Infinite Population of decks of 52 cards - PowerPoint Presentation

Slide1

Infinite

Population

of decks of 52 cards

(Assume each is fair)

Random Sample

n

= 5

(w/ or w/o replacement)

Random Variable

X =

# Spades in sample

(

x

= 0, 1, 2, 3, 4, 5)

Outcomes of “X = 2”

(0, 0, 0, 1, 1)

(0, 0, 1, 0, 1)

(0, 0, 1, 1, 0)

(0, 1, 0, 0, 1)

(0, 1, 0, 1, 0)

(0, 1, 1, 0, 0)(1, 0, 0, 0, 1)(1, 0, 0, 1, 0)(1, 0, 1, 0, 0)(1, 1, 0, 0, 0)

Hypergeometric Distribution

Binary

r.v

.

10 combinations

Calculate

P

(

X

= 2)

Binomial distribution



= P(Spades) = 13/52

Binary

r.v

.

Outcomes of “X = 2”

(0, 0, 0, 1, 1)(0, 0, 1, 0, 1)(0, 0, 1, 1, 0)(0, 1, 0, 0, 1)(0, 1, 0, 1, 0)(0, 1, 1, 0, 0)(1, 0, 0, 0, 1)(1, 0, 0, 1, 0)(1, 0, 1, 0, 0)(1, 1, 0, 0, 0)

Random Sample

n

= 5(w/ or w/o replacement)

Random Samplen = 5(w/ replacement)

Infinite Populationof decks of 52 cards (Assume each is fair)

Calculate

P

(

X

= 2)

Finite

Population

N

= 52

(Assume fair)

Hypergeometric Distribution

10 combinations

Random Variable

X =

# Spades in sample

(x = 0, 1, 2, 3, 4, 5)

Binomial distribution



=

P

(Spades) = 13/52

Outcomes of “X = 2”(0, 0, 0, 1, 1)(0, 0, 1, 0, 1)(0, 0, 1, 1, 0)(0, 1, 0, 0, 1)(0, 1, 0, 1, 0)(0, 1, 1, 0, 0)(1, 0, 0, 0, 1)(1, 0, 0, 1, 0)(1, 0, 1, 0, 0)(1, 1, 0, 0, 0)

Random Samplen = 5(w/ replacement)

Hypergeometric Distribution

Finite

Population

N

= 52

(Assume fair)

Calculate

P

(X = 2)

10 combinations

Random Variable

X =

# Spades in sample

(x = 0, 1, 2, 3, 4, 5)

Binomial distribution = P(Spades) = 13/52

Random Sample

n = 5(w/o replacement)

Binary

r.v

.

Random Samplen = 5(w/o replacement)

?

?

?

?

?

?

Outcomes of “X = 2”

(0, 0, 0, 1, 1)

(0, 0, 1, 0, 1)

(0, 0, 1, 1, 0)

(0, 1, 0, 0, 1)

(0, 1, 0, 1, 0)(0, 1, 1, 0, 0)(1, 0, 0, 0, 1)(1, 0, 0, 1, 0)(1, 0, 1, 0, 0)(1, 1, 0, 0, 0)

Hypergeometric Distribution

Finite

Population

N

= 52

(Assume fair)

Calculate

P

(X = 2)

= 0.27428

Random Variable

X =

# Spades in sample

(

x

= 0, 1, 2, 3, 4, 5)

Binomial distribution



=

P

(Spades) = 13/52

Binary

r.v

.

Sample SpaceAll combinations of 5 cards from 52

#

Hypergeometric Distribution

Finite

Population

N

= 52

(Assume fair)

Calculate

P

(

X = 2)

Random Samplen = 5(w/o replacement)

# combinations of 2

Spades

from 13

Spades

# combinations of

3

Non-Spades

from 39

Non-Spades

= 0.27428



R command:

dhyper

(2, 13, 39, 5)

Random Variable

X =

# Spades in sample

(x = 0, 1, 2, 3, 4, 5)

x, s, N – s, n

Hypergeometric Distribution

Finite

Populationof size N

s = # Successes

 N – s = # Failures

Random Sample

(without replacement) of size n  N/10

Discrete random variableX = # Successes in sample(x = 0, 1, 2, 3, …,, n)

Then for any x = 0, 1, 2,…, the pmf p(x) is given by the following…

See textbook for

 and  2.

