WELCOME To all PGDM

Students

FromDushyant tyagi

{M.Sc.(Statistics), M.Phil.}

PermutationsThe different arrangement can be made

without replacement & with regard of order of a given no. of things by taking some or all at a time are called permutations.

P (n,r) = n! / (n-r)!E.g. permutations made with letters a,b,c by

taking two at a time:P(3,2)=6ab, ba, ac, ca, bc, cb

CombinationsThe combination of ‘n’ different objects taken

‘r’ at a time is a selection of ‘r’ out of ‘n’ objects with no attention given to order of arrangement

C (n,r) = n!/r!(n-r)!e.g. From 5 boys & 6 girls a group of 3 is to be

formed having 2 boys & 1 girl is C(5,2) x C(6,1) = 60 ways

ProbabilityProbability is a concept which numerically

measures the degree of uncertainty or certainty of the occurrence of any event. i.e. the chance of occurrence of any event.

The probability of an event A is No. of Favorable cases

P(A)= Total No. of Cases

ProbabilityIf P(A)=0, Impossible EventIf P(A)=1, Sure Event0≤P(A)≤1P(A)= Probability of occurrence P(Ā)= Probability of Non-occurrenceP(A) + P(Ā) = 1

Some KeywordsEqually Likely Events: When the chance of

occurrence of all the events are same in an experiment.

Mutually Exclusive Events: If the occurrence of any one of them prevents the occurrence of other in the same experiment.

Sample Space: the set of all possible outcomes.

Some KeywordsIndependent Events: If two or more events

occur in such a way that the occurrence of one does not effect the occurrence of other.

Dependent Events: If the occurrence of one event influences the occurrence of the other.

Classical or Priori ProbabilityIf a trial result in ‘n’ exhaustive, mutually

exclusive and equally likely cases and ‘m’ of them are favorable to the happenings of an event E, then the probability ‘P’ of happening of E is given by:

P(E) = m / n

Empirical or Posteriori ProbabilityThe classical def requires that ‘n’ is finite and

that all cases are equally likely. This condition is very restrictive and can not

cover all situations.The above conditions are not necessarily

active in this case.

Example A coin is tossed three times. Find the

probability of getting:i) Exactly one headii) Exactly two headiii) One or two head

Example One card is randomly drawn from a pack of

52 cards. Find the probability thati) Drawn card is redii) Drawn card is an aceiii) Drawn card is red and kingiv) Drawn card is red or king

Example A bag contains 3 red, 6 white and 7 blue

balls. Two balls are drawn at random. Find the probability that

i) Both the balls are white.ii) Both the balls are blue.iii) One ball is red & other is white.iv) One ball is white & other is blue.

Addition TheoremFor any two event A and B the probability for

the occurrence of A or B is given by:P(AUB)= P(A) + P(B) – P(AПB)If A & B are mutually Exclusive then P(AПB)=0 P(AUB)= P(A) + P(B)

Multiplication or Conditional ProbabilityThe probability of an event B when it is

known that the event A has occurred already: P(B/A)= P(A B) / P(A) ;if P(A)>0

ie. P(A B)= P(A).P(B/A) If A and B are Independent event:

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

Example A bag contains 25 numbered from 1 to 25.

Two balls are drawn at random from the bag with replacement. Find the probability of selecting:

i) Both odd numbers.ii) One odd & one even.iii) No odd numbers.iv) At least one odd. v) Both even numbers.

ExampleFive men in a company of 20 are graduate. If

3 men are picked up at random, what is the probability that they are all graduate? What is the probability that at least one is graduate.

ExampleThe probability that A hits a target is 1/3 and

the probability that B hits the target is 2/5. What is the probability that the target will be hit, if each one of A and B shoots at the target.

Bayes’ TheoremThe Bayes’ theorem is useful to determine

the revised (Posterior) probability for Prior probability of known outcomes after getting additional information about these outcomes.

1Let A1, A2, ... , An be a set of mutually exclusive events

that together form the sample space S. Let B be any event from the same sample space, such that P(B) > 0. Then, 

P( Ak | B ) =  P( Ak ∩ B )

P( A1 ∩ B ) + P( A2 ∩ B ) + . . . + P( An ∩ B )

Note: Invoking the fact that P( Ak ∩ B ) = P( Ak )P( B | Ak ), Baye's theorem can also be expressed as 

P( Ak | B ) =   P( Ak ) P( B | Ak )

P( A1 ) P( B | A1 ) + P( A2 ) P( B | A2 ) + . . . + P( An ) P( B | An )

ExampleIn a bolt factory machines A,B and C

manufacture respectively 25 percent, 35 percent and 40 percent of the total production. Of the total of their output 5, 4 and 2 percent are defective bolts, A bolt is drawn at random and is found to be defective. What is the probability that it was manufactured by machine A.

ExampleAn organization wants to introduce a new product in the

market. Based on its past experience, it has a 65 percent chance of being successful and 35 percent of being not successful. To make a decision, whether to introduce or not, it decides to get additional information on consumer attitude toward the product. For this purpose organization decides on a survey. In the past when a product of this type was successful, survey yield favourable indication 85 percent of the time, whereas unsuccessful products received favourable survey indications 30 percent of the time. Determine the posterior probability of the product being successful given the favourable survey information.

Random variableRandom variable is a numerical valued

function based on sample space(S) of a random experiment.

Example: If we toss two coins ,and a random variable ‘X’ is define as numbers of head.Then the possible values of X is0, 1, 2

Types of random variable• A random variable can be classified as being

either discrete or continuous depending on the numerical values it assumes.

