PROBABILITY

Dr. Manjula Gunarathna

INTRODUCTION

Probability theory was originated from gambling theory.

HISTORY OF PROBABILITY

Galileo (1564-1642) an Italian mathematician- first man to attempt

quantitative measure of probability

B. Pascal (1623-1662) and Pierre de Fermat (1601-1665) two

French mathematicians-systematic and scientific foundation of the

mathematical theory of probability

James Bernoulli (1654-1705) Swiss mathematician- treatise on

probability published

De Moivre (1667-1754) Dotcrines of Charles published in 1718

Thomas Bayes (1702-1761) - Inverse Probability

Pierre-Simon de Laplace (1749-1812)- Theory of Analytical

Probability

R.A. Fisher and Von Mises - empirical approach to probability

Chebychev (1821-1894) and A. kolmogorov. - Russian

mathematicians - modern theory of probability

THE UTILITY AND IMPORTANCE OF PROBABILITY IN

ECONOMICS

Predictions for future

It is very much used in economic decision making

It is extensively used in economic situations characterized by

uncertainty (viz, investment problem, inventory problem,

problem of introducing new product and so on)

It is the base of the fundamental laws of economics i.e. decision

theory

DEFINITION OF PROBABILITY

The probability when defined in the simplest way is chance or

occurrence of a certain event when expressed quantitatively.

The probability is defined in four different ways though its

approaches

1. Subjective (personalistic) approach

2. Classical (a priori) approach

3. Statistical (empirical) approach

4. Axiomatic (modern) approach

SUBJECTIVE (PERSONALISTIC) APPROACH

This approach is used to determine the probability of events which

have either not occurred at all in the past or which occur only once

or where experiment cannot be performed repeatedly under

identical conditions. J.M. Keynes and L.J. Savage have identified the

subjective probability as a measure of one’s confidence in the

occurrence of a particular event.

CLASSICAL (A PRIORI) APPROACH

If an experiment has n mutually exclusive, equally likely and

exhaustive cases, out of which m are favorable to the happening of

event A, then the probability of the happening of A is denoted by P

(A) and is defined as;

P(A) = m/n

P(A) = No. of favorable to A/Total number of cases

STATISTICAL (EMPIRICAL) APPROACH

Von Mises has give the following statistical definition.

“if the experiment be repeated a large number of times under

essentially identical conditions, the limiting values of the ratio of the

number of times the event E happens to the total number of trials of

the experiment as the number of trials increases indefinitely is called

the probability of happening of the E”

AXIOMATIC (MODERN) APPROACH

The Russian mathematician A.N. Kolmogorv introduced this new

modern approach through the theory of sets in 1983. The modern

definition of probability includes both the classical and the statistical

definitions as particular cases overcomes the deficiencies of each of

them. It is based on certain axioms. The advantage of the axiomatic

theory is that it narrates all situations irrespective of whether the

outcomes of an experiment are equally likely or not.

(i) for all i 0 ≤ p(si) ≤ 1

(Probability for simple event is 0 to 1)

(ii) ∑p (si) = 1

(sum of probability is equal one)

SOME IMPORTANT TERMS AND CONCEPTS

Experiment

The term experiment refers to processes which result in different

possible out come or observation.

Ex; tossing a coin, or throwing a dice

SAMPLE SPACE (S)

A set of all possible outcomes from an experiment is called a Sample

Space. Let us toss a coin, the result is either head or tail. Let 1 denote

head and 0 denote tail.

S={0"1}

throwing a dice

S={1"2"3"4"5"6}

Mark the point 0, 1 on a Straight line. These Points are called Sample

Points or Event Points.

For a given experiment there are different possible outcomes and

hence different sample points. The collection of all such sample

points is a Sample Space.

DISCRETE SAMPLE SPACE

A sample Space whose elements are finite or infinite but countable

is called a discrete Sample Space.

For example, if we toss a coin as many times as we require for

turning up one head, then the sequence of points S1=(1), S

2 = (0,1), S

3

= (1,0,0), S4 = (0,0,0,1) etc. , is a discrete Sample Space.

