1. Chapter 3 Discrete Random Variables and Probability Distributions
3.2 - Probability Distributions for Discrete
Random Variables
3.3 - Expected Values
3.4 - The Binomial Probability Distribution
3.5 - Hypergeometric and Negative
Binomial Distributions
3.6 - The Poisson Probability Distribution
… and Multinomial Distrib

2. Hypergeometric Distribution
Random Sample
n = 5
(w/ or w/o replacement)
Infinite Population
of decks of 52 cards (Assume each is fair)
Random Variable
X = # Spades in sample
(x = 0, 1, 2, 3, 4, 5)
Calculate P(X = 2)
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)
Binary r.v.
10 combinations

3. Hypergeometric Distribution
Random Sample
n = 5
(w/ or w/o replacement)
Random Sample
n = 5
(w/ replacement)
Infinite Population
of decks of 52 cards (Assume each is fair)
Finite Population
N = 52
(Assume fair)
Random Variable
X = # Spades in sample
(x = 0, 1, 2, 3, 4, 5)
Calculate P(X = 2)
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)
Binary r.v.
10 combinations

4. Hypergeometric Distribution
Random Sample
n = 5
(w/o replacement)
Random Sample
n = 5
(w/ replacement)
Finite Population
N = 52
(Assume fair)
Random Variable
X = # Spades in sample
(x = 0, 1, 2, 3, 4, 5)
Calculate P(X = 2)
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)
Binary r.v.
10 combinations

5. Hypergeometric Distribution
Random Sample
n = 5
(w/o replacement)
Finite Population
N = 52
(Assume fair)
Random Variable
X = # Spades in sample
(x = 0, 1, 2, 3, 4, 5)
Calculate P(X = 2)
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)
Binary r.v. ? ? ? ? ? ? =

6. Hypergeometric Distribution
Random Sample
n = 5
(w/o replacement)
Finite Population
N = 52
(Assume fair)
Random Variable
X = # Spades in sample
(x = 0, 1, 2, 3, 4, 5)
Calculate P(X = 2)
# combinations of Spades from 13 Spades
# combinations of
3 Non-Spades from 39 Non-Spades 
Sample Space
All combinations of 5 cards from 52 = #
R command:
dhyper(2, 13, 39, 5)
x, s, N – s, n

7. Hypergeometric Distribution
Random Sample
(without replacement)
of size n  N/10
Finite Population
of size N
s = # Successes
Discrete random variable
X = # Successes in sample
(x = 0, 1, 2, 3, …,, n)
 N – s = # Failures
Then for any x = 0, 1, 2,…, the pmf p(x) is given by the following…
# combinations of x Successes out of s Successes
# combinations of n – x Failures out of N – s Failures
See textbook for  and  2.
# combinations of n out of N

8. Hypergeometric Distribution
POPULATION
N = 100
Random Sample
(w/o replacement), n = 12
Discrete random variable
X = # defectives in sample
(x = 0, 1, 2, 3, 4)
s = 4 defectives

9. Hypergeometric Distribution
POPULATION
N = 100
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)
s = 4 defectives

10. ∞ N < ∞ N < ∞, n  N/10 X = # Successes (x) in n trials
(x = 0, 1, 2, …, n)
Classical Discrete Model
Population Size
Sampling: replacement?
Bernoulli trials?2
pmf p(x)
P(X = x)
Binomial
X ~ Bin(n, )
dbinom(x, n, ) ∞
with or without
yes
N < ∞
with
Poisson1
X ~ Pois()
dpois(x, )
Hypergeometric
X ~ Hyp(x, s, N, n)
dhyper(x, s, N – s, n)
N < ∞,
n  N/10
without no
1 for rare events ONLY, i.e., small, n large
2 independent outcomes, with constant  = P(Success) in population

11. Negative Binomial Distribution
Infinite Population
of “Successes” and “Failures”
Random Sample
(w/ or w/o replacement)
P(Success) = 
P(Failure) = 1 – 
Sample size NOT specified!
Discrete random variable
X = # trials for s Successes
(x = s, s+1, s+2, s+3, …)
Then for any x = s, s+1, s+2,…, the pmf p(x) is…
See textbook for  and  2.
dnbinom(x–s, s, )
(In R, X counts the # Failures before s Successes)

12. Negative Binomial Distribution
Random Variable
X = # trials for s = 6 Spades
(x = 6, 7, 8, 9, 10,…)
 = P(Spades) = 13/52
Random Sample
(w/ replacement)

13. Negative Binomial Distribution
Random Variable
X = # trials for s = 1 Spades
(x = 1, 2, 3, 4, 5,…)
Random Variable
X = # trials for s = 6 Spades
(x = 6, 7, 8, 9, 10,…)
 = P(Spades) = 13/52
Random Sample
(w/ replacement)
SPECIAL CASE OF NEGATIVE BINOMIAL: s = 1