# combinations of x Successes out of s Successes

# combinations of n – x Failures out of N – s Failures

# combinations of

n

out of

N

Hypergeometric Distribution

POPULATIONN = 100

s = 4 defectives

Random Sample(w/o replacement), n = 12

Discrete

random variable

X

= #

defectives

in sample

(

x

= 0

, 1, 2, 3

, 4)

Hypergeometric Distribution

POPULATIONN = 100

s = 4 defectives

Random Sample(w/o replacement), n = 12

Discrete

random variable

X

= #

defectives

in sample

(

x

= 0

, 1, 2, 3

, 4)

dhyper

(0:4, 4, 96, 12)

0 0.594684

1 0.335822

2 0.064431

3 0.004937

4 0.000126

X = # Successes (x) in n trials(x = 0, 1, 2, …, n) Classical Discrete ModelPopulation SizeSampling: replacement?Bernoulli trials?2pmf p(x)P(X = x)BinomialX ~ Bin(n, )dbinom(x, n, )∞with or withoutyesN < ∞withyesPoisson1X ~ Pois()dpois(x, )∞with or withoutyesHypergeometricX ~ Hyp(x, s, N, n)dhyper(x, s, N – s, n)N < ∞,n  N/10withoutno

1 for rare events ONLY, i.e., small, n large2 independent outcomes, with constant  = P(Success) in population

Negative Binomial Distribution

Discrete

random variable

X = # trials for s Successes(x = s, s+1, s+2, s+3, …)

See textbook for

 and  2.

Random Sample(w/ or w/o replacement)

Infinite Populationof “Successes” and “Failures”

Then for any x = s, s+1, s+2,…, the pmf p(x) is…

P(Success) = P(Failure) = 1 – 

Sample size NOT specified!

dnbinom(x–s, s, )

(In R,

X

counts the #

Failures

before

s

Successes)

Random Sample

(w/ replacement)



=

P

(Spades) = 13/52

Negative Binomial Distribution

Random Variable

X =

# trials for

s

= 6 Spades

(

x

= 6, 7, 8, 9, 10,…)

Random Sample

(w/ replacement)



=

P

(Spades) = 13/52

Negative Binomial Distribution

Random Variable

X =

# trials for s = 6 Spades(x = 6, 7, 8, 9, 10,…)

SPECIAL CASE OF NEGATIVE BINOMIAL: s

= 1

Random Variable

X = # trials for s = 1 Spades(x = 1, 2, 3, 4, 5,…)

Random VariableX = # trials for s = 1 Spades(x = 1, 2, 3, 4, 5,…)

Random Sample

(w/ replacement)



=

P

(Spades) = 13/52

Negative Binomial Distribution

SPECIAL CASE OF NEGATIVE BINOMIAL: s

= 1

(

x

– 1) Failures

1 Success!

Geometric Distribution

dgeom

(x–1, )

(In R,

X

counts the #

Failures

before

1

Success)

Random VariableX = # trials for s = 1 Spades(x = 1, 2, 3, 4, 5,…)

Random Sample

(w/ replacement)



=

P

(Spades) = 13/52

Negative Binomial Distribution

SPECIAL CASE OF NEGATIVE BINOMIAL: s

= 1

Geometric Distribution

Random VariableX = # trials for s = 1 Spades(x = 1, 2, 3, 4, 5,…)

Random Sample(w/ replacement)



= P(Spades) = 13/52

Negative Binomial Distribution

Geometric Distribution

x

1

2

34567p(x).250.188.141.105.079.059.044

ExerciseGraph cdf

F(x)

Random VariableX = # trials for s = 1 Spades(x = 1, 2, 3, 4, 5,…)

Random Variables X1 = # Spades X2 = # Clubs X3 = # Hearts X4 = # DiamondsFor i = 1, 2, 3, 4 xi = 0, 1, 2,…,10 with x1 + x2 + x3 + x4 = 10.

