Principles of Statistics
Chapter 2
Elements of Probability
Basic Concepts:
Statistical Experiment: Process of collecting data
Sample Space (S): Collection of all outcomes ofthe experiment.
Event: Any subset of the sample space
Probability of the Event: Measure on the event
Probability Concept:
There are different approaches to defineprobability. These approaches are:
(1) Subjective probability:
The main idea of subjective probability is to letthe probability of an event reflect the personalbelief in the chance of the occurrence of theevent. For example, one may use personalprobability when evaluating the chance of raintomorrow.
(2) Statistical probability:
probability is the relative frequency of occurrence of anevent in a large number of trials.
P(A)= limit n(A)/N as N gets large, where n(A) is number ofelements in A and N the number of elements in S.
(3) Classical probability:
If the sample space of an experiment consists of finitesample points which are equally likely, then for anyevent we define the probability of as:
P(A)=n(A)/N, where n(A) is the number of outcomes in Aand 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 bysimply counting the number of different ways that acertain event occurs. Combinatorial methods play animportant role in computing probabilities wheneverequally likely outcomes can be identified and samplespace is finite. Now, we give a review of some countingmethods.
•A very basic technique that is frequently used incombinatorial problems is called the Multiplicationprinciple which states that if a first operation can beperformed in any of n1 ways, and and a secondoperation can be performed in any of n2 ways then bothoperations 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 samenumber
= n(A)/n(S)
= 0.093
Assume that we have an urn containing balls ofdifferent colors. We are interested in drawing a samplefrom this urn. There are different procedures to obtaina sample. We will describe some of these procedures.
(1) The balls may be drawn either sequentially, that is, oneat a time, or simultaneously, that is, all at a time.
(2) If the balls are drawn sequentially, then they may bedrawn with replacement or without replacement. Inthis case we obtain an ordered sample, that is, there isa first ball, a second ball, and so on.
(3) If the sampling is done simultaneously, then we obtainan unordered sample.
A permutation is an ordered arrangement of all orpart of a set of objects. The number ofpermutations of n distinguishable or differentobjects is
n!=n(n-1)(n-2) . . . (3)(2)(1)
Note that 0! = 1.
One may also be interested in the number ofpermutations of distinguishable objects taken r ata time. That is an ordered arrangements of r of aset n distinguishable objects. The number is calledn 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 havethe same birthday. Then
P(A)= n(A)/n(S)
= (365)(364) . . . (356)/(365^10)
Example: Two balls are drawn without replacementfrom an urn containing 6 balls, of which 4 arewhite 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 ofdifferent 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 ballsdrawn will be white.
Note that D = complement of B
P(D)= 1 – P(B)
=1 – 2/30
= 28/30
= 0.933