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

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

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

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

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. ? ? ? ? ? ? = 0.27428

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 2 Spades from 13 Spades # combinations of 3 Non-Spades from 39 Non-Spades  Sample Space All combinations of 5 cards from 52 = 0.27428 # R command: dhyper(2, 13, 39, 5) x, s, N – s, n

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

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

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) 0 0.594684 1 0.335822 2 0.064431 3 0.004937 4 0.000126 s = 4 defectives

∞ 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

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)

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)

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

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)

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

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)

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)

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

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

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

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

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 .

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))