distribution lecture 15 - purdue university

Negative BinomialDistribution

Motivation


15.1

Lecture 15Negative Binomial DistributionText: A Course in Probability by Weiss 5.7

STAT 225 Introduction to Probability ModelsFebruary 23, 2014

Whitney HuangPurdue University


Motivation


15.2

Agenda

1 Motivation

2 Negative Binomial Distribution


Motivation


15.3

Review

So far, we have covered

Bernoulli and Binomial distribution: number of successesin an independent trials (sampling with replacement) withfixed sample size

Hypergeometric distribution: number of successes in adependent trials (sampling without replacement) with fixedsample sizePoisson distribution: number of successes (events)occurring in a fixed interval of time and/or space withoutfixed sample size

In some cases, we want to know the sample size necessary toget a certain number of successes

Geometric distribution: number of trials until the 1stsuccess (including the success)Negative Binomial distribution: number of trials until the rthsuccess (including the rth success)In both Geometric and Negative Binomial distribution, thetrials are independent


Motivation


15.3

Review


Bernoulli and Binomial distribution: number of successesin an independent trials (sampling with replacement) withfixed sample sizeHypergeometric distribution: number of successes in adependent trials (sampling without replacement) with fixedsample size

Poisson distribution: number of successes (events)occurring in a fixed interval of time and/or space withoutfixed sample size




Motivation


15.3

Review


Bernoulli and Binomial distribution: number of successesin an independent trials (sampling with replacement) withfixed sample sizeHypergeometric distribution: number of successes in adependent trials (sampling without replacement) with fixedsample sizePoisson distribution: number of successes (events)occurring in a fixed interval of time and/or space withoutfixed sample size




Motivation


15.3

Review




Geometric distribution: number of trials until the 1stsuccess (including the success)

Negative Binomial distribution: number of trials until the rthsuccess (including the rth success)In both Geometric and Negative Binomial distribution, thetrials are independent


Motivation


15.3

Review




Geometric distribution: number of trials until the 1stsuccess (including the success)Negative Binomial distribution: number of trials until the rthsuccess (including the rth success)

In both Geometric and Negative Binomial distribution, thetrials are independent


Motivation


15.3

Review






Motivation


15.4

Negative Binomial Distribution:

Characteristics of the Negative Binomial Distribution:Let X be a Negative Binomial r.v.

The definition of X : The number of trials it takes to get ther st success

The support: x = r , r + 1, r + 2, · · ·Its parameter(s) and definition(s): r : the number ofsuccess of interest p: the probability of success in a singletrialThe probability mass function (pmf):pX (x) =

(x−1r−1

)pr (1 − p)x−r for x = r , r + 1, r + 2, · · ·

The expected value: E[X ] = rp

The variance: Var(X ) = r(1−p)p2


Motivation


15.4



The definition of X : The number of trials it takes to get ther st successThe support: x = r , r + 1, r + 2, · · ·

Its parameter(s) and definition(s): r : the number ofsuccess of interest p: the probability of success in a singletrialThe probability mass function (pmf):pX (x) =

(x−1r−1

)pr (1 − p)x−r for x = r , r + 1, r + 2, · · ·




Motivation


15.4



The definition of X : The number of trials it takes to get ther st successThe support: x = r , r + 1, r + 2, · · ·Its parameter(s) and definition(s): r : the number ofsuccess of interest p: the probability of success in a singletrial

The probability mass function (pmf):pX (x) =

(x−1r−1

)pr (1 − p)x−r for x = r , r + 1, r + 2, · · ·




Motivation


15.4



The definition of X : The number of trials it takes to get ther st successThe support: x = r , r + 1, r + 2, · · ·Its parameter(s) and definition(s): r : the number ofsuccess of interest p: the probability of success in a singletrialThe probability mass function (pmf):pX (x) =

(x−1r−1

)pr (1 − p)x−r for x = r , r + 1, r + 2, · · ·




Motivation


15.5

Example 37

Pat is required to sell candy bars to raise money for the 6th

grade field trip. There is a 40% chance of him selling a candybar at each house. He has to sell 5 candy bars in all. Let X beof number of houses it takes

1 Name distribution (with parameter(s)) of X

2 What is the probability he sells his last candy bar at the11th house?

3 What is the probability of Pat finishing on or before the 8th

house?


Motivation


15.5

Example 37



1 Name distribution (with parameter(s)) of X2 What is the probability he sells his last candy bar at the

11th house?

3 What is the probability of Pat finishing on or before the 8th

house?


Motivation


15.5

Example 37



1 Name distribution (with parameter(s)) of X2 What is the probability he sells his last candy bar at the

11th house?3 What is the probability of Pat finishing on or before the 8th

house?


