Chapter 17
Probability Models
Binomial Probability Models
Poisson Probability Models
Binomial Random Variables
Binomial Probability Distributions
Binomial Random
Variables
Through 2/25/2014 NC State’s free-throw
percentage is 65.1% (315th out 351 in Div. 1).
 If in the 2/26/2014 game with UNC, NCSU
shoots 11 free-throws, what is the probability
that:

NCSU makes exactly 8 free-throws?
NCSU makes at most 8 free throws?
NCSU makes at least 8 free-throws?
“2-outcome” situations are very
common
Heads/tails
 Democrat/Republican
 Male/Female
 Win/Loss
 Success/Failure
 Defective/Nondefective

Probability Model for this Common
Situation

Common characteristics
◦ repeated “trials”
◦ 2 outcomes on each trial

Leads to Binomial Experiment
Binomial Experiments

n identical trials
◦ n specified in advance

2 outcomes on each trial
◦ usually referred to as “success” and “failure”
p “success” probability; q=1-p “failure”
probability; remain constant from trial to
trial
 trials are independent

Binomial Random Variable
The binomial random variable X is
the number of “successes” in the n
trials
 Notation: X has a B(n, p)
distribution, where n is the number
of trials and p is the success
probability on each trial.

Binomial Probability Distribution
n trials, p  success probability on each trial
probability distribution:
p ( x)  n Cx p q
x
n x
, x  0,1, 2,
E ( x)   xp( x)   x 
n
x 0
n
x 0
n
x
p q
Var ( x)  E ( x      npq

,n
x
n x
 np
Rationale for the Binomial
Probability Formula
P(x) =
n!
•
(n – x )!x!
Number of
outcomes with
exactly x
successes
among n trials
px •
n-x
q
Binomial Probability
Formula
P(x) =
n!
•
(n – x )!x!
Number of
outcomes with
exactly x
successes
among n trials
px •
n-x
q
Probability of x
successes
among n trials
for any one
particular order
Graph of p(x); x binomial n=10 p=.5;
p(0)+p(1)+ … +p(10)=1
The sum of all the
areas is 1
Think of p(x) as the area
of rectangle above x
p(5)=.246 is the area
of the rectangle above 5
Binomial Probability Histogram: n=100, p=.5
0.09
0.08
0.07
0.06
0.05
0.04
0.03
0.02
0.01
70
68
66
64
62
60
58
56
54
52
50
48
46
44
42
40
38
36
34
32
30
0
Binomial Probability Histogram: n=100, p=.95
0.18
0.17
0.16
0.15
0.14
0.13
0.12
0.11
0.1
0.09
0.08
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0
70
72
74
76
78
80
82
84
86
88
90
92
94
96
98
100
Binomial Distribution Example:
Tennis First Serves
A tennis player makes 60% of her
first serves and attempts 4 first-serves.
 Assume the outcomes of first-serves are
independent.
 What is the probability that exactly 3 of her
first serves are successful?

n  4;success=successful first serve; p  .60
p(3)  4 C3 (.60)3 (.40)1  4(.216)(.40)  .3456
Binomial Distribution Example
•
•
1.
2.
Shanille O’Keal is a WNBA player who makes 25% of her 3point attempts.
Assume the outcomes of 3-point shots are independent.
If Shanille attempts 7 3-point shots in a game, what is the
expected number of successful 3-point attempts?
Shanille’s cousin Shaquille O’Neal makes 10% of his 3-point
attempts. If they each take 12 3-point shots, who has the
smaller probability of making 4 or fewer 3-point shots?
E ( X )  np  7(.25) 1.75
Shaq: n  12, p  .10; P( X  4)  .996
Shanille: n  12, p  .25; P( X  4)  .842
Shanille has the smaller probability.
15
Using binomial tables; n=20, p=.3
9, 10, 11, … , 20
P(x  5) = .4164
 P(x > 8) = 1- P(x  8)= 1- .8867=.1133
=P(x 8)
 P(x < 9) = ? 8, 7, 6, … , 0
 P(x  10) = ? 1- P(x  9) = 1- .9520
 P(3  x  7)=P(x  7) - P(x  2)
.7723 - .0355 = .7368

Binomial n = 20, p = .3 (cont.)
P(2 < x  9) = P(x  9) - P(x  2)
= .9520 - .0355 = .9165
 P(x = 8) = P(x  8) - P(x  7)
= .8867 - .7723 = .1144