CONTINUOUS SAMPLE SPACE

A sample space whose elements are infinite and uncountable or

assume all the values on a real line R or on an interval of R is called a

Continuous Sample Space. In this case the sample points build up a

continuum, and the sample space is said to be continuous.

Let us toss a coin, and throw a dice the result is (sample space)

S = { H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

EVENT

A sub-collection of a number of sample points under a definite rule

or law called an event.

For example, let us take a dice. Let its faces 1,2,3,4,5,6 be

represented by E1,E

2,E

3,E

4,E

5,E

6 respectively. Then all the E

i’s are

sample points. Let E be the event of getting an even number on the

dice. Obviously, E= {E2, E

4, E

6}, which is a subset of the set {E

1, E

2,

E3,

E4, E

5, E

6}

NULL EVENT

An event having no sample point is called a null event

SIMPLE EVENT

An event consisting of only one sample point of a sample space is

called a simple event.

For example, let a dice be rolled once and A be the event that face

number 5 is turned up, then A is a Simple event.

COMPOUND EVENTS

When an event is decomposable into a number of simple events,

then it is called a compound event.

For example, the sum of the two numbers shown by the upper faces

of the two dice is seven in the simultaneous throw of the two

unbiased dice, is a compound event as it can be decomposable.

EQUALLY PROBABLE EVENTS.

If in an experience all possible outcomes have equal chances of

occurrence, then such events are said to be equally probable events.

For example, in throwing a coin, the events head and tail have equal

chances of occurrence, therefore, they are equally probable events.

FAVORABLE CASES

The cases which ensure the occurrence of an event are said to be

favorable to the event.

INDEPENDENT AND DEPENDENT EVENTS

Two or more events are said to be independent if the happening of

any one does not depend on the happening of the other.

Events which are not independent are called dependent events.

QUESTIONS

1. Two dice are thrown together. Find the probability that we get a

total of 9

2. Out of a sample of 200 items, 20 items are found to be defective;

find the probability that an item chosen at random from the sample

is not defective.

3. A DISCRETE RANDOM VARIABLE X HAS THE

PROBABILITY FUNCTION GIVEN BELOW;

X 0 1 2 3 4 5 6 7 8

p(x) a 3a 5a 7a 9a 11a 13a 15a 17a

(i) Find the value of a

(ii) Find P(x < 3)

(iii) Find P(0 < x < 3)

(iv) Find P (x ≥ 3) and

(V) FIND THE DISTRIBUTION FUNCTION OF X

4. A RANDOM VARIABLE X ASSUMES THE VALUES

-3, -2, -1, 0, 1, 2, 3 AND

P(X=-3) = P(X=-2) = P(X=-1)

P(X=3) = P(X=2) = P(X=1)

P(X=0) = P(X>0) = P(X<0)

Find

(i) P(x < 3)

(ii) P(0 < x < 3)

(iii) P (x ≥ 3)

5. A discrete random variable X has the probability function given

below

Find

(i) The value of b

(ii) p( x< 6), p (x ≥ 6) and p (0<x<4)

(iii) The distribution function of X

6. A bag contains 6 white balls, 9 black balls. What is the probability

of drawing a black ball?

7. FIND THE PROBABILITY THAT

1. A leap year has 53 Sundays?

2. A non leap year has 53 Sundays?

3. A leap year has 53 Sundays or Mondays?

4. A non leap year has 53 Sundays or Mondays?

5. A year chosen at random has 53 Sundays?

6. A year chosen at random has 53 Sundays or Mondays?

MUTUALLY EXCLUSIVE EVENTS

If in an experiment the occurrence of an event prevents or rules out

the happening of all other events in the same experiment, then these

events are said to be Mutually Exclusive Events.

xi 0 1 2 3 4 5 6 7

P(xi) b b 2b 3b 3b 4b

2

4b2

4b2

+b

For example, in tossing a coin the event head and tail are mutually

exclusive, because if the outcome is head, then the possibility of

getting a tail in the same trial is ruled out.

ADDITION THEOREM OR THEOREM ON TOTAL PROBABILITY

If n events are mutually exclusive, then the probability of happening

of any one of them is equal to the sum of probabilities of the

happening of the separate events.

