Introduction to Probability Introduction to Probability and Statisticsand Statistics

Chapter 5

Discrete Distributions

Discrete Random Discrete Random VariablesVariables

• Discrete random variables take on only a finite or countable many of values.

Number of heads in 1000 trials of coin tossing

Number of cars that enter UNI in a certain day

Number of heads in 1000 trials of coin tossing

Number of cars that enter UNI in a certain day

Binomial Random Variable

• The coin-tossing experimentcoin-tossing experiment is a simple example of a binomial random variable. binomial random variable. Toss a fair coin n = 3 times and record

• x = number of heads.

x p(x)

0 1/8

1 3/8

2 3/8

3 1/8

ExampleExample• Toss a coin 10 times

• For each single trial, probability of getting a head is 0.4

• Let x denote the number of heads

The Binomial ExperimentThe Binomial Experiment1. The experiment consists of nn identical trials. identical trials.2. Each trial results in one of two outcomesone of two outcomes,

success (S) or failure (F).3. Probability of success on a single trial is p and

remains constantremains constant from trial to trial. The probability of failure is q = 1 – p.

4. Trials are independentindependent.5. Random variable xx, the number of successes , the number of successes

in in n n trials. trials.

x x – Binomial random variable with – Binomial random variable with parameters parameters nn and and p p

Binomial or Not?Binomial or Not?

• A box contains 4 green M&Ms and 5 red ones

• Take out 3 with replacement

• x denotes number of greens

• Is x binomial? Yes,3 trials are independent with same probability of getting a green.

m

m

mmm m

Binomial or Not?Binomial or Not?

• A box contains 4 green M&Ms and 5 red ones

• Take out 3 without replacement

• x denotes number of greens

• Is x binomial? NO, when we take out the second M&M, the probability of getting a green depends on color of the first.3 trials are dependent.

m

m

mmm m

Binomial or Not?Binomial or Not?• Very few real life applications

satisfy these requirements exactly.

• Select 10 people from the U.S. population, and suppose that 15% of the population has the Alzheimer’s gene.• For the first person, p = P(gene) = .15• For the second person, p P(gene) = .15,

even though one person has been removed from the population…

• For the tenth person, p P(gene) = .15Yes, independent trials with the same probability of success

Binomial Random VariableBinomial Random Variable

• Example: Example: A geneticist samples 10 people and x counts the number who have a gene linked to Alzheimer’s disease.

• Success:Success:

• Failure:Failure:

• Number ofNumber of trials:trials:• Probability of Probability of

SuccessSuccess

Has gene

Doesn’t have gene

n = 10

p = P(has gene) = 0.15

• Rule of Thumb: Sample size n; Population size N;If n/N < .05, the experiment is Binomial.

ExampleExample• Toss a coin 10 times• For each single trial, probability of getting a

head is 0.4• Let x denote the number of heads

Find probability of getting exactly 3 heads. i.e. P(x=3).Find probability distribution of x

SolutionSolution

• Simple events:

73 )6.0((0.4)73 )6.0((0.4)

103C

103C

73103 )6.0((0.4)C

73103 )6.0((0.4)C

• Event A: {strings with exactly 3 H’s};

• Probability of getting a given string in A:

• Probability of event A. i.e. P(x=3)

• Number of strings in A

Strings of H’s and T’s with length 10

HTTTHTHTTT TTHHTTTTHT…

HTTTHTHTTT

A General ExampleA General Example

• Toss a coin n times; For each single trial, probability of getting a head is p;

• Let x denote the number of heads;

Find the probability of getting exactly k heads. i.e. P(x=k)

Find probability distribution of x.

Binomial Probability Binomial Probability DistributionDistribution

• For a binomial experiment with n trials and probability p of success on a given trial, the probability of k successes in n trials is

nk

qpCkxP knknk

,...2,1,0for

)(

nk

qpCkxP knknk

,...2,1,0for

)(

.1!0 and 1)2)...(2)(1(!with

)!(!

! Recall

nnnn

knk

nC n

k

.1!0 and 1)2)...(2)(1(!with

)!(!