• A discrete random variable may assume either a finite number of values or an infinite sequence of values.

• A continuous random variable may assume any numerical value in an interval or collection of intervals.

Expected Value of ProbabilityLet X be the random variable with the

following distribution:X : x1 x2 x3………..

P(X) :P(x1) P(x2) P(x3)……..

Expected Value is given by:E(X) = Σ xi . P (xi)

ExampleA player tossed two coins. If two heads show

he wins Rs. 4. if one head shows he wins Rs. 2, but if two tails show he pays Rs. 3 as penalty. Calculate the expected value of the game to him.

Solution:E(X)= (-3) ¼ + (2) ½ + (4) ¼ =1.25

ExampleAn insurance company sells a particular life

insurance policy with a face value of Rs. 1000 and a yearly premium of Rs. 20. If 0.2% of the policy holder can be expected to die in the course of a year, what would be the company’s expected earning per policy holder per year.

E(X)= (-980) 0.002 + (20) 0.998=18

Probability Distributions• The probability distribution for a random variable

describes how probabilities are distributed over the values of the random variable.

• The probability distribution is defined by a probability function which provides the probability for each value of the random variable.

Binomial DistributionThe probability of ‘x’ success in ‘n’ trials

where ‘p’ is the probability of success and ‘q’ is the probability of failure is given by:P(x) = C(n,x) px qn-x x=0,1,2..

Mean= npVariance= npq

Assumptions of Binomial DisttThe repeated trials are independent.Each trial has only two possible outcome.The probability of success or failure in each

trails are constant.P and q are the two parameters of BDMean of BD is np.Standard Deviation of BD is √npq.

ExampleIn tossing 10 coins. What is the probability of

having exactly 3 heads.

Example If mean & variance of a binomial

distribution are 4 and 2 respectively. Find the probability of

i) Exactly two successesii) Less than two successesiii) More than 6 successesiv) At least 2 successes

Poisson Probability DistributionIt is another discrete probability distribution

having specific use. It is a limiting case of binomial distribution, when ‘n’ is large and ‘p’ is small binomial distribution tends towards Poisson distribution. It is defined as

p (x)= e-λ λx / x! X=0,1,2…

Poisson Probability DistributionConsider a case when n=1000 and p=1/100

then and x=3p(3)= C(1000,3) (1/100)3 (99/100)997

It is very difficult to calculate this expression.

Properties- Poisson DistributionStatistical independence is assumed.When p, probability of occurrence is very

small and n is very large.The only parameter is λ=npThe mean and variance of PD is λ.Each trial has only two possible outcome.It is a limiting case of Binomial distt.

Some example of Poisson DisttNo. of defective products in a large box.No. of air accidents in India in a year.No. of telephone calls received at a particular

exchange in a unit time.

ExampleSuppose that the chance of an individual

coal-miner being killed in a mine accident during a year is 1/1400. Find the probability that in the mine employing 350 miners, there will be at least one fatal accident in a year. (given e-0.25=0.78)

Sol. λ=350/1400=0.25Probability = 1-P(0)=0.22

ExampleA manufacturer of blades knows that 5% of

his product is defective. He sells blades in boxes of 100, and guarantees that not more than 4 blades will be defective. Find the probability that a box will meet the guaranteed quality.

(given e-5=0.0067)Sol. λ=np=100*0.05=5

P(x≤4)=p(0)+p(1)+p(2)+p(3)+p(4)=.438

Normal DistributionIt is a continuous probability distribution in

which the relative frequencies of a continuous variable are distributed according to the normal probability law. It is a symmetrical distribution in which frequencies are distributed evenly about the mean of the distribution.

Normal DistributionThe normal distribution of a variable, when

represented graphically, takes the shape of a symmetrical curve, known as normal curve. This curve is asymptotic on either side.

It is discovered by De-Movire & Friedrich Gauss.

The Normal Probability DistributionGraph of the Normal Probability Density

Function

xx

f f ((x x ))

Properties Normal Distt or curveThe normal curve is symmetrical about the

vertical axis through mean.Mean(µ) & SD(σ) are known as the

parameter of the distribution.The curve is Asymptotic to X-axis.The problems related to Normal distt can be

solved by using the properties of Normal Curve.

Properties Normal Distt or curveThe random variable X should be transform

to Standard Normal Variable z usingZ = (X-µ)/σ

After the transformation the probability (area) can be find using Normal distribution table.

The total are under the normal curve is 1.

The table will give this probability

Given any Given any positive value for positive value for zz, the table will give us the , the table will give us the following probability following probability

Given positive z

The probability that we find using the table is the The probability that we find using the table is the probability of having a standard normal variable between probability of having a standard normal variable between

0 and the given positive z.0 and the given positive z.

Example 2000 students appeared in an examination.

Distribution of marks is assumed to be normal with mean=30 and SD=6.25. How many students are expected to get marks.

i) Between 20 and 40ii) Less than 35.

ExampleAssuming that the mean height of students is

an exactly normal distribution be 68.22 inches and variance of 10.8 inches2. how many students in a college of 1000 students would you expect to be over six feet tall?

ExampleIn a distribution exactly normal, 7% of the

items are under 35 and 89% are under 63. What are the mean and standard deviation of the distribution.

ExampleFrom a large group of men, 5% are under 60

inches in height and 40% are between 60 and 65 inches. Assuming a normal distribution, find the mean height & standard deviation.