Download Chapter 4 - University of South Alabama

Chapter 4
Nutan S. Mishra
Department of Mathematics and Statistics
University of South Alabama
Experiments outcomes and
sample 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 examples
Experiment: Roll a die
S = { 1,2,3,4,5,6}
Experiment: Toss two coins
S = { 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 student
S = (0,4)
An event
An event is a collection of one or more
outcomes of an experiment
And 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
compound events
Experiment: toss two coins
S = {HH, TT, HT, TH}
Define event A = occurrence of two tails
Then A = { TT}
We say that A has taken place whenever TT shows up
A 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 Example
Experiment : Roll a die
S = { 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 number
Then B = { 1,3,5}
B is a compound event.
Solution to 4.2
a. 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 die
S ={H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6}
Solution to 4.3
• Experiment : drawing two items from
the box containing three items
• S = { AB, AC, BC}
Calculating probability of an event
Probability 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 1
2. The sum of the probabilities of all simple event
of an experiment is always 1.
Solution to 4.4 , 4.10
Denote a student by M type if suffers from
math anxiety and type N if does not suffer
from math anxiety.
Experiment : selecting two students
S = { MM, MN, NM, NN}
4.10(a) event A = {MM} simple event
4.10(b) event B = {MN, NM} compound event
4.10(c) event C ={NM} simple event
4.10(d) event D = {NN} simple event
Two properties of probability
First property can be described as
0  P( Ei )  1
0  P( A)  1
Example of first property : In tossing a coin P(H) = .5, P(T) = .5
Second property can be described as
 P(E )  P(E )  P(E )  P(E )  ...  1
i
1
2
3
Example of second property: in tossing a coin P(H)+P(T) = 1
Example of two properties
Experiment consists of rolling a fair die
Then
P(1) =1/6, P(2) = 1/6, P(3)=1/6, P(4) =
1/6, P(5)=1/6, P(6) = 1/6
P(1)+P(2)+P(3)+P(4)+P(5)+P(6) = 1
Classical 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.
1
P( Ei) 
total number of outcomes of the experiment
P( A) 
number of outcomes favorable to A
total number of outcomes of the experiment
Example: tossing a coin, total number of outcomes
=2 . P(H ) = ½
Example: classical probability rule
Experiment: rolling a die
Define event E1 as occurrence of a 6
E1 = {6}
P(E1) = 1/total number of outcomes = 1/6
Define event A = occurrence of an even
number
A = { 2,4,6}
P(A) = 3/6
Relative 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
Note
• 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 numbers
If 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, 4.20
-.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 4.21
Experiment: 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 4.31
Experiment : selecting an answer
S = {w,w,w,w, c}
w = selecting a wrong answer
c = selecting a correct answer
Out of the five multiple choices there are four
wrong answers and one correct answer
P(w) = 4/5 and P(c) = 1/5
Sum 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 4.32
Experiment: 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 selected
P(F) = #outcomes favorable to F/total number of outcomes in S
=105/320
P(M) =#outcomes favorable to M/total number of outcomes in S
= 215/320
Exercise 4.35
Denote by FHF = free health fitness centers
NHF= no free health fitness centers
Expt: pick a company is check if it has FHF or not
S = { FHF,NHF}
This expt was repeated 400 times (relative
frequency approach)
P(FHF) = 130/400
P(NHF) = 270/400
The sum of these two probabilities is 1 because
these two are the only two distinct outcomes of
S.
Exercise 4.37
Experiment: 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 Rule
Rule: 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 outcomes
Example: 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 rule
Rule : 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
probabilities
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
total
Male
25
35
60
female
10
30
40
total
35
65
100
Marginal 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 probability
Question asked: what the probability of a female smoking?
Yes
No
total
Yes
No
total
Male
25
35
60
Male
25
35
60
Female
10
30
40
Female
10
30
40
Total
35
65
100
Total
35
65
100
Thus given the condition that the adult chosen is a female then
what the probability that she smokes.
This conditional probability denoted by
P(Yes|Female) = (YesFemale)/# Females = 10/40
Question: what is the probability of a smoker being female?
P(Female|Yes) = (YesFemale)/# Yes = 10/35