PROBABILITY

The Probability of a given event is an expression of likelihood of occurrence of an event. A probability is a number which ranges from 0 (zero) to 1 (0ne) -- zero for an event which cannot occur and 1 for an event which is certain to occur.

DEFINITION

APPROACHES TO PROBABILITY

THE CLASSICAL APPROACH

RELATIVE FREQUENCY THEORY

SUBJECTIVE APPROACH

AXIOMATIC APPROACH

THE CLASSICAL APPROACH

The definition of probability given by disciples of the classical school runs as follows : “ Probability” it is said, is the ratio of the number of “favourable’ cases to the total number of equally likely cases.

The term “equally likely”, though undefined, conveys the notion that each outcome of an experiment has the same chance of appearing as any other.

THE CLASSICAL APPROACH

If probability of occurrence of A is denoted by P(A), then by the definition , we have : P(A)= Number of favourable cases

Total number of qually likely cases

For example, If a coin is tossed, there are two equally likely cases, a head or a tail, hence the probability of a head is ½ and the probability of toss is ½.

RELATIVE FREQUENCY THEORY

If an experiment be repeated a large number of times under essentially identical conditions, the limiting value of the ratio of the number of times an event A occurs to the number of times the experiment is conducted as the number of trials of the experiment increases indefinitely, is called the probability of occurrence of A

Thus if n is the number of times an experiment is conducted and m is the number of times an event A occurs, then probability of occurrence of the event A is given by:

P(A)= lim m n→∞ n

SUBJECTIVE APPROACH

The Subjective School of thought is also known as the personalistic school of probability. The Subjective probability is defines as the probability assigned to an event by an individual based on the beliefs of the person making the probability statement.

For Example, if a teacher wants of find probability of Mr. X toping in M.Com. Examination in Delhi University this year, he may assign a value between his degree of beliefs for possible occurrence. He may take into account the such factors as the past academic performance, the view of his other colleagues.

Axiomatic approach

The Axiomatic approach introduced by Russian mathematician A.M. Kalmogorov in the twentieth Century.

Definition: Let S be the sample space. S is called a Probability space It for every event A in S, there is number P(A) with the following three Properties:(i) 0 < P(A)(ii) P(S) = 1(iii) P( AUB)= P(A)+P(B) Whenever A and B are

two events.

IMPORTANCE OF PROBABILITY

PROBABILITY THEORY HAS BEEN DEVELOPED AND EMPLOYED TO TREAT AND SOLVE MY WEIGHTED PROBLEMS.

PROBABILITY IS THE FOUNDATION OF THE CLASSICAL DECISION PROCEDURES OF ESTIMATION AND TESTING.

PROBABILITY MODELS CAN BE VERY USEFUL FOR MAKING ANY PREDICTIONS.

MUTUALLY EXCLUSIVE EVENTS

TWO EVENTS ARE SAID TO BE MUTUALLY EXCLUSIVE EVENTS OR DISJOINT EVENTS WHEN EVENTS CANNOT HAPPENS SIMULTANEOUSLY IN A SINGLE TRAIL.

EXAMPLE :-IF A SINGLE COIN IS TOSSED EITHER HEAD CAN BE UP OR TAIL CAN BE UP, BOTH CANNOT BE UP AT THE SAME TIME.

INDEPENDENT AND DEPENDENT EVENTS.

TWO EVENTS ARE SAID TO BE INDEPENDENT WHEN THE OUTCOME OF ONE DOES NOT AFFECT OTHER AND IS NOT AFFECTED BY THE OTHER. EX:- IF A COIN IS TOSSED TWICE, THE RESULT OF SECOND THROW WOULD IN NO WAY BE AFFECTED BY FIRST THROW.

DEPENDENT EVENTS ARE THOSE IN WHICH THE OCCURRENCE OR NON-OCCURRENCE OF ONE EVENT IN ANY TRIAL AFFECTS THE PROBABILITY OF OTHER EVENT IN OTHER TRIALS.

EX:- IF A CARD IS DRAWN FROM A PACK OF PLAYING CARDS AND IS NOT REPLACED THEN THIS WILL ALTER THE PROBABILITY OF SECOND CARD DRAWN.

EQUALLY LIKELY EVENTS.

EVENTS ARE SAID TO BE EQUALLY LIKELY WHEN ONE DOES NOT OCCUR MORE OFTEN THAN OTHERS.

EX:- IF AN UNBIASED COIN OR DIE IS THROWN, EACH FACE OR NO. IS EXPECTED TO BE OCCUR SAME NUMBER OF TIME IN THE LONG RUN.

SIMPLE AND COMPOUND EVENTS

IN SIMPLE EVENTS, WE SIMPLY CONSIDER THE PROBABILITY OF HAPPENING OR NON-HAPPENING OF A SINGLE EVENT.

