Chapter 4Chapter 4

Nutan S. MishraNutan S. MishraDepartment of Mathematics and Department of Mathematics and

StatisticsStatisticsUniversity of South AlabamaUniversity of South Alabama

Experiments outcomes and Experiments outcomes and sample spacesample space

An experiment is a process when performed results into a unique outcome out of several possible outcomes.

For example when experiment consists of tossing a coin, there are two possible outcomes head and tail. When we perform this experiment, only one face will show up either head or tail.

A sample space of an experiment is the set of all possible outcomes. We denote such a set by S

Example: in the above experiment S = {H, T}

More examplesMore examplesExperiment: Roll a dieS = { 1,2,3,4,5,6}

Experiment: Toss two coinsS = { HH, HT, TH, TT}The first letter indicates outcome of first toss and

second letter indicates outcome of second toss.

Experiment: recording GPA of a studentS = (0,4)

An eventAn eventAn event is a collection of one or more

outcomes of an experimentAnd event may be simple or compound

Event is called simple event if it consists of only one final outcome of the experiment

Event is called compound if it consists of more than one out comes of an experiment

Example of simple and Example of simple and compound eventscompound events

Experiment: toss two coinsS = {HH, TT, HT, TH}Define event A = occurrence of two tailsThen A = { TT}We say that A has taken place whenever TT shows upA is a simple event.Define event B = occurrence of one head Then B = { HT, TH}We say that B took place whenever either HT or TH

shows up. Thus B is a compound event.

More ExampleMore ExampleExperiment : Roll a dieS = { 1,2,3,4,5,6}

Define event A = occurrence of square number. Then A = {1,4}

A is a compound event

Define event B = occurrence of an odd numberThen B = { 1,3,5}B is a compound event.

Solution to 4.2Solution to 4.2a. One roll of a die S = {1,2,3,4,5,6}b. Three tosses of a coin S = {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}c. One toss of a coin and one roll of a dieS ={H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6}

Solution to 4.3Solution to 4.3• Experiment : drawing two items from

the box containing three items• S = { AB, AC, BC}

Calculating probability of an eventCalculating probability of an eventProbability is the likelihood of occurrence of an

event. It’s the numerical measure of the likelihood that a

specific event will occur.Notation: denote by Ei all the simple events and by

A a compound event Denote by P(Ei) and P(A) corresponding

probabilities.1. Probability of an event always lies in the range

0 to 12. The sum of the probabilities of all simple event

of an experiment is always 1.

Solution to 4.4 , 4.10Solution to 4.4 , 4.10Denote a student by M type if suffers from

math anxiety and type N if does not suffer from math anxiety.

Experiment : selecting two studentsS = { MM, MN, NM, NN}4.10(a) event A = {MM} simple event4.10(b) event B = {MN, NM} compound event4.10(c) event C ={NM} simple event4.10(d) event D = {NN} simple event

Two properties of probabilityTwo properties of probability

1)(0

1)(0

AP

EP i

First property can be described as

Example of first property : In tossing a coin P(H) = .5, P(T) = .5

Second property can be described as

1...)()()()( 321 EPEPEPEP i

Example of second property: in tossing a coin P(H)+P(T) = 1

Example of two propertiesExample of two properties

Experiment consists of rolling a fair dieThenP(1) =1/6, P(2) = 1/6, P(3)=1/6, P(4) = 1/6,

P(5)=1/6, P(6) = 1/6P(1)+P(2)+P(3)+P(4)+P(5)+P(6) = 1

Classical probability RuleClassical probability Rule

Two outcomes that have equal probability of occurrence are called equally like outcomes.

Under the classical probability rule we assume that all outcomes of the experiment are equally likely.

Example: tossing a coin, total number of outcomes =2 . P(H ) = ½

experiment theof outcomes ofnumber total

1)( EiP

experiment theof outcomes ofnumber total

A tofavorable outcomes ofnumber )( AP

Example: classical probability ruleExample: classical probability rule

Experiment: rolling a dieDefine event E1 as occurrence of a 6 E1 = {6}P(E1) = 1/total number of outcomes = 1/6Define event A = occurrence of an even

number A = { 2,4,6}P(A) = 3/6

Relative frequency approachRelative frequency approach

If an experiment is repeated n times and an event A is observed f times, then according relative frequency concept of probability P(A) = f/n

Example: At an assembly plant 500 cars are selected and 10 are found lemon

Experiment : inspecting a car and declaring it lemon or good

This experiment is repeated 500 times and lemon cars found ten times then

P(lemon) = 10/500 and p(good) = 490/500

NoteNote• Under classical approach the

probability assigned to an event is exact probability

• Under relative frequency approach the probability assigned to an event is approximate probability.

