probability

introduction to probability

1. PROBABILITY
BY UNSA SHAKIR

2. WhatisaPROBABILITY?
-Probabilityisthechancethatsomeeventwillhappen
-Itistheratioofthenumberofwaysacertainevent
canoccurtothenumberofpossibleoutcomes
Probability of Simple Events

3. Probability as a Numerical Measure
of the Likelihood of Occurrence
0 1.5
Increasing Likelihood of Occurrence
Probability
The event
is very
unlikely
to occur.
Theoccurrence
oftheeventis
justaslikelyas
itisunlikely.
The event
is almost
certain
to occur.

4. An Experiment and Its Sample Space
An experimentis any process that generates
well-defined outcomes.
The sample spacefor an experiment is the set of
all experimental outcomes.
An experimental outcome is also called a sample
point.

5. An eventis a collection of sample points.
The probability of any eventis equal to the sum of
the probabilities of the sample points in the event.
If we can identify all the sample points of an
experiment and assign a probability to each, we
can compute the probability of an event.
Events and Their Probabilities

6. An Experiment and Its Sample Space
Experiment
Toss a coin
Inspection a part
Conduct a sales call
Roll a die
Play a football game
Experiment Outcomes
Head, tail
Defective, non-defective
Purchase, no purchase
1, 2, 3, 4, 5, 6
Win, lose, tie

7. Using Simple Events
•The probability of an event Ais equal to
the sum of the probabilities of the simple
events contained in A
•If the simple events in an experiment are
equally likely, you can calculateevents simple ofnumber total
Ain events simple ofnumber
)( 
N
n
AP
A

8. The Probability of an Event
•P(A) must be between 0 and 1.
–If event A can never occur, P(A) = 0. If
event A always occurs when the experiment
is performed, P(A) =1.
•The sum of the probabilities for all simple
events in S equals 1.

9. Example
•The die toss:
•Simple events: Sample space:
1
2
3
4
5
6
E
1
E
2
E
3
E
4
E
5
E
6
S ={E
1, E
2, E
3, E
4, E
5, E
6}
S
•E
1
•E
6
•E
2
•E
3
•E
4
•E
5

10. Basic Concepts
•An eventis a collection of one or more simple
events.
•The die toss:
–A: an odd number
–B: a number > 2
S
A ={E
1, E
3, E
5}
B ={E
3, E
4, E
5,E
6}
B
A
•E
1
•E
6
•E
2
•E
3
•E
4
•E
5

11. –Suppose that 10% of the U.S. population has
red hair. Then for a person selected at random,
Finding Probabilities
•Probabilitiescanbefoundusing
–Estimatesfromempiricalstudies
–Commonsenseestimatesbasedon
equallylikelyevents.
P(Head) = 1/2
P(Red hair) = .10
•Examples:
–Toss a fair coin.

12. Example 1
Toss a fair coin twice. What is the probability
of observing at least one head?
H
1st Coin 2nd Coin E
i P(E
i)
H
T
T
H
T
HH
HT
TH
TT
1/4
1/4
1/4
1/4
P(at least 1 head)
= P(E
1) + P(E
2) + P(E
3)
= 1/4 + 1/4 + 1/4 = 3/4

13. Example 2
A bowl contains three M&Ms
®
, one red, one
blue and one green. A child selects two M&Ms
at random. What is the probability that at least
one is red?
1st M&M 2nd M&M E
i P(E
i)
RB
RG
BR
BG
1/6
1/6
1/6
1/6
1/6
1/6
P(at least 1 red)
= P(RB) + P(BR)+ P(RG)
+ P(GR)
= 4/6 = 2/3
m
m
m
m
m
m
m
m
m
GB
GR

14. Example 3
The sample space of throwing a pair of dice is

15. Example 3
Event Simple events Probability
Dice add to 3(1,2),(2,1) 2/36
Dice add to 6(1,5),(2,4),(3,3),
(4,2),(5,1)
5/36
Red die show 1(1,1),(1,2),(1,3),
(1,4),(1,5),(1,6)
6/36
Green die show 1(1,1),(2,1),(3,1),
(4,1),(5,1),(6,1)
6/36

