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

2. Discrete random variable X Examples: shoe size, dosage (mg), # cells,…
Recall…
POPULATION
Discrete random variable X Examples: shoe size, dosage (mg), # cells,…
Pop values x
Probabilities
p(x)
Cumul Probs
F (x) x1
p(x1) x2
p(x2)
p(x1) + p(x2) x3
p(x3)
px1) + p(x2) + p(x3) ⋮ 1
Total X
Total Area = 1
Mean
Variance

3. Classical Discrete Probability Distributions
Binomial ~ X = # Successes in n trials, P(Success) = 
Poisson ~ As above, but n large,  small, i.e., Success RARE
Negative Binomial ~ X = # trials for k Successes, P(Success) = 
Geometric ~ As above, but specialized to k = 1
Hypergeometric ~ As Binomial, but  changes between trials
Multinomial ~ As Binomial, but for multiple categories, with 1 + 2 + … + last = 1 and x1 + x2 + … + xlast = n

4. ~ The Binomial Distribution ~
Used only when dealing with binary outcomes (two categories: “Success” vs. “Failure”), with a fixed probability of Success () in the population.
Calculates the probability of obtaining any given number of Successes in a random sample of n independent “Bernoulli trials.”
Has many applications and generalizations, e.g., multiple categories, variable probability of Success, etc.

5. For x = 0, 1, 2, 3, …, n X = # Successes in sample x p(x) p(0) 1 p(1)
For any randomly selected individual, define a binary random variable:
POPULATION
Discrete random variable
X = # Successes in sample
(x = 0, 1, 2, 3, …, n)
RANDOMSAMPLE
size n x
p(x)
p(0) 1
p(1) … n
p(n)
F(x)
F(0)
F(1) … 1
For x = 0, 1, 2, 3, …, n

6. Reformulate this as n independent coin tosses.
For any randomly selected individual, define a binary random variable:
POPULATION
Discrete random variable
X = # Heads in sample
(x = 0, 1, 2, 3, …, n)
Reformulate this as n independent coin tosses.
RANDOMSAMPLE
size n
.… etc….
.… etc….
Each such sequence has probability
There are such sequences of n tosses, with x Heads.

7. X = # Successes in sample
For any randomly selected individual, define a binary random variable:
POPULATION
“Success” vs.
“Failure”
RANDOM SAMPLE
of n “Bernoulli trials.”
Discrete random variable
X = # Successes in sample
(x = 0, 1, 2, 3, …, n)
independent, with constant probability () per trial
Then X is said to follow a Binomial distribution, written X ~ Bin(n, ), with “probability mass function”
p(x) = , x = 0, 1, 2, …, n.

8. Example: Blood Type probabilities, revisited
Rh Factor
Blood Type + – O
.384
.077
.461 A
.323
.065
.388 B
.094
.017
.111 AB
.032
.007
.039
.833
.166
.999
Check:
1. Independent outcomes?
Reasonably assume that outcomes “Type O” vs. “Not Type O” between two individuals are independent of each other. 
2. Constant probability  ?
Suppose n = 10 individuals are to be selected at random from the population.
Probability table for X = #(Type O)
From table,  = P(Type O) = .461 throughout population. 
Binomial model applies?

9. Example: Blood Type probabilities, revisited
p(x) = (.461)x (.539)10 – x
Example: Blood Type probabilities, revisited
R: dbinom(0:10, 10, .461)
Rh Factor
Blood Type + – O
.384
.077
.461 A
.323
.065
.388 B
.094
.017
.111 AB
.032
.007
.039
.833
.166
.999 x
p(x)
F (x)
(.461)0 (.539)10 = 1
(.461)1 (.539)9 = 2
(.461)2 (.539)8 = 3
(.461)3 (.539)7 = 4
(.461)4 (.539)6 = 5
(.461)5 (.539)5 = 6
(.461)6 (.539)4 = 7
(.461)7 (.539)3 = 8
(.461)8 (.539)2 = 9
(.461)9 (.539)1 = 10
(.461)10 (.539)0 =
Suppose n = 10 individuals are to be selected at random from the population.
Probability table for X = #(Type O)
Binomial model applies.
X ~ Bin(10, .461)
Also, can show mean  =  x p(x) =
and variance  2 =  (x – ) 2 p(x) = n
= 4.61
= (10)(.461)
n (1 – )
= 2.48

10. Example: Blood Type probabilities, revisited
p(x) = (.461)x (.539)10 – x
Example: Blood Type probabilities, revisited
R: dbinom(0:10, 10, .461)
Rh Factor
Blood Type + – O
.384
.077
.461 A
.323
.065
.388 B
.094
.017
.111 AB
.032
.007
.039
.833
.166
.999 x
p(x)
F (x)
(.461)0 (.539)10 = 1
(.461)1 (.539)9 = 2
(.461)2 (.539)8 = 3
(.461)3 (.539)7 = 4
(.461)4 (.539)6 = 5
(.461)5 (.539)5 = 6
(.461)6 (.539)4 = 7
(.461)7 (.539)3 = 8
(.461)8 (.539)2 = 9
(.461)9 (.539)1 = 10
(.461)10 (.539)0 =
Suppose n = 10 individuals are to be selected at random from the population.
Probability table for X = #(Type O)
Binomial model applies.
X ~ Bin(10, .461)
Also, can show mean  =  x p(x) =
and variance  2 =  (x – ) 2 p(x) = n
= 4.61
n (1 – )
= 2.48