P(A or B) = P (A) + P (B)

P (A B) = P(A) + P (B)

EXAMPLE: A dice is rolled. What is the probability that a number 1

or 2 may appear on the upper face

P (A) = The probability of appearing the number 1 on the upper face

P (B) = The probability of appearing the number 1 on the upper face

P (A) = 1/6

P (B) = 1/6

P (AB) = P (A) + P (B) (By addition rule)

= 1/6 + 1/6 = 1/3

ADDITION THEOREM FOR COMPATIBLE EVENTS

The probability of the occurrence of at least one of the events A and

B (not mutually exclusive) is given by

P ( AB) = P (A) + P (B) – P (AB)

P ( A or B) = P (A) + P (B) – P (A and B)

INDEPENDENTS EVENTS

If two events say A and B are independent, then

P (A and B) = P (A). P (B)

P (AB) = P (A). P (B)

COMPLEMENTARY EVENTS

The events ‘A occurs’ and the event ‘A does not occur’ are called

complementary events.

P (A)’ + P (A) = 1

P (A)’ = 1 – P (A)

CONDITIONAL PROBABILITY

The probability of the happening of an event B, when it is known that

A has already happened, is called the conditional probability of B and

is denoted by p (B/A)

P (B/A) = P (A B) / P (A)

QUESTION

Let A and B be events with P(A) = 1/3, P(B) = 1/4, P(A B ) =

1/12,

Find P (A/B)

P (B/A)`

FACTORIAL N

Factorial n is the continued product of first n natural numbers.

Factorial n is symbolically written as n!

n! = n (n-1) (n -2) ………….. 3.2. 1

by definition

0!=1

5! = 5.4.3.2.1 = 120

4! 3! = (4.3.2.1) (3.2.1) = 144

n! = n (n-1) !

5! = 5.4! = 120

QUESTION

In how many ways can the letters in the word: STATISTICS be

arranged?

APPLICATION OF PERMUTATION AND COMBINATION

A Permutation is an arrangement of items in a particular order.

To find the number of Permutations of n items chosen r at a time,

you can use the formula

A Combination is an arrangement of items in which order does

not matter.

Since the order does not matter in combinations, there are fewer

combinations than permutations. The combinations are a

"subset" of the permutations.

To find the number of Combinations of n items chosen r at a time,

you can use the formula

. 0 where nrrn

nrpn

)!(

!

. 0 where nrrnr

n

rC

n

)!(!

!

QUESTIONS:

1. A committee of 4 persons is to be appointed from 7 men and 3

women. What is the probability that the committee contains

(i) exactly two women and

(ii) at least one woman

2. A committee including 3 boys and 4 girls is to be formed from a

group of 10 boys and 12 girls. How many different committee can be

formed from the group?

3. There are 9students in a class: 5 boys and 4 girls.

If the teacher picks a group of 4 at random, what is the

probability that everyone in the group is a boy?

4. What is the total number of possible 4-letter arrangements of the

letters

m, a, t, h, if each letter is used only once in each arrangement?

5. Christopher is packing his bags for his vacation. He has 8 unique

shirts, but only 5 fit in his bag.

How many different groups of 5 shirts can he take?

MATHEMATICAL EXPECTATION OR EXPECTED VALUES

Mathematical expectation of a random variable is obtained by

multiplying each probable value of the variable by its

corresponding probability and then adding these products.

N

E(X) = ∑ XI P(XI) I=1

Variance

V(x) = var (x) = E(x2) – [E(x)]

2

N

E(X2) = ∑ XI

2 P(XI)

I=1

THEOREMS ON MATHEMATICAL EXPECTATION

1. Expected value of constant term is constant, that is, if C is

constant, then

E(C)= C

2. If C is constant, then

E(CX) = C.E(X)

3. If A and B are constants, then

E(aX ± b)= a.E(X) ± b

4. If a, b and c are constants, then

E {(aX+b)/c} = 1/c {a E(x) + b}

5. If X and Y are two random variables, then

E(X+Y) = E(X)+E(Y)

