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Principles of Statistics
Chapter 2
Elements of Probability
Basic Concepts:
Statistical Experiment: Process of collecting data
Sample Space (S): Collection of all outcomes of
the experiment.
Event: Any subset of the sample space
Probability of the Event: Measure on the event
Probability Concept:
There are different approaches to define
probability. These approaches are:
(1) Subjective probability:
The main idea of subjective probability is to let
the probability of an event reflect the personal
belief in the chance of the occurrence of the
event. For example, one may use personal
probability when evaluating the chance of rain
tomorrow.
(2) Statistical probability:
probability is the relative frequency of occurrence of an
event in a large number of trials.
P(A)= limit n(A)/N as N gets large, where n(A) is number of
elements in A and N the number of elements in S.
(3) Classical probability:
If the sample space of an experiment consists of finite
sample points which are equally likely, then for any
event we define the probability of as:
P(A)=n(A)/N, where n(A) is the number of outcomes in A
and N is the number of outcomes in S.
Example:
Experiment of tossing a coin twice.
S={HH, HT, TH, TT}
A=getting at least one head
P(A)=3/4=0.75
Example: Experiment of throwing two dice
S={(1,1),(1,2), . . ., (1,6)
(2,1), (2,2), . . ., (2,6)
...
(6,1), (6,2), . . ., (6,6)}
A=sum of the points=7 - - -> P(A) = 6/36=1/6
Probability Techniques:
•
Many problems in probability theory can be solved by
simply counting the number of different ways that a
certain event occurs. Combinatorial methods play an
important role in computing probabilities whenever
equally likely outcomes can be identified and sample
space is finite. Now, we give a review of some counting
methods.
•
A very basic technique that is frequently used in
combinatorial problems is called the Multiplication
principle which states that if a first operation can be
performed in any of n1 ways, and and a second
operation can be performed in any of n2 ways then both
operations can be performed in (n1 n2) ways.
Example: A die is rolled 5 times. Compute the
probability that no two dice show the same number of
spots.
Solution: The sample space for this experiment is:
S={(x1,x2,x3,x4,x5): xi=1,2,3,4,5,6, i=1,2,3,4,5}
n(S)=6*6*6*6*6=(6^5)
n(A)=6*5*4*3*2
Probability of A =Probability of no two dice have the same
number
= n(A)/n(S)
= 0.093

Assume that we have an urn containing balls of
different colors. We are interested in drawing a sample
from this urn. There are different procedures to obtain
a sample. We will describe some of these procedures.
(1) The balls may be drawn either sequentially, that is, one
at a time, or simultaneously, that is, all at a time.
(2) If the balls are drawn sequentially, then they may be
drawn with replacement or without replacement. In
this case we obtain an ordered sample, that is, there is
a first ball, a second ball, and so on.
(3) If the sampling is done simultaneously, then we obtain
an unordered sample.
 A permutation is an ordered arrangement of all or
part of a set of objects. The number of
permutations of n distinguishable or different
objects is
n!=n(n-1)(n-2) . . . (3)(2)(1)
Note that 0! = 1.
 One may also be interested in the number of
permutations of distinguishable objects taken r at
a time. That is an ordered arrangements of r of a
set n distinguishable objects. The number is called
n permutation r (denoted by nPr)
nPr= n(n-1) . . . (n-r+1) = n!/(n-r)!
Example: There are 10 students in a class. What is the
probability that no two students in the class have the same
birthday? Assuming that each student in the class can have
as his birthday any one of the 365 days in the year, so we
ignore the existence of leap year, and that each day of the
year is equally likely to be the person's birthday.
Let A be the event that no two students in the class have
the same birthday. Then
P(A)= n(A)/n(S)
= (365)(364) . . . (356)/(365^10)
Example: Two balls are drawn without replacement
from an urn containing 6 balls, of which 4 are
white and 2 are red. Find the probability that:
(a) both balls will be white
(b) both balls will be of the same color
(c) both balls will be of different color
(d) at least one of the balls will be white
Solution:
(1) Let A: two balls are white.
P(A)=4P2/6P2=(4)(3)/6(5)
= 12/30
= 0.4
(2) Let B be the event that both balls drawn are red.
Then
P(B) =2P2/6P2 = 2/30
= 0.067
Hence,
P(both balls will be of the same color) = P(A) + P(B)
= 0.4 + 0.067
= 0.467
(3) Let C be the event that both balls drawn are of
different color. Then
P(C) = [(2)(4)+(4)(2)]/30
= 16/30
= 0.533
(4) Let D be the event that at least one of the balls
drawn will be white.
Note that D = complement of B
P(D)= 1 – P(B)
=1 – 2/30
= 28/30
= 0.933