16. •I have 40candies
•26 areRed
•4 areBlue
•10 areYellow
•I’mgoingtotakeoutone.
Example:

17. •Q: What is the probability of getting a red candy?
•A: 13/20 or 0.65
•Q: What is the probability of getting a yellow
candy?
•A: ¼ or 0.25
•Q:What is the probability of getting a blue candy
•A: 1/10 or o.1
Question andAnswers

18. Example
•Arandomsampleofsizetwoistobeselectedfromthelist
ofsixcities,SanDiego,LosAngeles,SanFrancisco,
Denver,Paris,andLondon.
•A company has offices in six cities, San Diego, Los
Angeles, San Francisco, Denver, Paris, and London. A new
employee will be randomly assigned to work in one of these
offices.
Outcomes ?
Sample Space ?

19. Real World Example:
BestBuyishavinganIPODgiveaway.Theyputallthe
IPODShufflesinabag.Customersmaychoosean
IPODwithoutlookingatthecolor.Insidethebagare
4orange,5blue,6green,and5pinkIPODS.If
MariachoosesoneIPODatrandom,whatisthe
probabilityshewillchooseanorangeIPOD?
P(orange)=
4
/
20=
2
/
10=
1
/
5or20%
Probability of Simple Events

20. Example : Deck of Cards.
What is the probability of picking a heart?
# favorable outcomes 13 1
# possible outcomes 52 4
The probability of picking a heart is
1 out of 4 or .25 or 25%
What is the probability of picking a non heart?
# favorable outcomes 39 3
# possible outcomes 52 4
The probability of picking a heart is
3 out of 4 or .75 or 75%
Probability of Simple Events
P(heart) = = =
P(nonheart)= =

21. Real World Example:
Acomputercompanymanufactures2,500
computerseachday.Anaverageof100of
thesecomputersarereturnedwithdefects.
Whatistheprobabilitythatthecomputer
youpurchasedisnotdefective?
P(notdefective)=#notdefective=2,400=24
total#manufactured2,50025
Probability of Simple Events

22. Probability Examples
●Ajarcontains30redmarbles,12yellow
marbles,8greenmarblesand5bluemarbles.What
istheprobabilitythatyoudrawandreplacemarbles
3timesandyougetNOredmarbles?
•Thereare55marbles,25ofwhicharenotred
•P(gettingacolorotherthanred)=P(25/55)≈.455
•Probabilityofthishappening3timesinarowis
foundby.455*.455*.455≈.094

23. Multiplication Rule and
Independent Events
Multiplication Rule for Independent Events:Let A and B
be two independent events, then ( ) ( ) ( ).P A B P A P B
Examples:
•Flipacointwice.Whatistheprobabilityofobservingtwo
heads?
•Flipacointwice.Whatistheprobabilityofgettingahead
andthenatail?Atailandthenahead?Onehead?
Three computers are ordered. If the probability of getting a
“working” computer is .7, what is the probability that all three
are “working” ?.7 x .7 x .7=.7³ =.343 or 34.3%……

24. S
Event Relations
Thebeautyofusingevents,ratherthansimpleevents,isthatwe
cancombineeventstomakeothereventsusinglogicaloperations:
and,orandnot.
Theunionoftwoevents,AandB,istheeventthateitherAorB
orbothoccurwhentheexperimentisperformed.Wewrite
A B
A B

25. S
A
B
Event Relations
The intersectionof two events, Aand B, is
the event that both AandB occur when the
experiment is performed. We write A B.
•If two events A and B are mutually
exclusive, then P(A B) = 0.

