Lecture 8. Discrete Probability
Distributions
David R. Merrell
90-786 Intermediate Empirical
Methods for Public Policy and
Management
AGENDA
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Review Random Variables
Binomial Process
Binomial Distribution
Poisson Process
Poisson Distribution
Example 1. Flip Three Coins
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Sample space is HHH, HHT, HTH, THH,
HTT, THT, TTH, TTT
X = the number of heads, X = 0, 1, 2,
3
Probability Distribution:
x
0
1
2
3
P(x)
1/8
3/8
3/8
1/8
P(x)
3/8
2/8
1/8
0
1
2
3
Probability Tree Form
1/8
H
H
H
T
T
H
T
1/8
1/8
T
H
1/8
T
1/8
H
1/8
T
1/8
1/8
H
T
P(X=1) = 3/8
1/8
H
H
H
T
T
H
T
1/8
1/8
T
H
1/8
T
1/8
H
1/8
T
1/8
1/8
H
T
Expected Value
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The expected value (mean) of a
probability distribution is a weighted
average: weights are the probabilities
Expected Value: E(X) =  = xiP(xi)
Calculating Expected Value
E(X) =
0(1/8) + 1(3/8) + 2(3/8) + 3(1/8) = 1.5
Variance
V(X) = E(X-)2
Calculating Variance: 


x
(x-
(x- p(x)
0
(0-1.5)2
2.25(1/8)
1
2
0.25(3/8)
2
0.25(3/8)
2
2.25(1/8)
2
3
(1-1.5)
(2-1.5)
(3-1.5)
Example 2. Program Pilot--Bayes
Rule
Success
.3
.7
Failure
Good
.9
.1
Bad
Good
.9
P( Good) .27.63 .90
P(Success|Good) =
.03 P( Good|Success) P(Success) 
P( Good)
.63 (.9)(.3)
.30
.90
.27
.1
Bad
.07
Bernoulli RV
1
X=
0
P(X = 1) = P(A) inherits probabilities
Application: Survey of Employment
Discrimination
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Wall Street Journal, 1991 May 15
Pairs of equally qualified white and
black applicants for entry-level positions
Dichotomy: job offer or not
Results:
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28% of whites offered jobs
18% of blacks offered jobs
Bernoulli Probabilities
X  1 if White offered job
Y = 1 if Black offered job
P ( X  1) .28
P (Y  1) .18
Expected Value
1-p
p
0

1
  p  1  (1  p)  0
note long-run relative frequency interpretation
Variance of a Bernoulli RV
V(X) = p - p2 = p(1-p)
0.25
V(X)
0.20
0.15
0.10
0.05
0.00
0.0
0.5
p
1.0
Bernoulli Process
1
2
3
4
5
6
A sequence of independent Bernoulli trials each with
probability p of taking on the value 1
Application: Examine “abandoned” buildings to see
if they are in fact occupied
Binomial Distribution
Count occurrences in n trials
 n k
P(Y = k) =   p (1-p)n-k; k = 0,1,..., n
 k
Survey 1200 buildings.
How many are actually occupied?
Parameters
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Mean:
= np
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Variance:
 = n p q
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Standard Deviation:

Example 3. Racial Discrimination
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Stermerville Public Works Department charged
with racial discrimination in hiring practices
40% of the persons who passed the
department’s civil service exam were minorities
From this group, the Department hired 10
individuals; 2 of them were minorities.
What is the probability that, if the Department
did not discriminate, it would have hired 2 or
fewer minorities?
Example 3. Solution
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Success: a minority is hired
Probability of success: p = 0.4, if the
department shows no preferences in
regard to hiring minorities
Number of trials, n = 10
Number of successes, x = 2
P(x  2) = 0.12 + 0.04 + 0.006 = 0.166
Example 4. Probability Distribution
x
0
1
2
3
4
5
6
7
8
9
10
P(x)
0.006047
0.040311
0.120932
0.214991
0.250823
0.200658
0.111477
0.042467
0.010617
0.001573
0.000105
0
1
2
3
4
5
6
7
8
9
10
0.006047
0.040311
0.120932
0.214991
0.250823
0.200658
0.111477
0.042467
0.010617
0.001573
0.000105
Poisson Process
rate
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x x
0
x
time
Assumptions
time homogeneity
independence
no clumping
Application: Toll Booth
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Arrival times of cars
Mean arrival rate,  cars per minute
Busy:cars per minute
Slow: = 0.5 cars per minute
Poisson Distribution
Count in time period t
P(Y  y)
e
 t
(t )
; y0,1,
y!
y
Probability Calculation
  8, t  0.5
e 4 ( 4 ) 2
P ( X  2) 
 01465
.
2!
Poisson Mean and Variance
E (Y )  t
V (Y )  t
Note: For a Poisson RV, E(Y) = V(Y).
  4, t .5 
E (Y )  V (Y )  2
Next Time ...
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Continuous Probability Distributions
Normal Distribution