Law of large numbersLaw of large numbersIf an experiment is repeated again and

again, the probability of an event obtained by relative frequency approach approaches to exact (true) probability.

Solution to 4.19, 4.20Solution to 4.19, 4.20-.55 can not be probability of any event because

probability of an event is never negative, it always lies between 0 and 1.

1.56 can not be probability of any event because this value is larger than one and probability of an event is always less than or equal to 1.

(4.20)Same is true with 1.42 and 9/4.5/3 can not be probability because its larger than 1-2/7 can not be probability because it’s a negative

value.(4.20)Same is true with -.09 and -1/4.

Solution to 4.21Solution to 4.21Experiment: passenger passing through metal

detector.S={alarm goes off, alarm does not go off}The two outcomes are not equally likely because most

of people take off all their metallic belongings and put into scan trays. Thus the chances that alarm goes off is much smaller than chances that it does not.

Thus we would prefer the relative frequency approach to find the probability of these two events

Solution to 4.31Solution to 4.31Experiment : selecting an answerS = {w,w,w,w, c}w = selecting a wrong answer c = selecting a correct answerOut of the five multiple choices there are four wrong

answers and one correct answerP(w) = 4/5 and P(c) = 1/5Sum of these two probabilities is one because there

are only two possible outcomes of this experiment. According to second property, they must sum up to one.

Exercise 4.32Exercise 4.32Experiment: selecting a professor and recording gender.S = {105 F’s, 215 M’s}Out of total of 320 professors 105 are F and (320-105=) 215 are

M If we select a professor randomly from 320 When we select an item randomly from a group of items means

each one of them has equal chance of being selectedP(F) = #outcomes favorable to F/total number of outcomes in S

=105/320P(M) =#outcomes favorable to M/total number of outcomes in S

= 215/320

Exercise 4.35Exercise 4.35Denote by FHF = free health fitness centersNHF= no free health fitness centersExpt: pick a company is check if it has FHF or notS = { FHF,NHF}This expt was repeated 400 times (relative

frequency approach) P(FHF) = 130/400P(NHF) = 270/400The sum of these two probabilities is 1 because

these two are the only two distinct outcomes of S.

Exercise 4.37Exercise 4.37Experiment: recording the # cards held by an adult.S = { 0c, 1c, 2c, 3c,4c, ≥5c }S lists the possible outcomes of the experiment that is the adult may

hold zero cards or one card ,or two cards, or three cards or four cards or more than four cards.

The experiment was repeated 820 times (i.e. 820 adults were asked the question)

And data was collected on 820 adults as follows

#cards #adults(f) Relative f

0 80 80/820

1 116 116/820

2 94 94/820

3 77 77/820

4 43 43/820

≥5 410 410/820

P(3c) = 77/820 =.094

P(≥5c ) = 410/820 = .5

Counting RuleCounting RuleRule: If an experiment consists of two steps and first

step consists of n1 outcomes and second step consists of n2 outcomes then there are n1*n2 final outcomes

Example: tossing two coins consists of two steps. First step first coin two outcomes Second step- second coin two outcomes.

Then |S| = # members in S = 2*2 =4 final outcomesExample: tossing a coin and rolling a die. This

experiment consists of two steps. First step toss coin –two outcomes. Second step roll a die – 6 outcomes

Thus |S| = 2*6= 12 final outcomes

Counting ruleCounting ruleRule : If an experiment consists of three

steps with first step with n outcomes, second step with m outcomes and third step with k outcomes then total number of outcomes for the experiment are n*m*k

Example : toss three coins. Experiment consists of three steps with two outcomes at each step thus total number of outcomes of the experiment are 2*2*2 =8

Marginal and conditional Marginal and conditional probabilitiesprobabilities

Consider a two way classification of data :“Do you smoke?” the question was asked to 100

adults and the responses are as follows yes No

Male 25 35

female 10 30

total 35 65 100

total

60

40

Marginal probabilityMarginal probability

Yes No total

Male 25 35 60

Female 10 30 40

Total 35 65 100

The totals in the fourth column and fourth row are called marginal totals

P(Male) = #males/total number of adults = 60/100

Is called marginal probability. Other marginal probabilities are

P(Female) = 40/1000

P(Yes) = 35/100

P(No) = 65/100

Conditional probabilityConditional probabilityQuestion asked: what the probability of a female smoking?

Thus given the condition that the adult chosen is a female then what the probability that she smokes.

This conditional probability denoted byP(Yes|Female) = (YesFemale)/# Females = 10/40Question: what is the probability of a smoker being female?P(Female|Yes) = (YesFemale)/# Yes = 10/35

Yes No total

Male 25 35 60

Female 10 30 40

Total 35 65 100

Yes No total

Male 25 35 60

Female 10 30 40

Total 35 65 100