26. S
Event Relations
A
A
C
sometimes,wewanttoknowtheprobabilitythatanevent
willnothappen;aneventoppositetotheeventofinterestis
calledacomplementaryevent.
IfAisanevent,itscomplementisTheprobabilityofthe
complementisACorA
Example:Thecomplementofmaleeventisthefemale

27. Example
Select a student from the classroom and
record his/herhair color andgender.
–A: student has brown hair
–B: student is female
–C:student is male
What is the relationship between events Band C?
•A
C
:
•BC:
•BC:
Mutually exclusive; B = C
C
Student does not have brown hair
Student is both male and female = 
Student is either male and female = all students = S

28. Calculating Probabilities for Unions and
Complements
•There are special rules that will allow you to
calculate probabilities for composite events.
•The Additive Rule for Unions:
•For any two events, Aand B, the probability
of their union, P(A B), is)()()()( BAPBPAPBAP 
A B

29. Example: Additive Rule
Example: Suppose that there were 120 students
in the classroom, and that they could be
classified as follows: totalBrownNot Brown
Male 60 20 40
Female60 30 30
A:brown hair
P(A) = 50/120
B:female
P(B) = 60/120
P(AB) = P(A) + P(B) –P(AB)
= 50/120 + 60/120 -30/120
= 80/120 = 2/3 Check:P(AB)
= (20 + 30 + 30)/120

30. Example: Two Dice
A:red die show 1
B:green die show 1
P(AB) = P(A) + P(B) –P(AB)
= 6/36 + 6/36 –1/36
= 11/36

31. A Special Case
When two events A and B are
mutually exclusive, P(AB) = 0
andP(AB) = P(A) + P(B).
BrownNot Brown
Male20 40
Female30 30
A:male withbrown hair
P(A) = 20/120
B:female with brown hair
P(B) = 30/120
P(AB) = P(A) + P(B)
= 20/120 + 30/120
= 50/120
A and B are mutually
exclusive, so that

32. Example: Two Dice
A:dice add to 3
B:dice add to 6
A and B are mutually
exclusive, so that
P(AB) = P(A) + P(B)
= 2/36 + 5/36
= 7/36

33. Calculating Probabilities
for Complements
•We know that for any event A:
–P(A A
C
) = 0
•Since either Aor A
C
must occur,
P(A A
C
) =1
•so thatP(AA
C
) = P(A)+ P(A
C
) = 1
P(A
C
) = 1 –P(A)
A
A
C

34. Example
BrownNot Brown
Male20 40
Female30 30
A: male
P(A) = 60/120
B: female
P(B) = ?
P(B) = 1-P(A)
= 1-60/120 = 60/120
A and B are
complementary, so that
Select a student at random from
the classroom. Define:

35. Example
•Ajarcontains30redmarbles,12yellowmarbles,8
greenmarblesand5bluemarbles.Whatistheprobability
thatyoudrawandreplacemarbles3timesandyougetat
least1Red?
•It'seasiertocalculatetheprobabilityofgettingNOred
marbles,andsubtractthatfrom1(weusethecomplement
rule:P(AC)=1–P(C)
•Frompreviousexample,itis1-.094=.906

36. Calculating Probabilities for
Intersections
Inthepreviousexample,wefoundP(AB)directly
fromthetable.Sometimesthisisimpracticalor
impossible.TheruleforcalculatingP(AB)depends
ontheideaofindependentanddependentevents.
Twoevents,AandB,aresaidtobeindependentifthe
occurrenceornonoccurrenceofoneoftheeventsdoes
notchangetheprobabilityoftheoccurrenceofthe
otherevent.

37. Conditional Probabilities
TheprobabilitythatAoccurs,giventhatevent
Bhasoccurrediscalledtheconditional
probabilityofAgivenBandisdefinedas0)( if
)(
)(
)|( 

 BP
BP
BAP
BAP
“given”

38. Example 1
Toss a fair coin twice. Define
–A: head on second toss
–B: head on first toss
HT
TH
TT
1/4
1/4
1/4
1/4
P(A|B) = ½
P(A|not B) = ½
HH
P(A) does not
change, whether
B happens or
not…
A and B are
independent!