IN COMPOUND EVENTS ,WE CONSIDER THE JOINT OCCURRENCE OF TWO OR MORE EVENTS. EX:- IF A BAG CONTAINS 10 WHITE BALLS AND 6 BLACK BALLS AND TWO SUCCESSIVE DRAWS OF 3 BALLS ARE MADE , WE SHALL FIND THE PROBABILITY OF 3 WHITE BALLS IN FIRST DRAW AND 3 BLACK BALLS IN SECOND DRAW.

EXHAUSTIVE EVENTS

EVENTS ARE SAID TO BE EXHAUSTIVE EVENTS WHEN THEIR TOTALITY INCLUDES ALL THE POSSIBLE OUTCOMES.

EX:-WHLE THROWING A DIE, THE POSSIBLE OUTCOMES ARE 1,2,3,4,5,6 AND HENCE THE EXHAUSTIVE NUMBER OF CASES IS 6.

COMPLEMENTARY EVENTS

IF THERE ARE TWO EVENTS A AND B. A IS CALLED THE COMPLEMENTARY EVENT OF B ( and vice versa ) if A AND B ARE MUTUALLY EXCLUSIVEAND EXHAUSTIVE.

EX:- WHEN A DIE IS THROWN, OCCURRENCE OF AN EVEN NUMBER(2,4,6) AND ODD NUMBER( 1,3,5) ARE COMPLEMENTARY EVENTS.

ADDITION THEOREM

THE ADDITION THEOREM STATES THAT IF TWO EVENTS ARE MUTUALLY EXCLUSIVE THAN THE PROBABILITY OF OCCURRENCE OF EITHER A OR B IS THE SUM OF INDIVIDUAL PROBABILTY OF A AND B.

symbolically, P(A or B) = P(A)+P(B).

. IN CASE OF NON-MUTUALLY EXCLUSIVE EVENTS THE PROBABILITY OF OCCURRENCE OF EITHER A OR B WILL BE –

P(AorB)= P(A) +P(B)- P(A and B)

EX- if the probability of a man to buy shirt is 0.65 and that to a trouser is 0.30, the we cannot calculate the probability of buying either shirt or trousre because both events are not mutually exclusive events.

CONDITIONAL PROBABILITY

Is the probability of an event occurring given that another event has already occurred. The conditional probability of event B occurring, given that event A has occurred, is denoted by P(B|A) and is read as “probability of B, given A.”

CONDITIONAL PROBABILITY

Conditional Probability contains a condition that may limit the sample space for an event.

You can write a conditional probability using the notation

- This reads “the probability of event B, given event A”

)( ABP

INDEPENDENT AND DEPENDENT EVENTS

The question of the interdependence of two or more events is important to researchers in fields such as marketing, medicine, and psychology. You can use conditional probabilities to determine whether events are independent or dependent.

DEFINITIONTwo events are independent if the occurrence of one of the events does NOT affect the probability of the occurrence of the other event. Two events A and B are independent if:

P(B|A) = P(B) or if P(A|B) = P(A)

Events that are not independent are dependent.

CLASSIFYING EVENTS AS INDEPENDENT OR DEPENDENT

Decide whether the events are independent or dependent.

1. Selecting a king from a standard deck (A), not replacing it and then selecting a queen from the deck (B).

Solution: P(B|A) = 4/51 and P(B) = 4/52. The occurrence of A changes the probability of the occurrence of B, so the events are dependent.

THE MULTIPLICATION RULE

To find the probability of two events occurring in a sequence, you can use the multiplication rule.

The probability that two events A and B will occur in sequence is

P(A and B) = P(A) ● P(B)

If events A and B are independent, then the rule can be simplified to P(A and B) = P(A) ● P(B). This simplified rule can be extended for any number of events.

USING THE MULTIPLICATION RULE TO FIND PROBABILITIES

Two cards are selected without replacement, from a standard deck. Find the probability of selecting a king and then selecting a queen.

Solution: Because the first card is not replaced, the events are dependent.

P(K and Q) = P(K) ● P(Q|K)

So the probability of selecting a king and then a queen is about .0006

006.02652

16

51

4

52

4

USING THE MULTIPLICATION RULE TO FIND PROBABILITIES

A coin is tossed and a die is rolled. Find the probability of getting a head and then rolling a 6.

Solution: The events are independent

P(H and 6) = P(H) ● P(6)

So the probability of tossing a head and then rolling a 6 is about .0083

083.012

1

6

1

2

1

USING THE MULTIPLICATION RULE TO FIND PROBABILITIES

A coin is tossed and a die is rolled. Find the probability of getting a head and then rolling a 2.

P(H) = ½. Whether or not the coin is a head, P(2) = 1/6—The events are independent.

083.012

1

6

1

2

1)2()()2( PHPHandP

So, the probability of tossing a head and then rolling a two is about .083.