Color blindness
The frequency of color blindness (dyschromatopsia) in the
Caucasian American male population is estimated to be
about 8%. We take a random sample of size 25 from this population.
We can model this situation with a B(n = 25, p = 0.08) distribution.
 What
is the probability that five individuals or fewer in the sample are color blind?
Use Excel’s “=BINOMDIST(number_s,trials,probability_s,cumulative)”
P(x ≤ 5) = BINOMDIST(5, 25, .08, 1) = 0.9877
 What
is the probability that more than five will be color blind?
P(x > 5) = 1  P(x ≤ 5) =1  0.9877 = 0.0123
 What
is the probability that exactly five will be color blind?
P(x = 5) = BINOMDIST(5, 25, .08, 0) = 0.0329
30%
25%
20%
B(n = 25, p = 0.08)
15%
10%
5%
24
22
20
18
16
14
12
10
8
6
4
2
0%
0
P(X = x) P(X <= x)
12.44%
12.44%
27.04%
39.47%
28.21%
67.68%
18.81%
86.49%
9.00%
95.49%
3.29%
98.77%
0.95%
99.72%
0.23%
99.95%
0.04%
99.99%
0.01% 100.00%
0.00% 100.00%
0.00% 100.00%
0.00% 100.00%
0.00% 100.00%
0.00% 100.00%
0.00% 100.00%
0.00% 100.00%
0.00% 100.00%
0.00% 100.00%
0.00% 100.00%
0.00% 100.00%
0.00% 100.00%
0.00% 100.00%
0.00% 100.00%
0.00% 100.00%
0.00% 100.00%
P(X = x)
x
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Number of color blind individuals (x)
Probability distribution and histogram for
the number of color blind individuals among
25 Caucasian males.
What are the mean and standard deviation of the count of
color blind individuals in the SRS of 25 Caucasian American
males?
µ = np = 25*0.08 = 2
σ = √np(1  p) = √(25*0.08*0.92) = 1.36
What if we take an SRS of size 10? Of size 75?
µ = 10*0.08 = 0.8
µ = 75*0.08 = 6
σ = √(75*0.08*0.92) = 2.35
0.5
0.2
0.4
0.15
0.3
p = .08
n = 10
0.2
0.1
P(X=x)
P(X=x)
σ = √(10*0.08*0.92) = 0.86
p = .08
n = 75
0.1
0.05
0
0
0
1
2
3
4
5
Number of successes
6
0 1 2 3 4 5 6 7 8 9 10 11 12 13
Number of successes
Recall Free-throw
question


Through 2/25/14 NC State’s
free-throw percentage was
65.1% (315th in Div. 1).
If in the 2/26/14 game with
UNC, NCSU shoots 11 freethrows, what is the probability
that:
1.
2.
3.
NCSU makes exactly 8
free-throws?
NCSU makes at most 8
free throws?
NCSU makes at least 8
free-throws?
1.
n=11; X=# of made
free-throws; p=.651
p(8)= 11C8 (.651)8(.349)3
=.226
2. P(x ≤ 8)=.798
3.
P(x ≥ 8)=1-P(x ≤7)
=1-.5717 = .4283
Poisson Probability Models
The Poisson experiment typically models
situations where rare events occur over a
fixed amount of time or within a specified
region
 Examples

◦ The number of cellphone calls per minute
arriving at a cellphone tower.
◦ The number of customers per hour using an
ATM
◦ The number of concussions per game
experienced by the participants.
22
Poisson Experiment
◦ Properties of the Poisson
experiment
1) The number of successes (events) that occur
in a certain time interval is independent of the
number of successes that occur in another
time interval.
2) The probability of a success in a certain time
interval is
 the same for all time intervals of the same size,
 proportional to the length of the interval.
3) The probability that two or more successes
will occur in an interval approaches zero as
the interval becomes smaller.
24

The Poisson Random Variable
◦ The Poisson random variable X is the
number of successes that occur during a
given time interval or in a specific region

Probability Distribution of the
Poisson Random Variable.
e   x
p ( x) 
x  0,1, 2...
x!
E( X )  V ( X )  
where   0is the average number of occurences
in an interval of time or in a specified region.
25
Poisson Prob Dist =1
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
e 110
P( X  0)  p (0) 
 .3679
0!
e111
P( X  1)  p (1) 
 .3679
1!
e112
P( X  2)  p (2) 
 .1839
2!
4
5
Poisson Prob Dist =5
0.2
0.18
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0
1
2
3
 
            
0067
 
       
4

5
6
7
8
9
10
 
         
  
 

0337
         
 

Example
◦ Cars arrive at a tollbooth at a rate of 360 cars
per hour.
◦ What is the probability that only two cars will
arrive during a specified one-minute period?
 The probability distribution of arriving cars for any oneminute period is Poisson with  = 360/60 = 6 cars per
minute. Let X denote the number of arrivals during a
one-minute period.
e 6 62
P( X  2) 
 .0446
2!
28
◦ Example (cont.)
◦ What is the probability that at least four cars
will arrive during a one-minute period?
◦ P(X>=4) = 1 - P(X<=3) = 1 - .151 = .849
29