What is PROBABILITY?It is a quantitative measure of uncertainty.

ETYMOLOGYThe word PROBABILTY derives from PROBITY

HISTORY OF PROBABILITYMIDDLE OF 17TH CENTURY

RICHARD JEFFRYterm PROBABLE-means APPROVABLE

1774PIERRE-SIMON LAPLACE

Made the first attempt to deduce a rule for the combinations of observations.

INTERPRETATIONSFREQUENTISTS

-dealing with experiments that are random and well-defined.

BAYESIANS-individual’s degree of belief

APPLICATIONS OF PROBABILITYRISK ASSESSMENT

TRADE ON COMMODITY MARKETS

RELATION OF PROBABILITY ON RISK MANAGEMENT

Definition of TermsDefinition of Terms

Is any action or process that generates observations or set of data

Random Experiment Random Experiment

Is an experiment whose outcome cannot be predicted with certainty,but the experiment is such a nature that the collection of every possible outcomes can be described prior to its performance.

ExperimentExperiment

Sample SpaceSample SpaceIs the set of all possible outcomes of a

statistical experiment.It is collections of all conceivable outcome of an experiment.

ElementElement

Each outcome in sample space also called as sample point.

EventEventIs any collection of outcomes contained in the

sample space.

It consist exactly one outcome.

Simple EventSimple Event

Compound EventCompound EventIt consist more than one outcome.

ComplementComplementThe complement of an event A, A’ ,with respect

to the sample space,S,is the subset of all elements of S that are not in A.

The intersection of two events A and B,AΩB,is the event containing all elements that are common to both A and B.

IntersectionIntersection

Example:Let S={1,2,3,4,5,6,7,8,9,0} and let

R={2,4,6,8,0}R’=(1,3,4,7,9)

Example:Let S={1,2,3,4,5,6,7,8,9,0} and let

R={2,4,6,8,0}AΩB= (2,4,6,8,0)

Mutually Exclusive or DisjointMutually Exclusive or DisjointTwo events, A and B are mutually exclusive or

disjoint if A ΩB =ø,that is,if A and B have no elements in common.

The union of the two events A and B,A U B, is the event containing all the elements that belong to A or B or both .

UnionUnion

Example:Let S={1,3,5,7,9} and let R={2,4,6,8,0}

A ΩB =ø

Example:Let S={1,2,3,4,5,6,7,8,9,0} and let

R={2,4,6,8,0}A U B=(1,2,3,4,5,6,7,8,9,0)

Counting Sample PointsTechniques in counting sample points

Multiplication RuleMultiplication RuleIs the fundamental principle of countingTheorem #1If an operation in n1 ways,and if for each of these operation can

be performed in n 2 ways,then the two operations can be performed in n1 n 2 ways.

Example:How many sample points are in the sample space when a fair

coin is tossed twice?Solution:n1 n 2 =6(6)ways= 36 possible ways

Generalized Multiplication RuleGeneralized Multiplication RuleIt is used when involving operations is more than two.

Theorem #2If an operation in n1 ways,and if for each of these

operation can be performed in n 2 ways,and for each of the first two a third operation can be performed in n 3 ways and so forth then the sequence of k operations can be performed in n1 n 2 n 3 …….ways.

Example:

How many 3 digits can be formed from the digits 1, 2, 5, 6, and 9 if each no. can only used once?

Solution:

n1 n 2 n 3 =5(4)(3)ways =60 possible ways

PermutationPermutationIt is the arrangement of all or part of a set of objects in

a definite order.

Order is important aspect of permutation.

Theorem #3The number of permutation of n distinct objects is n!

Example:

Four people are to be arrange in a row to have their picture taken. In how many ways can this done?

Solution:

n= 4 people to be arrange in definite order

n!= 4!=4(3)(2)(1)=24 ways

PermutationPermutationTheorem #4

The number of permutation of n distinct objects taken r at a time is

n Pr = n!

(n-r)!

Example:

Five applicants enter an interview room in which there are seven seats.In how many ways may they be seated?

Solution:

n= 7;r=5

7P5 = 7! = 2520 ways

(7-5)!

Circular Circular PermutationPermutationPermutations that occur by arranging objects

in a circle are called circular permutation .Theorem # 5The number of permutation of n distinct

objects arranged in a circle is (n-1)!.