39. Example 2
A bowl contains five M&Ms
®
, two red and three
blue. Randomly select two candies, and define
–A: second candy is red.
–B: first candy is blue.
P(A|B) =P(2
nd
red|1
st
blue)= 2/4 = 1/2
P(A|not B) = P(2
nd
red|1
st
red) = 1/4
P(A) does change,
depending on
whether B happens
or not…
A and B are
dependent!

40. Example 3: Two Dice
Toss a pair of fair dice. Define
–A: red die show 1
–B: green die show 1
P(A|B) = P(A and B)/P(B)
=1/36/1/6=1/6=P(A)
P(A) does not
change, whether
B happens or
not…
A and B are
independent!

41. Example 3: Two Dice
Toss a pair of fair dice. Define
–A: add to 3
–B: add to 6
P(A|B) = P(A and B)/P(B)
=0/36/5/6=0
P(A) does change
when B happens
A and B are dependent!
In fact, when B happens,
A can’t

42. Problem .
Blood
Group
Males Females Total
O
A
B
AB
20
17
8
5
20
18
7
5
40
35
15
10
Total 50 50 100

43. Problem .
IllNot IllTotal
Ate Barbecue
Did Not Eat Barbecue
90
20
30
60
120
80
Total 110 90 200
An outbreak of food poisoning occurs in a group
of students who attended a party

44. Conditional probabilities
Example:Ifasubjectwasselectedrandomlyand
foundtobefemalewhatistheprobabilitythatshe
hasabloodgroupO
Herethetotalpossibleoutcomesconstitutea
subset(females)ofthetotalnumberofsubjects.
ThisprobabilityistermedprobabilityofOgivenF
P(O\F)=20/50
=0.40

45. Example:
probability of being male &
belong to blood group AB
P(M and AB) = P(M∩AB)
= 5/100
= 0.05
∩= intersection

46. Example
The joint probability of being male and having blood
type O
Toknowthattwoeventsareindependentcompute
themarginalandconditionalprobabilitiesofoneof
themiftheyareequalthetwoeventsare
independent.Ifnotequalthetwoeventsare
dependent
P(O)=40/100=0.40
P(O\M)=20/50=0.40
Thenthetwoeventsareindependent
P(O∩M)=P(O)P(M)=(40/100)(50/100)
=0.20

47. Example:
The joint probability of being ill and eat barbecue
P(Ill)=110/200=0.55
P(Ill\EatB)=90/120=0.75
Thenthetwoeventsaredependent
P(Ill∩EatB)=P(EatB)P(Ill\EatB)
=(10/200)(90/120)
=0.45

48. Example:
The probability of being either blood type O or
blood type A
P(OUA) = P(O) + P(A)
= (40/100)+(35/100)
= 0.75

49. Two events are not mutually exclusive
(male gender and blood type O).
P(M OR O) = P(M)+P(O) –P(M∩O)
= 0.50 + 0.40 –0.20
= 0.70
Example:

50. Excercises
1.Iftuberculousmeningitishadacasefatalityof20%,
(a)Findtheprobabilitythatthisdiseasewouldbefatalin
tworandomlyselectedpatients(thetwoeventsare
independent)
(b)Iftwopatientsareselectedrandomlywhatisthe
probabilitythatatleastoneofthemwilldie?
(a)P(firstdieandseconddie)=20%20%=0.04
(b)P(firstdieorseconddie)
=P(firstdie)+P(seconddie)-P(bothdie)
=20%+20%-4%
=36%

51. 2.Inanormallydistributedpopulation,theprobabilitythata
subject’sbloodcholesterollevelwillbelowerthan1SD
belowthemeanis16%andtheprobabilityofbeingblood
cholesterollevelhigherthan2SDabovethemeanis
2.5%.Whatistheprobabilitythatarandomlyselected
subjectwillhaveabloodcholesterollevellowerthan1SD
belowthemeanorhigherthan2SDabovethemean.
P(bloodcholesterollevel<1SDbelowthemeanor2SD
abovethemean)=16%+2.5%
=18.5%