Motivation


15.6

Example 37 cont’d

Solution.

1 X ∼ NB(r = 5,p = 0.2)

2 P(X = 11) =(10

4

)(0.4)5(1 − 0.4)6 = 0.1003

3 P(X ≤ 8) = P(5 ≤ X ≤ 8) =∑8x=5

(10x

)(0.4)x(1 − 0.4)11−x = 0.1737


Motivation


15.6

Example 37 cont’d

Solution.

1 X ∼ NB(r = 5,p = 0.2)2 P(X = 11) =

(104

)(0.4)5(1 − 0.4)6 = 0.1003

3 P(X ≤ 8) = P(5 ≤ X ≤ 8) =∑8x=5

(10x

)(0.4)x(1 − 0.4)11−x = 0.1737


Motivation


15.7

Example 38

Bob is a high school basketball player who has a 70% freethrow percentage. Assume all free throw attempts areindependent of one another (i.e. there is no such thing as a hothand).

1 What is the probability it takes more than 3 shots to get hisfirst made free throw?

2 What is the probability his first made free throw is on thethird shot?

3 What is the probability that his third made free throw is onhis fifth shot?

4 What is the probability that his 100th made free throw is onhis 123rd shot?


Motivation


15.8

Example 38 cont’d

Solution.

Let X , Y , and Z represent respectively Geo(p = 0.7),NB(r = 3,p = 0.7) and NB(r = 100,p = 0.7)

1 P(X > 3) = (1 − .7)3 = 0.027

2 P(X = 3) = .7(1 − .7)2 = 0.0633 P(Y = 5) =

(42

)(0.7)3(0.3)2 = 0.1852

4 P(Z = 123) =(122

99

)(0.7)100(0.3)23 = 0.0012


Motivation


15.8

Example 38 cont’d

Solution.


1 P(X > 3) = (1 − .7)3 = 0.0272 P(X = 3) = .7(1 − .7)2 = 0.063

3 P(Y = 5) =(4

2

)(0.7)3(0.3)2 = 0.1852

4 P(Z = 123) =(122

99

)(0.7)100(0.3)23 = 0.0012


Motivation


15.8

Example 38 cont’d

Solution.


1 P(X > 3) = (1 − .7)3 = 0.0272 P(X = 3) = .7(1 − .7)2 = 0.0633 P(Y = 5) =

(42

)(0.7)3(0.3)2 = 0.1852

4 P(Z = 123) =(122

99

)(0.7)100(0.3)23 = 0.0012


Motivation


15.9

Example 39

The Minnesota Twins are having a bad year. Suppose theirability to win any one game is 42% and games areindependent of one another.

1 What is the probability it takes 14 games for them to wintheir fourth game?

2 What is the expected value and variance of the number ofgames it will take them to win their fortieth game?

3 What is the expected value and variance of the number ofgames it will take them to win their first game?

4 Knowing they got their 49th win with 5 games remaining inthe season, what is the probability that they do not get 50or more wins?


Motivation


15.10

Example 39 cont’d

Solution.

Let X , Y , and Z represent respectively NB(r = 4,p = 0.42),NB(r = 40,p = 0.42) and Geo(p = 0.42)

1 P(X = 14) =(13

3

)(0.42)4(1 − 0.42)10 = 0.0383

2 E[Y ] = 400.42 = 95.2381 Var(Y ) = 40(1−0.42)

0.422 = 131.5193

3 E[Z ] = 10.42 = 2.3810 Var(Z ) = (1−0.42)

0.422 = 3.28804 P(Z > 5) = (1 − 0.42)5 = 0.0656


Motivation


15.10

Example 39 cont’d

Solution.


1 P(X = 14) =(13

3

)(0.42)4(1 − 0.42)10 = 0.0383

2 E[Y ] = 400.42 = 95.2381 Var(Y ) = 40(1−0.42)

0.422 = 131.5193

3 E[Z ] = 10.42 = 2.3810 Var(Z ) = (1−0.42)

0.422 = 3.2880

4 P(Z > 5) = (1 − 0.42)5 = 0.0656


Motivation


15.10

Example 39 cont’d

Solution.


1 P(X = 14) =(13

3

)(0.42)4(1 − 0.42)10 = 0.0383

2 E[Y ] = 400.42 = 95.2381 Var(Y ) = 40(1−0.42)

0.422 = 131.5193

3 E[Z ] = 10.42 = 2.3810 Var(Z ) = (1−0.42)

0.422 = 3.28804 P(Z > 5) = (1 − 0.42)5 = 0.0656

distribution lecture 15 - purdue university

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