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Probability Probability Distributions Distributions

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Probability Dist Random Variables

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Page 1: Probability Dist Random Variables

Probability Probability DistributionsDistributions

Page 2: Probability Dist Random Variables

Random VariableRandom Variable• Random variable

– Outcomes of an experiment expressed numerically

– e.g.: Throw a die twice; Count the number of times the number 6 appears (0, 1 or 2 times)

Page 3: Probability Dist Random Variables

Discrete Random VariableDiscrete Random Variable

Page 4: Probability Dist Random Variables

Probability Distribution Values Probability

0 1/4 = .25

1 2/4 = .50

2 1/4 = .25

Discrete Probability Discrete Probability Distribution ExampleDistribution Example

Event: Toss 2 Coins. Count # Tails.

T

T

T T

Page 5: Probability Dist Random Variables

Discrete Probability Distributions

Binomial Poisson

Page 6: Probability Dist Random Variables

Binomial Distribution- AssumptionsBinomial Distribution- Assumptions• “n” Identical & finite trials

– e.g.: 15 tosses of a coin; 10 light bulbs taken from a warehouse

• Two mutually exclusive outcomes on each trial– e.g.: Heads or tails in each toss of a coin; defective or

not defective light bulb

• Trials are independent of each other– The outcome of one trial does not affect the outcome of

the other.

• Constant probability for each trial– e.g.: Probability of getting a tail is the same each time a

coin is tossed

Page 7: Probability Dist Random Variables

Binomial DistributionBinomial Distribution• A random variable X is said to follow

Binomial distribution if it assumes only non negative values and its function is given by

• x: No. of “successes” in a sample, x=0,1,2..,n.

• n: Number of trials.• p: probability of each success.• q: probability of each failure (q=1-p)

( ) ( ) ( ) xnxX pp

xnx

nxP −−

−= 1

!!

!

•Notation: X ~ B (n, p).

Page 8: Probability Dist Random Variables

Binomial Binomial Distribution CharacteristicsDistribution Characteristics

• Mean

EX:

• Variance and standard deviation-

Ex:

( )E X npµ = =

( )5 .1 .5npµ = = =n = 5 p = 0.1

0.2.4.6

0 1 2 3 4 5X

P(X)

( ) ( ) ( )1 5 .1 1 .1 .6708np pσ = − = − =

( )( )

2 1

1

np p

np p

σ

σ

= −

= −

Page 9: Probability Dist Random Variables

ExampleExample• The experiment: Randomly draw red ball with

replacement from an urn containing 10 red balls and 20 black balls.

• Use S to denote the outcome of drawing a red ball and F to denote the outcome of a black ball.

• Then this is a binomial experiment with p =1/3.

• Q: Would it still be a binomial experiment if the balls were drawn without replacement?

Page 10: Probability Dist Random Variables

Questions for practiceQuestions for practice• Ten coins are thrown simultaneously. Find the

probability of getting at least seven heads.

• A & B play a game in which their chances of winning are in the ratio 3:2. find A’s chance of winning at least 3 games out of the 5 games played.

• Mr. Gupta applies for a personal loan of Rs 150,000 from a nationalized bank to repair his house. The loan offer informed him that over the years bank has received about 2920 loan applications per year and that the prob. Of approval was on average, about 0.85. Mr. Gupta wants to know the average and standard deviation of the number of loans approved per year.

Page 11: Probability Dist Random Variables

• The incidence of a certain disease in such that on the average 20% of workers suffer from it. If 10 workers are selected at random, find the probability that

1. Exactly 2 workers suffer from the disease.

2. Not more than 2 workers suffer from the disease.

• Bring out the fallacy, if any1. The mean of a binomial distribution is 15

and its standard deviation is 5.2. Find the binomial distribution whose

mean is 6 and variance is 4.

Page 12: Probability Dist Random Variables

• The probability that an evening college student will be graduate is 0.4.Determine the probability that out of 5 students

• None will be graduate.• One will be graduate.• At least one will be graduate.• Multiple choice test consists of 8 questions and

three answers to each question (of which only one is correct). A student answers each question by rolling a balanced die and marking the first answer if he gets 1 or 2, the second answer if he gets 3 or 4 and the third answer if he gets 5 or 6. To get a distinction, student must secure at least 75% correct answers.If there is no negative marking, what is the probability that student secures a distinction?

Page 13: Probability Dist Random Variables

Q. Assuming that half the population are Consumers of rice and assuming that 100 investigators can take sample of 10 individuals to see whether they are rice consumers , how many investigations would you expect to report that three people or less were consumers of rice?

Q. Five coins are tossed 3200 times. Find the frequencies of the distribution of heads and tails and tabulate the result. Also calculate the mean number of successes and S.D.

Page 14: Probability Dist Random Variables

Credit Card ExampleCredit Card Example• Records show that 5% of the customers in a

shoe store make their payments using a credit card.

• This morning 8 customers purchased shoes. • Use the binomial table to answer the following

questions.1. Find the probability that exactly 6 customers

did not use a credit card.