Example:Seven children join hands and form a circle. In

how many ways could the circle be formed?Solution:n=7(n-1)!= (7-1)!= 6! = 720 ways

Distinct Distinct PermutationPermutationTheorem # 6The number of distinct permutation of n things of which n 1 are

the same kind, n 2 of a second kind……., n k of a second kind,…..n k of a kth kind is

n! n 1 ! n 2!....... n k

Example:Eight books are to be arrange on a shelf. There are 2 math

identical books,3 identical English books and 3 identical Physics books. How many distinct arrangement are possible?

Solution:n= 8 booksn 1 =2 math books

n 2 =3 english books

n 3 =3 physics books

P= 8! = 560 arrangements 2! 3! 3!

CombinationCombinationIt is arrangement of number of ways of selecting r

objects from n without regard to order.Theorem # 7The number of combination of n distinct objects taken r

at a time is

n C r= n!

r!(n-1)!

Example:In how many ways a committee of 3 can be chosen from

a group of 8?Solution:n= 8 ; r= 3

8 C 3= 8! = 56 different ways

3!(8-1)!

Probability of an event

P(A) = The Number Of Ways Event A Can Occur

The Total Number Of Possible Outcomes

P(A)= n__ S

Basic Properties of Probability Certainty- this property states that the probability

that the event is on the sample space is one.P(S)=1 Non-negativity- this property states that every

event has a non-negativity probability. The lowest probability of an event is 0 or no chance and the highest probability of an event is 1 or certain.

P(E)≤0, for all E c S Union- the probability of the union of mutually

exclusive events is the sum of each event’s probability.

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

Example 1:A fair dice is tossed once. What is the probability that a. ) 5 will appearb.) an even number will appear?Solution:S={1,2,3,4,5,6}= 6a.) If A is the event that a 5 will appearA= {5}n(A)= 1P(A)= n(A) = 1

S 6b.) If B is the event that an even number will appearB= {2,4,6}n(B)= 3P(A)= n(A) =3 = 1

S 6 2

Example 2:Suppose that a card is drawn at random from an ordinary

deck of playing cards, What is the probability of drawing

a. ) a heartb.) a red cardSolution:S=52P(heart)= n(heart) = 13 = 1

S 52 4P(red card)= n(red card) = 26 = 1

S 52 2

ADDITIVE RULESADDITIVE RULESADDITIVE RULESADDITIVE RULES

ADDITIVE RULES• It is applicable to union of events

THEOREM

• If A and B are any two events, then P(AυB) = P(A)+P(B)-P(AΩ B)

•For three events A, B, and C, P(AυBυC) = P(A)+P(B)+P(C)-P(A Ω B)-P(A Ω C)-P(B Ω C)+P(AυBυC)

•If A and B are complementary events, then P(A)+P(A’)=1

EXAMPLES:ADDITIVE RULE1. The probability that Mel passes Calculus is 2/3 and

the probability that she passes statistics is 4/9. If the probability of passing both courses is ¼, what is the probability that Mel will pass at least one of these subjects?

2. If the probabilities are 0.9, 0.15, 0.21, and 0.23 that a person purchasing a new automobile will choose the color green, white, red and blue, what is the probability that a given buyer will purchase a new automobile that comes in one of those colors? (mutually exclusive)

Solution #1• Let A be the event that Mel passes Calculus• Let B the event that Mel passes Statistics• P(A U B)=P(A) + P(B) -P(A Ω B)

2 + 4 -13 9 4=31 36

CONDITIONAL PROBABILITY

CONDITIONAL PROBABILITY

• The probability of an event B occurring when it is known that some event A has occurred is called conditional probability and is denoted by P(B/A).

• The conditional probability of A, given B, denoted by P(A/B), P(A/B) = P(B Ω A)

P(B)If P(B)>ø

• Two events A and B are independent if and only if P(B/A) = P(B) and P(A/B) = P(A). Otherwise, A and B are dependent.

EXAMPLES:CONDITIONAL PROBABILITY1. A fair die is tossed once. Find the probability that a 4

appear, when it is known that a number greater than 2 results in the toss of the die.

2. The probability that a regularly scheduled flight departs on time is P(A)= 0.83, the probability that it arrives on time is P(B) = 0.82; and the probability that it departs and arrive on time is P(A ΩB) = 0.78. Find the probability that a planea) arrives on time given that it departed on time.b) departed on time given that it has arrived on time.

