

a P0A# P0 ) P0# ."8 ) .JJ .),

b P0#:

P0# 1 ) P0AA# 1 ) .41 .$D

P0# P0 L P0#


23/24

Chapter 4: Probability "

Therefore, P0# P0# ) P0 .$D ) ."8 .(&

c P0# P0#2P0 .JJ2."8 .$$

d P0AA# P0AA#2P0A# .412.@D .*&-$

where: P0A# 1 ) P0# 1 ) .1 .@D

4.4@ =et # #arabia

T Tran < J8> .)-&4

f P0A%AT P0A%

AT2P0AT O1 ) P0%

T2P0AT

01 ) .4"@J2.>$ .,##(

4.4D

7vent Prior Conditional


24/24


P07i

P0A#7iP0A#7i

P07iA##oup .>J .@" .4"8J ",4*#

&rea6fast

eats ."$ .1@ .J$D$ "&&4-

ot Bogs .J$ .41 .JJ$ "$)-# .$18J

4.$J =et ? ?ood health

appy marriage

;? ;aith in ?od

P0? .D P0 .1 P0;? .4J

a P0;? P0 L P0;? ) P0;?

but P0;? .JJJJP0;? P0 L P0;? .1 L .4J "#&

b P0;?? P0 L P0;? L P0?

.D L .1 L .4J "-$$$

c P0;?? "$$$$

The categories are mutually e-clusive.

The respondent could not select more than oneanswer.

d P0neither ;? nor ? nor

1 ) P0;?? 1 ) .DJJJ "&$$$

25083436 Chapter 4 Probability

Documents

Transcript of 25083436 Chapter 4 Probability