6. If X and Y are two independent random variables, then

E(XY) = E(X).E(Y)

THEOREMS ON VARIANCE OF A RANDOM VARIABLE

1. If c is constant then,

V(CX) = C2V(X)

2. Variance of constant is zero

V(C) = 0

3. If X is a random variable and C is a constant then,

V(X+C)= V(X)

4. IF A AND B ARE CONSTANTS THEN

V(AX+B) = A2 V(X)

5. IF X AND Y ARE TWO INDEPENDENT RANDOM

VARIABLES, THEN

V(X+Y) = V(X)+V(Y)

V(X-Y) = V(X)+V(Y)

QUESTION:

THE PROBABILITY DISTRIBUTION OF A RANDOM

VARIABLE X IS GIVEN BELOW. FIND

1. E(X)

2. V(X)

3. E (2X-3)

4. V(2X-3)

X -2 -1 0 1 2

P(X) 0.2 0.1 0.3 0.3 0.1

BAYES’ THEOREM (INVERSE PROBABILITY THEOREM)

British mathematician thomas bayes (1702-1769)

Let A1, A

2….A

k be the set of n mutually exclusive and exhaustive

events whose union is the random sample space S, of an experiment.

If B be any arbitrary event of the sample space of the above

experiment with P(B) ǂ 0, then the probability of event Ak, when the

event B has actually occurred is given by P(Ak/B), where

P(Ak/B)= P(B∩A

k) = P(B/A

k)P(A

k)

p(B) {P(B/A1)P(A

1)+ P(B/A

2)P(A

2)… P(B/A

k)P(A

k}

QUESTIONS:

1. A desk lamp produced by The Luminar Company was found to be

defective (D). There are three factories (A, B, C) where such desk lamps

are manufactured. A Quality Control Manager (QCM) is responsible for

investigating the source of found defects. This is what the QCM knows

about the company's desk lamp production and the possible source of

defects:

The QCM would like to answer the following question: If a randomly

selected lamp is defective, what is the probability that the lamp was

manufactured in factory B?

Factory % of total

production Probability of

defective lamps

A 0.50 = P(A) 0.02 = P(D | A)

B 0.40 = P(B) 0.04 = P(D | B)

C 0.10 = P(C) 0.05 = P(D | C)

2. In a factory which manufactures bolts, machine A, B and C

manufacture respectively 25%, 35% and 40% of the bolts. Of their

outputs 5, 4 and 2 per cent are respectively defective bolts. A bolt

is drawn at random from the product and is found to be defective.

What is the probability it is manufactured by the (i) machine A,

(ii) machine B, (iii) machine C, (iv) manufactured by machine B

or C.

Event Prior

Probability

Conditional

Probability

Joint

Probability

Posterior

(revised)

Probability

(1) (2) (3) (4) = (2) x

(3)

(5) = (4) / P

(D)

A P (A) =

0.25

P (D/A) =

0.05

0.0125 P (A/D) =

0.36

B P (B) = 0.35 P (D/B) =

0.04

0.0140 P (B/D) =

0.41

C P (C) =

0.40

P (D/C) =

0.02

0.0080 P (C/D) =

0.23

Total 1.0 P (D) =

0.0345

1.0

3. Suppose there is a school with 60% boys and 40% girls as its

students. The female students wear trousers or skirts in equal

numbers; the boys all wear trousers. An observer sees a (random)

student from a distance, and what the observer can see is that this

student is wearing trousers. What is the probability this student is a

girl?

4. Marie is getting married tomorrow, at an outdoor ceremony in the

desert. In recent years, it has rained only 5 days each year.

Unfortunately, the weatherman has predicted rain for tomorrow.

When it actually rains, the weatherman correctly forecasts rain 90%

of the time. When it doesn't rain, he incorrectly forecasts rain 10% of

the time. What is the probability that it will rain on the day of Marie's

wedding?

Binomial distribution

The binomial distribution describes the behavior of a count variable X if

the following conditions apply:

1: The number of observations n is fixed.

2: Each observation is independent.

3: Each observation represents one of two outcomes ("success" or

"failure").

4: The probability of "success" p is the same for each outcome.