52. 3.Inastudyoftheoptimumdoseoflignocainerequiredtoreduce
painoninjectionofanintravenousagentusedforinductionof
anesthesia,fourdosinggroupswereconsidered(groupA
receivednolignocaine,whilegroupsB,C,andDreceived0.1,
0.2,and0.4mg/kg,respectively).Thefollowingtableshowsthe
patientscross-classifiedbydoseandpainscore:
Pain
score
Group Total
A B C D
0
1
2
3
49
16
8
4
73
7
5
1
58
7
6
0
62
8
6
0
242
38
25
5
Total7786 71 76310
Compute the following probabilities for a
randomly selected patient:
1.being of group D and experiencing
no pain
2.belonging to group B or having a
pain score of 2
3.having a pain score of 3 given that
he belongs to group A
4.belonging to group C

53. Nightlights and Myopia
Assumingthesedataarerepresentativeofalargerpopulation,
whatistheapproximateprobabilitythatsomeonefromthat
populationwhosleepswithanightlightinearlychildhood
willdevelopsomedegreeofmyopia?
Note: 72 + 7 = 79 of the 232 nightlight users developed some
degree of myopia. So the probability to be 79/232 = 0.34.

54. Defining Independence
•We can redefine independence in terms of
conditional probabilities:
Two events A and B are independentif and only if
P(A|B) = P(A)orP(B|A) = P(B)
Otherwise, they are dependent.
•Onceyou’vedecidedwhetherornottwoevents
areindependent,youcanusethefollowingrule
tocalculatetheirintersection.

55. The Multiplicative Rule for Intersections
•For any two events, Aand B, the probability that
both Aand Boccur is
P(A B) = P(A) P(B given that A occurred)
= P(A)P(B|A)
•IftheeventsAandBareindependent,thenthe
probabilitythatbothAandBoccuris
P(A B) = P(A) P(B)

56. Example 1
Inacertainpopulation,10%ofthepeoplecanbe
classifiedasbeinghighriskforaheartattack.Three
peoplearerandomlyselectedfromthispopulation.
Whatistheprobabilitythatexactlyoneofthethreeare
highrisk?
Define H: high riskN: not high risk
P(exactly one high risk) = P(HNN) + P(NHN) + P(NNH)
= P(H)P(N)P(N) + P(N)P(H)P(N) + P(N)P(N)P(H)
= (.1)(.9)(.9) + (.9)(.1)(.9) + (.9)(.9)(.1)= 3(.1)(.9)
2
= .243

57. Example 2
Supposewehaveadditionalinformationinthe
previousexample.Weknowthatonly49%ofthe
populationarefemale.Also,ofthefemalepatients,8%
arehighrisk.Asinglepersonisselectedatrandom.What
istheprobabilitythatitisahighriskfemale?
Define H: high riskF: female
From the example,P(F) = .49 andP(H|F) = .08.
Use theMultiplicative Rule:
P(high risk female) = P(HF)
= P(F)P(H|F) =.49(.08) = .0392

58. Counting Rules
•Sample space of throwing 3 dice has 216 entries,
sample space of throwing 4 dice has 1296 entries,
…
•At some point, we have to stop listing and start
thinking …
•We need some counting rules

59. The mn Rule
•If an experiment is performed in two stages,
with mways to accomplish the first stage and
nways to accomplish the second stage, then
there are mnways to accomplish the
experiment.
•This rule is easily extended to kstages, with
the number of ways equal to
n
1 n
2 n
3 … n
k
Example: Toss two coins. The total number of
simple events is:
2 2 = 4

60. Examples
Example:Tossthreecoins.Thetotalnumberof
simpleeventsis: 2 2 2 = 8
Example:TwoM&Msaredrawnfromadish
containingtworedandtwobluecandies.Thetotal
numberofsimpleeventsis:
6 6 = 36
Example:Tosstwodice.Thetotalnumberofsimple
eventsis:
4 3 = 12
Example:Tossthreedice.Thetotalnumberof
simpleeventsis:
6 6 6 = 216

61. Permutation:
Example:Alockconsistsoffivepartsand
canbeassembledinanyorder.Aquality
controlengineerwantstotesteachorderfor
efficiencyofassembly.Howmanyordersare
there?120)1)(2)(3)(4(5
!0
!5
5
5 P
The order of the choice is
important!