Solution #1• S=(1,2,3,4,5,6)• Let A be the event that 4 appear in a single toss.• A=(4)• Let B the event that a number is greater that 2 appear.• B=(3,4,5,6)• A ΩB= (4)

• 1 • P(A/B)= 6 = 1• 4 4 • 6

RANDOM VARIABLES

(r.v) is a function whose value is a real number determined by each element in a sample space.

DISCRETE SAMPLE SPACEIf the sample space contains a finite number

of possibilities or a n unending sequence with the elements as there are whole numbers.

CONTINUOUS SAMPLE SPACE• If a sample space contains an infinite number

possibilities equal to the number of points on a line segment.

DISCRETE RANDOM VARIABLEIf the set of possible outcome is countable

and it refers to the count data.

CONTINUOUS RANDOM VARIABLE

• if it takes the values of on a continuous scale and it refers to measured data.

Example:

1. A shipment of 8 similar microcomputers to a retail outlet contains 3 that are defective. If a school makes a random purchase of 2 of these computers, find the probability distribution for the number of defectives.

Solution:Let X be a random variable whose values are the possible

numbers of defective computers purchased by the school. Then X can be any of the numbers 0,1, and 2.

f(0)= P(X=0) = 3 C 0 * 5 C 2 = 5

8 C 2 14

f(1)= P(X=1) = 3 C 1 * 5 C 1 = 15

8 C 2 28

f(2)= P(X=2) = 3 C 2 * 5 C 0 = 3

8 C 2 28Therefore,the probability distribution of X is

We can write P(X=x)= f(x)

X 0 1 2

P(X=x) 5/14 15/28 3/28

Mathematical Expectation

Average no. of Heads =

ExperimentNo Heads - 4One Head - 7Two Heads - 5

16

0(4)+1(7)+2(5

)16

• Average Value is referred to as the Mean or Expected Value or the Mean of the Random variable of x or the Mean of the probability of distribution of x.

Mean or Expected Value of X is

µ=E(x)= ∑xf(x) if x is discrete

µ=E(x)=∫ xf(x)dx if x is continuous

x∞

-∞

Mean of the Function on Random Variable

Let x =be the random variableThen the variance of x is

σ 2 = ∑(x-µ) 2 f(x), if x is discrete

σ 2 = ∫ (x-µ) 2 f(x)dx, if x is continuous

x∞

-∞

Variance of the Random Variable

Let g(x) =be the function of random variableThen the expected value of x is

E (g(x))= ∑ g(x) f(x), if x is discrete

E (g(x))= ∫ g(x) f(x) dx, if x is continuous

x∞

-∞

Example 1If the probability mass function of x is given by:

f(x)= 1/3, for x=1,2,30, elsewhere

Find the mean and variance of x.

Solution:

µ=E(x)= ∑xf(x) = [1(1/3)+2(2/3)+3(1/3) µ = 2

σ 2 = ∑(x-µ)2 f(x) =(1-2)2(1/3)+(2-2)2(1/3)+(3-2)2(1/3) =1/3+0+1/3 σ 2 =2/3

Example 2Suppose that the probability density function of x is given by:

f(x)= 1/2, for -1<x<10, elsewhere

Find the mean and variance of x.Solution:

µ=E(x)= ∫ xf(x)dx

= ∫ x(1/2)dx

= x2/4] µ = 0

σ 2 = ∫ (x-µ) 2 f(x)dx

= ∫ (x-0)2(0)dx + ∫ (x-0)2(1/2)dx + ∫ (x-0)2(0)dx

= x3/6]

=1/6-(-1/6) σ 2 =1/3

∞

-∞

∞

-∞

1 -1

1

-1

1

-1

1

-1

∞

1

-1

-∞

Example 3

If the probability mass function of x is given by:

f(x)= 1/6, for x=1,2,30, elsewhere

Let g(x)= x2

Find the mean of g(x).

Solution:

E (g(x))= ∑ g(x) f(x)

= ∑ x2 (x/6)

=[(1)2(1/6)+(2)2(2/6)+(3)2(3/6)]

=1/6+8/6+27/6

E (g(x))= 6

Example 4

Given the probability density function of x :

f(x)= 2(1-X), for 0<x<10, elsewhere

Let g(x)= x2

Find the mean of g(x).

Solution:

E (g(x))= ∫ g(x) f(x) dx

E (x2) = ∫ x 2 [2(1-x)]dx

= ∫ (2x 2-2x 3)dx

=2x 3-2x4 3 4

= 2/3-1/2

E (x2) =1/6

1

0

∞

-∞

1

0

1

0