14. Negative Binomial Distribution Geometric Distribution
Random Variable
X = # trials for s = 1 Spades
(x = 1, 2, 3, 4, 5,…)
 = P(Spades) = 13/52
Random Sample
(w/ replacement)
SPECIAL CASE OF NEGATIVE BINOMIAL: s = 1
(x – 1) Failures
1 Success!
dgeom(x–1, )
(In R, X counts the # Failures before 1 Success)

15. Negative Binomial Distribution Geometric Distribution
Random Variable
X = # trials for s = 1 Spades
(x = 1, 2, 3, 4, 5,…)
 = P(Spades) = 13/52
Random Sample
(w/ replacement)
SPECIAL CASE OF NEGATIVE BINOMIAL: s = 1

16. Negative Binomial Distribution Geometric Distribution
Random Variable
X = # trials for s = 1 Spades
(x = 1, 2, 3, 4, 5,…)
 = P(Spades) = 13/52
Random Sample
(w/ replacement) x 1 2 3 4 5 6 7
p(x)
.250
.188
.141
.105
.079
.059
.044
Exercise
Graph cdf
F(x)

17. Multinomial Distribution Geometric Distribution
Random Variable
X = # trials for s = 1 Spades
(x = 1, 2, 3, 4, 5,…)
Random Variables
X1 = # Spades
X2 = # Clubs
X3 = # Hearts
X4 = # Diamonds
For i = 1, 2, 3, 4
xi = 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
 = P(Spades) = 13/52
Random Sample
(w/ replacement)
Random Sample
n = 10
(w/ replacement)

18. Discrete random variable
RECALL…
BINARY POPULATION
of “Successes” vs. “Failures”
P(Success) = 
P(Failure) = 1 – 
Discrete random variable
X = # “Successes” in sample
(n – X = # “Failures” in sample)
(0, 1, 2, 3, …, n)
RANDOM SAMPLE
of n “Bernoulli trials”
Then X is said to follow a Binomial distribution, written
X ~ Bin(n, ), with “probability mass function” p(x) = …
x = 0, 1, 2, …, n

19. Discrete random variable
RECALL…
BINARY POPULATION
of “Successes” vs. “Failures”
P(Success) = 
P(Failure) = 1 – 
OR…
Discrete random variable
X = # “Successes” in sample
(n – X = # “Failures” in sample)
(0, 1, 2, 3, …, n)
RANDOM SAMPLE
of n “Bernoulli trials”
Then X is said to follow a Binomial distribution, written
X ~ Bin(n, ), with “probability mass function” p(x) = …
x = 0, 1, 2, …, n

20. Discrete random variables Discrete random variable
RECALL…
BINARY POPULATION
of “Successes” vs. “Failures”
P(Success) = 1
P(Failure) = 2
1 + 2 = 1
OR…
Discrete random variables
X1 = # “Successes” in sample
X2 = # “Failures” in sample = n – X1
(0, 1, 2, 3, …, n)
Discrete random variable
X = # “Successes” in sample
(n – X = # “Failures” in sample)
(0, 1, 2, 3, …, n)
RANDOM SAMPLE
of n “Bernoulli trials”
Then X is said to follow a Binomial distribution, written
X ~ Bin(n, ), with “probability mass function” p(x) = …
x = 0, 1, 2, …, n

21. Discrete random variables
RECALL…
BINARY POPULATION
of “Successes” vs. “Failures”
P(Success) = 1
P(Failure) = 2
1 + 2 = 1
OR…
Discrete random variables
X1 = # “Successes” in sample
X2 = # “Failures” in sample = n – X1
(0, 1, 2, 3, …, n)
RANDOM SAMPLE
of n “Bernoulli trials”
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

22. Discrete random variables
POPULATION of k categories
P(Category 1) = 1
P(Category 2) = 2
P(Category k) = k
BINARY POPULATION
of “Successes” vs. “Failures”
P(Success) = 1
P(Failure) = 2
1 + 2 = 1
1 + 2 + … + k = 1
Discrete random variables
X1 = # “Successes” in sample
X2 = # “Failures” in sample = n – X1
(0, 1, 2, 3, …, n)
Discrete random variables
X1 = # Category 1 in sample
X2 = # Category 2 in sample
Xk = # Category k in sample
RANDOM SAMPLE
of n “Bernoulli trials”
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

23. Multinomial Distribution
Random Variables
X1 = # Spades
X2 = # Clubs
X3 = # Hearts
X4 = # Diamonds
xi = 0, 1, 2,…,10 with
x1 + x2 + x3 + x4 = 10.
Random Variable
X = # trials for s = 6 Spades
(x = 6, 7, 8, 9, 10,…)
1 = P(Spades) = 13/52
2 = P(Clubs) = 13/52
3 = P(Hearts) = 13/52
4 = P(Diamonds) = 13/52
Random Sample
(w/ replacement)
Random Sample
n = 10
(w/ replacement)
 = P(Spades) = 13/52
dmultinom(c(1,2,3,4), 10,
c(.25,.25,.25,.25))