Geometric Distribution

Random Sample(w/ replacement)



=

P

(Spades) = 13/52

Multinomial Distribution



1

=

P

(Spades) = 13/52

2 = P(Clubs) = 13/523 = P(Hearts) = 13/524 = P(Diamonds) = 13/52

Random Samplen = 10(w/ replacement)

Discrete random variableX = # “Successes” in sample(n – X = # “Failures” in sample)(0, 1, 2, 3, …, n)

RANDOM SAMPLE of n “Bernoulli trials”

BINARY POPULATIONof “Successes” vs. “Failures” P(Success) =  P(Failure) = 1 – 

Then

X is said to follow a Binomial distribution, written X ~ Bin(n, ), with “probability mass function” p(x) = … x = 0, 1, 2, …, n

RECALL…

Discrete random variableX = # “Successes” in sample(n – X = # “Failures” in sample)(0, 1, 2, 3, …, n)

BINARY POPULATIONof “Successes” vs. “Failures” P(Success) =  P(Failure) = 1 – 

Then X is said to follow a Binomial distribution, written X ~ Bin(n, ), with “probability mass function” p(x) = … x = 0, 1, 2, …, n

RECALL…

RANDOM SAMPLE

of

n “Bernoulli trials”

OR…

Discrete random variableX = # “Successes” in sample(n – X = # “Failures” in sample)(0, 1, 2, 3, …, n)

Discrete random variablesX1 = # “Successes” in sampleX2 = # “Failures” in sample = n – X1(0, 1, 2, 3, …, n)

BINARY POPULATIONof “Successes” vs. “Failures” P(Success) = 1 P(Failure) = 2

1 + 2 = 1

RECALL…

OR…

Then

X is said to follow a Binomial distribution, written X ~ Bin(n, ), with “probability mass function” p(x) = … x = 0, 1, 2, …, n

RANDOM SAMPLE

of

n

“Bernoulli trials”

BINARY POPULATION

of “Successes” vs. “Failures” P(Success) = 1 P(Failure) = 2

1 + 2 = 1

RECALL…

RANDOM SAMPLE of n “Bernoulli trials”

OR…

Discrete random variablesX1 = # “Successes” in sampleX2 = # “Failures” in sample = n – X1(0, 1, 2, 3, …, n)

Then

X1 and X2 “jointly” follow a Binomial distribution, written X1 ~ Bin(n, 1), X2 ~ Bin(n, 2), with “probability mass function”… for any x1 = 0, 1, 2, …, n, x2 = 0, 1, 2,…, n, with .

x1 + x2 = n

Then the components of

X = (X1, X2,…, Xk) “jointly” follow a Multinomial distribution, written X ~ Multi(n, 1, 2, …, k), with “probability mass function” p(x1,…, xk) = for any xi = 0, 1, 2, …, n, with .



1

+ 2 = 1

BINARY POPULATIONof “Successes” vs. “Failures” P(Success) = 1 P(Failure) = 2

1 + 2 + … + k = 1

POPULATION

of k categories P(Category 1) = 1 P(Category 2) = 2 P(Category k) = k

Discrete

random variablesX1 = # “Successes” in sampleX2 = # “Failures” in sample = n – X1(0, 1, 2, 3, …, n)

Discrete

random variablesX1 = # Category 1 in sampleX2 = # Category 2 in sampleXk = # Category k in sample

RANDOM SAMPLE

of

n

“Bernoulli trials”

Multinomial Distribution

Random Sample(w/ replacement)



=

P

(Spades) = 13/52

Random Variable

X =

# trials for

s = 6 Spades(x = 6, 7, 8, 9, 10,…)

Random Variables

X

1

= # Spades X2 = # Clubs X3 = # Hearts X4 = # Diamondsxi = 0, 1, 2,…,10 with x1 + x2 + x3 + x4 = 10.

1 = P(Spades) = 13/522 = P(Clubs) = 13/523 = P(Hearts) = 13/524 = P(Diamonds) = 13/52

Random Samplen = 10(w/ replacement)

dmultinom

(c(1,2,3,4), 10,

c(.25,.25,.25,.25))