2. What is the probability that at least 3 customers used a credit card?

Page 15: Probability Dist Random Variables

3. What is the expected number of customers who used a credit card?

4. What is the standard deviation of the number of customers who used a credit card?

Credit Card ExampleCredit Card Example

Page 16: Probability Dist Random Variables

Parking ExampleParking Example• Sarah drives to work everyday, but does not own a parking

permit. She decides to take her chances and risk getting a parking ticket each day. Suppose – A parking permit for a week (5 days) cost $ 30. – A parking fine costs $ 50. – The probability of getting a parking ticket each day is

0.1. – Her chances of getting a ticket each day is independent

of other days. – She can get only 1 ticket per day.

• What is her probability of getting at least 1 parking ticket in one week (5 days)?

• What is the expected number of parking tickets that Sarah will get per week?

• Is she better off paying the parking permit in the long run?

Page 17: Probability Dist Random Variables
Page 18: Probability Dist Random Variables

Binomial Distribution-Binomial Distribution- Different situations Different situations

Random experiments and random variable

Page 19: Probability Dist Random Variables

Situations Contd..Situations Contd..

Random experiments and random variables

Page 20: Probability Dist Random Variables

Poisson Poisson DistributionDistribution

Page 21: Probability Dist Random Variables

AssumptionsAssumptions

Applicable when-

1) No. of trials is indefinitely large n→∞2) Probability of success p for each trial

is very small. p →03) Mean is a finite number given by np = λ

Page 22: Probability Dist Random Variables

Poisson Distribution FunctionPoisson Distribution Function

( )

( )!

: probability of "successes" given

: number of "successes" per unit

: expected (average) number of "successes"

: 2.71828 (base of natural logs)

XeP X

XP X X

X

e

λλ

λ

λ

= X =0,1,2…

X ~ P(λ)

Page 23: Probability Dist Random Variables

Poisson Distribution Poisson Distribution CharacteristicsCharacteristics

( )

( )1

N

i ii

E X

X P X

µ λ

=

= =

= ∑

• Mean

• Variance & Standard deviation 2 σ λ σ λ= =

Page 24: Probability Dist Random Variables

ExampleArrivals at a bus-stop follow a Poisson distribution with an average of 4.5 every quarter of an hour.

(assume a maximum of 20 arrivals in a quarter of an hour) and calculate the probability of fewer than 3 arrivals in a quarter of an hour.

Page 25: Probability Dist Random Variables

The probabilities of 0 up to 2 arrivals can be calculated directly from the formula

()!

xep x

x

λλ−

=

4.504.5(0)

0!

ep

=

with λ =4.5

So p(0) = 0.01111

Page 26: Probability Dist Random Variables

Similarly p(1)=0.04999 and p(2)=0.11248

So the probability of fewer than 3 arrivals is

0.01111+ 0.04999 + 0.11248 =0.17358

Page 27: Probability Dist Random Variables

Questions for practiceQuestions for practice• Suppose on an average 1 house in 1000 in a

certain district has a fire during a year.If there are 2000 houses in that district,what is the probability that exactly 5 houses will have a fire during the year?

• If 3% of the electric bulbs manufactured by a company are defective, find the probability that in a sample of 100 bulbs

a) 0 b) 5 bulbs c) more than 5 d) between 1 and 3 e) less than or equal to 2 bulbs are defective

Page 28: Probability Dist Random Variables

• Comment on the following for a Poisson distribution with Mean = 3 and s.D = 2

• In a Poisson distribution if p(2)= 2/3 p(1). Find

i) mean ii) standard deviation iii)P(3) iv) p(x > 3)

• A car hire firm has two cars, which it hires out day by day.The number of demands for a car on each day is distributed as a Poisson distribution with mean 1.5. Calculate the proportion of days on which

i) Neither car is used.ii) Some demand is refused.

Page 29: Probability Dist Random Variables

• A manufacturer who produces medicine bottles,finds that 0.1% of the bottles are defective. The bottles are packed in boxes containing 500 bottles.A drug manufacturer buys 100 boxes from the producer of bottles.Find how many boxes will contain

i) no defectivesii) at least two defectives

• The following table gives the number of days in a 50 day period during which automobile accidents occurred in a city-

No. of accidents: 0 1 2 3 4No. of days : 21 18 7 3 1Fit a Poisson distribution to data.

Page 30: Probability Dist Random Variables

Note:In most cases, the actual number of trials is not known, only the average chance of occurrence based on the past experience can help in constructing the whole distribution .

For ex: The number of accidents in any Particular period of time : the distribution is based on the mean value of occurrence of an event , which may also be obtained based on Past Experience .

Page 31: Probability Dist Random Variables

• In a town 10 accidents took place in a span of 50 days. Assuming that the number of accidents per day follow Poisson distribution, find the probability that there will be 3 or more accidents in a day.

HINT:λ = 10/50 = 0.2Average no. of accidents per day

Page 32: Probability Dist Random Variables

• Case-Let• An insurance company insures 4000

people against loss of both eyes in a car accident.Based on previous data,the rates were computed on the assumption that on the average 10 persons in 1,00,000 will have car accident each year that result in this type of injury.What is the probability that more than 4 of the insured will collect on a policy in a given year?

Page 33: Probability Dist Random Variables

Poisson type SituationsPoisson type Situations• Number of deaths from a disease.• No. of suicides reported in a city.• No. of printing mistakes at each page of a

book.• Emission of radioactive particles.• No. of telephone calls per minute at a small

business.• No. of paint spots per new automobile.• No. of arrivals at a toll booth• No. of flaws per bolt of cloth