Bayes’ Theorem

The Bayes’ theorem was given by the British mathematician Thomas Bayes(1702-61)

Bayes’ theorem can be explained as, “A theorem describing how the conditional probability of each of a set of possible causes for a given observed outcome can be computed from knowledge of the probability of each cause and the conditional probability of the outcome of each cause.”

It is given by the formula :-

P(Ai/B)= P(B/Ai).P(Ai)

Σ P(B/Ai).P(Ai) K

i=1

The formula of Bayes’ theorem given in the previous slide is based on the formula of conditional probability. Let :A1, A2 ,..., An are n non empty events which constitute a partition of sample space S, i.e. A1, A2, ..., An are pairwise disjoint and A1 A∪ 2 ... A∪ ∪ n = S, andB is any event of nonzero probability, then

P(An|B) =

= (by Multiplication rule of Probability)

= where I = 1, 2, …. , n

The following terminology is generally used when Bayes' theorem is applied :-The events A1, A2, ..., An are called hypotheses.The probability P(Ai) is called the priori probability of the hypothesis Ai.The conditional probability P(Ai|B) is called a posteriori probability of the hypothesis Ai.

P(B ∩ Ai) P(B)P(Ai).P(B|Ai )

P(B)

P(Ai).P(B|Ai )

Σ P(B/Ai).P(Ai) n

i=1

ProblemQ. Two different suppliers, A and B, provide a

manufacturer with the same part. All supplies of this part are kept in a large

bin.  in the past, 5% of the parts supplied by A and 9% of the parts supplied by B have been defective. 

A supplies four times as many parts as B

Suppose you reach into the bin and select a part, and find it is nondefective.  What is the probability that it was supplied by A?

SolutionLet E1 = Part supplied by A , E2 = Part supplied by B; and

A = Nondefective partNow, 5% of the parts supplied by A and 9% of the parts

supplied by B have been defective.

P(A|E1) = 0.95

P(A|E2) = 0.91

A supplies four times as many parts as B

So, P(E1) = 0.8

P(E2) = 0.2 Suppose you reach into the bin and select a part, and

find it is non-defective.  What is the probability that it was supplied by A?

So, we have to find P(E1|A)

Solution (cont.)

P(A|E1).P(E1)

P(A|E1).P(E1) + P(A|E2).P(E2)

P (A|E1) = 0.95 ; P(E1) = 0.8

P(A|E2) = 0.91 ; P(E2) = 0.2

(0.95).(0.8) (0.95).(0.8)+(0.91).(0.2) = 0.76 0.76 + 0.182 So, P(E1|A) = 0.807 (approx)

Random Variable and Probability Distribution

A variable whose value is determined by the outcome of a random experiment is called a random variable. It is also known as a chance variable or Stochastic variable.

In terms of symbols if a variable X can assume discrete set of values X1, X2, …., Xn with respective probabilities p1, p2, …., pn where p1 + p2 + …. + pn =1, we say that discrete probability distribution for X has been defined. The function P(X) which has the respective values p1, p2, …., pn for X = X1, X2, …., Xn is called the probability function or frequency function of X.

Here is an example which shows the probability distribution of a pair of fair dice tossed, :-

Mathematical Expectation

The mathematical expectation, also called the expected value, of a random variable is the weighted arithmetic mean of the variable, where, the weights used to find the mathematical expectation are all the respective probabilities of the values that the variable can possibly assume.

If X denotes a discrete random variable which can assume the values X1, X2,….., Xn, with respective probabilities p1, p2,….., pn, where p1+p2+…..+pn=1, the mathematical expectation of X denoted by E(X) is defined as : E(X) = p1X1 +p2X2 + ….. + pnXn.

Thus, the expected value equals the sum of each particular value within the set (X) multiplied by the probability that X equals that particular value. E(X) = Σpixi, where i=1, 2,…..,n

PROBLEM :A dealer in refrigerators estimates from his past experience the probabilities of his selling refrigerators in a day. These are as follows :

No. of refrigerators sold in a day(X) : 0 1 2 3 4 5 6Probability P(X) : 0.3 0.2 0.23 0.25 0.12 0.1 0.07

Find the mean no. of refrigerators sold in a day.

SOLUTION :

Now, Mean no. of refrigerators sold, i.e., Mathematical Expectation = Σxi.pi(xi), where i =1, 2, 3….

So, E(X) = 0*0.3 + 1*0.2 +2*0.23 + 3*0.25 +4*0.12 +5*0.1 +6*0.07 = 0 + 0.2 + 0.46 + 0.75 + 0.48 +0.5 + 0.42 = 2.81

Hence, Mean no. of refrigerators sold in a day is 3

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Made by: Sachin

Kumar (2071) Krishan

Sharma 2004) Mohit

Ostwal (2096) Yugal Gupta (2003)