! Recall

nnnn

knk

nC n

k

Binomial Mean, Variance and Binomial Mean, Variance and Standard DeviationStandard Deviation

• For a binomial experiment with n trials and probability p of success on a given trial, the measures of center and spread are:

npq

npq

np

:Deviation Standard

:Variance

:Mean2

npq

npq

np

:Deviation Standard

:Variance

:Mean2

n = p = x =success =

ExampleExampleA marksman hits a target 80% of the time. He fires 5 shots at the target. What is the probability that exactly 3 shots hit the target?

333)3( nn qpCxP

5 .8hit # of hits

353 )2(.)8(.!2!3

!5

2048.)2(.)8(.10 23

ExampleExample

What is the probability that more than 3 shots hit the target?

55555

45454

)5()4()3(

qpCqpC

PPxP

0514 )2(.)8(.!0!5

!5)2(.)8(.

!1!4

!5

7373.)8(.)2(.)8(.5 54

ExampleExample• x = number of hits. • What are the mean and standard

deviation for x? (n=5,p=.8)

89.)2)(.8(.5

:Deviation Standard

4)8(.5 :Mean

npq

np

89.)2)(.8(.5

:Deviation Standard

4)8(.5 :Mean

npq

np

Cumulative ProbabilityCumulative Probability

You can use the cumulative probability cumulative probability tablestables to find probabilities for selected binomial distributions.

Binomial cumulative probability:

P(x k) = P(x = 0) +…+ P(x = k)

Binomial cumulative probability:

P(x k) = P(x = 0) +…+ P(x = k)

Key ConceptsKey ConceptsI. The Binomial Random VariableI. The Binomial Random Variable

1. Five characteristics: the experiment consists of n identical trials;each resulting in either success S or failure F; probability of success is p and remains constant;all trials are independent; x is the number of successes in n trials.

2. Calculating binomial probabilities

a. Formula:

b. Cumulative binomial probability P(x k).3. Mean of the binomial random variable:4. Variance and standard deviation:

np np

npq2 npq2 npq npq

knknk qpCkxP )(

knknk qpCkxP )(

ExampleExampleAccording to the Humane Society of theUnited States, there are approximately 40% ofU.S. households own dogs. Suppose 15households are selected at random. Find

1. probability that exactly 8 households own dogs?2. probability that at most 3 households own dogs?3. probability that more than 10 own dogs?4. the mean, variance and standard deviation of

x, the number of households that own dogs.

n = p = x =success =

ExampleExampleAccording to the Humane Society of the United States, there are approximately 40% of U.S. households own dogs. Suppose 15 households are selected at random. What is probability that exactly 8 households own dogs?

8158158 )6(.)4(.)8( CxP

15 .4own dog # households that own dog118.)6(.)4(.6435 78

ExampleExampleWhat is the probability that at most 3 households own dogs?

091.

0634.0219.0047.0005.

)6(.)4(.)6(.)4(.)6(.)4(.)6(.)4(.

)3()2()1()0()3(12315

313215

214115

115015

0

CCCC

PPPPxP

ExampleExample

6)4(.15 np 6)4(.15 np

90.16.32 90.16.32

What are the mean, variance and standard deviation of random variable x? (n=15, p=.4)

6.3)6)(.4(.152 npq 6.3)6)(.4(.152 npq

Binomial ProbabilityBinomial Probability• Probability distribution for Binomial random

variable x with n=15, p=0.4

x

P(x

)

1614121086420

0.20

0.15

0.10

0.05

0.00

Probability Distribution of Binomial n=15, p=.4

ExampleExample6)4(.15 np 6)4(.15 np

)8.9,2.2(

8.9)9.1(262

2.2)9.1(262

)8.9,2.2(

8.9)9.1(262

2.2)9.1(262

90.16.32 90.16.32

1. What are the mean, variance and standard deviation of random variable x?

2. Calculate interval within 2 standard deviations of mean. What values fall into this interval?

3. Find the probability that x fall into this interval.

6.3)6)(.4(.152 npq 6.3)6)(.4(.152 npq

9,8,7,6,5,4,3 9,8,7,6,5,4,3

939.)9(...)4()3()93( PPPxP 939.)9(...)4()3()93( PPPxP