11. n = 10
p = .461
pmf = function(x)(dbinom(x, n, p))
N =
x = 0:10
bin.dat = rep(x, N*pmf(x))
hist(bin.dat, freq = F, breaks = c(-.5, x+.5), col = "green")
axis(1, at = x)
axis(2)

12. Example: Blood Type probabilities, revisited
Rh Factor
Blood Type + – O
.384
.077
.461 A
.323
.065
.388 B
.094
.017
.111 AB
.032
.007
.039
.833
.166
.999
Rh Factor
Blood Type + – O
.384
.077
.461 A
.323
.065
.388 B
.094
.017
.111 AB
.032
.007
.039
.833
.166
.999
Therefore,
p(x) =
x = 0, 1, 2, …, 1500.
RARE EVENT!
Suppose n = 10 individuals are to
be selected at random from the population.
Probability table for X = #(Type AB–)
n = 1500 individuals are to
Binomial model applies.
X ~ Bin(10, .461)
X ~ Bin(1500, .007)
Also, can show mean  =  x p (x) =
and variance  2 =  (x – ) 2 p(x) = n
= 10.5
n (1 – )
2.48 =

13. Example: Blood Type probabilities, revisited
Rh Factor
Blood Type + – O
.384
.077
.461 A
.323
.065
.388 B
.094
.017
.111 AB
.032
.007
.039
.833
.166
.999
Rh Factor
Blood Type + – O
.384
.077
.461 A
.323
.065
.388 B
.094
.017
.111 AB
.032
.007
.039
.833
.166
.999
Therefore,
p(x) =
x = 0, 1, 2, …, 1500.
Is there a better alternative?
RARE EVENT!
RARE EVENT!
Suppose n = 10 individuals are to
be selected at random from the population.
Probability table for X = #(Type AB–)
n = 1500 individuals are to
Long positive skew as x  1500
…but contribution  0
Binomial model applies.
X ~ Bin(10, .461)
X ~ Bin(1500, .007)
Also, can show mean  =  x p (x) =
and variance  2 =  (x – ) 2 p(x) = n
= 10.5
n (1 – )
2.48 =

14. Example: Blood Type probabilities, revisited
Rh Factor
Blood Type + – O
.384
.077
.461 A
.323
.065
.388 B
.094
.017
.111 AB
.032
.007
.039
.833
.166
.999
Rh Factor
Blood Type + – O
.384
.077
.461 A
.323
.065
.388 B
.094
.017
.111 AB
.032
.007
.039
.833
.166
.999
Therefore,
p(x) =
x = 0, 1, 2, …, 1500.
Is there a better alternative?
Poisson distribution
x = 0, 1, 2, …,
where mean and variance are
 = n and  2 = n
RARE EVENT! 
Suppose n = 10 individuals are to
be selected at random from the population.
Probability table for X = #(Type AB–)
n = 1500 individuals are to
= 10.5
Binomial model applies.
X ~ Bin(1500, .007)
X ~ Poisson(10.5)
Also, can show mean  =  x p(x) =
and variance  2 =  (x – ) 2p(x) = n
= 10.5
Notation: Sometimes the symbol  (“lambda”) is used instead of  (“mu”).
n (1 – ) =

15. Example: Blood Type probabilities, revisited
Rh Factor
Blood Type + – O
.384
.077
.461 A
.323
.065
.388 B
.094
.017
.111 AB
.032
.007
.039
.833
.166
.999
Rh Factor
Blood Type + – O
.384
.077
.461 A
.323
.065
.388 B
.094
.017
.111 AB
.032
.007
.039
.833
.166
.999
Therefore,
p(x) =
x = 0, 1, 2, …, 1500.
Is there a better alternative?
Poisson distribution
x = 0, 1, 2, …,
where mean and variance are
 = n and  2 = n
RARE EVENT!
Suppose n = 1500 individuals are to
be selected at random from the population.
Probability table for X = #(Type AB–)
= 10.5
X ~ Poisson(10.5)
Ex: Probability of exactly X = 15 Type(AB–) individuals = ?
Poisson:
Binomial:
(both ≈ .0437)

16. Example: Deaths in Wisconsin

17. Example: Deaths in Wisconsin
Assuming deaths among young adults are relatively rare, we know the following:
Average deaths per year
λ =
Mortality rate (α) seems constant.
Therefore, the Poisson distribution can be used as a good model to make future predictions about the random variable X = “# deaths” per year, for this population (15-24 yrs)… assuming current values will still apply.
Probability of exactly X = 600 deaths next year
P(X = 600) =
0.0131
R: dpois(600, 584)
Probability of exactly X = 1200 deaths in the next two years
Mean of 584 deaths per yr  Mean of 1168 deaths per two yrs,
so let λ = 1168:
P(X = 1200) =
Probability of at least one death per day:
λ =
= 1.6 deaths/day
P(X ≥ 1) =
P(X = 1) + P(X = 2) + P(X = 3) + …
True, but not practical.
P(X ≥ 1) =
1 – P(X = 0)
= 1 –
= 1 – e–1.6 =
0.798