62. Example of Combinations
•AboxcontainssixM&Ms
®
,fourred
andtwogreen.AchildselectstwoM&Msat
random.Whatistheprobabilitythatexactly
oneisred?
The order of
the choice is
not important!Ms.&M 2 choose toways
15
)1(2
)5(6
!4!2
!6
6
2 C M.&Mgreen 1
choose toways
2
!1!1
!2
2
1
C M.&M red 1
choose toways
4
!3!1
!4
4
1 C
4 2 =8 ways to
choose 1 red and 1
green M&M.
P(exactly one
red) = 8/15

63. Example
A deck of cards consists of 52 cards, 13 "kinds"
each of four suits (spades, hearts, diamonds, and
clubs). The 13 kinds are Ace (A), 2, 3, 4, 5, 6, 7,
8, 9, 10, Jack (J), Queen (Q), King (K). In many
poker games, each player is dealt five cards from
a well shuffled deck. hands possible
960,598,2
1)2)(3)(4(5
48)49)(50)(51(52
)!552(!5
!52
are There
52
5 

C

64. Example
Fourofakind:4ofthe5cardsarethesame
“kind”.Whatistheprobabilityofgettingfourof
akindinafivecardhand?
and
Thereare13possiblechoicesforthekindof
whichtohavefour,and52-4=48choicesfor
thefifthcard.Oncethekindhasbeen
specified,thefourarecompletelydetermined:
youneedallfourcardsofthatkind.Thusthere
are13×48=624waystogetfourofakind.
Theprobability=624/2598960=.000240096

65. Example
Onepair:twoofthecardsareofonekind,the
otherthreeareofthreedifferentkinds.
Whatistheprobabilityofgettingonepairina
fivecardhand? kind that of cardsfour theof
twoof choices possible 6 are there
choice, given the pair; a have which toof
kind for the choices possible 13 are There
4
2C

66. Example
Thereare12kindsremainingfromwhichtoselect
theotherthreecardsinthehand.Wemustinsist
thatthekindsbedifferentfromeachotherand
fromthekindofwhichwehaveapair,orwe
couldendupwithasecondpair,threeorfourofa
kind,orafullhouse.

67. Example422569.
989601098240/25 y probabilit The
1,098,240. 64220613
is hands pair" one" ofnumber theTherefore
three.all of suits for the choices 64 4 of
totala cards, three thoseofeach ofsuit for the
choices 4 are There cards. threeremaining the
of kinds pick the to ways220 are There
3
12
3




C

68. Key Concepts
I. Experiments and the Sample Space
1. Experiments, events, mutually exclusive events,
simple events
2. The sample space
II. Probabilities
1. Relative frequency definition of probability
2. Properties of probabilities
a. Each probability lies between 0 and 1.
b. Sum of all simple-event probabilities equals 1.
3. P(A), the sum of the probabilities for all simple events in A

69. Key Concepts
III. Counting Rules
1. mnRule; extended mnRule
2. Permutations:
3. Combinations:
IV. Event Relations
1. Unions and intersections
2. Events
a. Disjoint or mutually exclusive: P(AB) 0
b. Complementary: P(A) 1 P(A
C
) )!(!
!
)!(
!
rnr
n
C
rn
n
P
n
r
n
r





70. Key Concepts
3. Conditional probability:
4. Independent and dependent events
5. Additive Rule of Probability:
6. Multiplicative Rule of Probability:)(
)(
)|(
BP
BAP
BAP

 )()()()( BAPBPAPBAP  )|()()( ABPAPBAP 