Example A manufacturer has a 5% rate of defect when making
thermostats, which are produced in batches of 4. Let’s assume that
production involves independent events. That is, the failure of any
individual thermostat does affect the probability of failure for any
other thermostat.
Number of trials is fixed (4)
The trials are independent. (according to given assumption).
Each trial has two categories of outcome: the thermostat is
manufactured successfully or it is a failure.
The probabilities of failure (0.05) remain constant for different
thermostats.
S
D
D
D
S
S
D
D
D S
D S
0 1
1
D S
2 1
2
S
D
S
D
S
D S D S
D S D S
D
S
2
2
3
4
3 1 2
S
3 2
3
# Successes Arrangements
4
1
3
4
2
6
1
4
0
1
0
1
2
Probability
(0.95)4 = 0.8415
4(0.95)3(0.05)1 = 0.1715
6(0.95)2(0.05)2 = 0.0135
4(0.95)1(0.05)3 = 0.0005
(0.05)4 = 0.0000
1.0000
3
4
Example
Suppose in the previous example, batches of size 5 were
being examined, and the overall defect rate was 12%. Show
Defects
0
Prob of D’s
.12
Non Defects
5
Prob of ND’s
.88
Arrangement
1
1
0
.12
2
1
4
5
.88
5
.12
3
2
3
4
.88
10
.12
4
3
2
3
.88
10
.12
5
4
1
2
.88
5
.12
5
0
1
.88
0
1
P(0) = 0.527732, P(1) = 0.359817, P(2) = 0.098132,
P(3) = 0.013382, P(4) = 0.000912, P(5) = 0.000025
A binomial experiment meets all the following requirements
1. The experiment must have a fixed number of trials
2. The trials must be independent. (The outcome of any individual
trial doesn’t affect the probabilities in the other trials).
3. Each trial must have all outcomes classified into two categories
(even though the sample space may have more than two simple
events).
4. The probabilities must remain constant for each trial.
The Binomial Probability Formula
n
denotes the fixed number of trials
x
denotes a specific number of successes in n trials, so that x
can be any whole number between O and n
p
denotes the probability of success in one of the n trials
1-p
denotes the probability of failure in one of the n trials
P(x)
denotes the probability of getting exactly x successes
among the n trials.
n!
n x
x
P x  
p 1  p
 n  x! x !
Example
A manufacturer has a 15% rate of defect when making microchips,
which are produced in batches of 10. In a QC inspection plan batches
are considered acceptable if there are fewer than 2 defect per batch.
Let X denote the number of defect per batch of 10,
and p = 0.15 the defect rate.
10!
10  x
x
P x  
(0.15) 0.85
10  x! x!
P(0) = 0.197, and P(1) = 0.347,
so that the probability of accepting a batch equals
P(0) + P(1) = 0.544
The Binomial Probability Mass Function
Example: A cosmetic salesperson who calls potential customers to
sell her products has determined that 30% of her telephone calls
result in a sale. Determine the probability distribution for her next
three calls.
3!
3 0
P x  0 
(0.30) 0 0.70
3  0! 0!
 0.343
P x  1 
3!
31
(0.30)1 0.70
3  1!1!
 0.441
P x  2  
3!
3 2
(0.30) 2 0.70  0.189
 3  2! 2 !
3!
3 3
P x  3 
(0.30) 3 0.70  0.027
3  3! 3!
Determine the cumulative distribution for her next three calls.
The cumulative distribution function (CDF) for any discrete random
variable is defined as Prob(x < k), where k defines the sample space.
0.45
P x  0  P x  0
0.40
0.35
 0.343
0.30
PDF
0.25
P x  1  P x  0  P x  1
0.20
0.15
0.10
 0.343  0.441
0.05
0.00
0
1
2
3
X
P x  2  P x  1  P x  2
1.0
0.9
 0.784  0.189
0.8
CDF
 0.784
 0.973
0.7
0.6
P x  3  0.973  0.027
0.5
0.4
0.3
0
1
2
X
3
 1.0
Example: The records of a department store show that 20% of their
customers who make a purchase return the merchandise in order to
exchange it. What is the probability that in the next six purchases
(a) Exactly three customers return the merchandise?
6!
6 3
P x  3 
(0.20) 3 0.80
6  3! 3!
 0.08192
(b) At least three customers return the merchandise?
P x  3  1  P(0)  P(1)  P(2)
 1.262 .393.246
 1.901
Random Variables And Expectations
Take the discrete Random Variable X with probability mass
function P(xi)
Define the expectation of X as
E{X}   xi P xi 
i
Define the Variance of X as
Var{X}    xi  EX P xi 
2
i
A Bernoulli experiment, is an experiment in which the outcome can
be classified as a “success” (coded 1) or “failure” (coded 0) as
follows..
P{X = 0} = 1-p
P{X = 1} = p
where p, 0<p<1, is the probability that a trial is a “success”.
O
1
Expected Value and Variance of a Bernoulli RV
E{X}   xi P xi 
i
 1. P( X  1)  0. P( X  0)
 1. p  0.(1  p)
p
Var{X}    xi  EX P xi 
2
i
   xi  p2 P xi 
i
 (1  p) 2 p  (0  p) 2 (1  p)
 p(1  p)
Rules for Expectations of RV’s
If a and b are scalars,
E{aX + bY} = aE{X} + bE{Y}
(in general)
Var{aX} = a2Var{X}
(in general)
and
and
Var{X + Y} = Var{X} + Var{Y}
(for independent
variables X and Y)
Expected Value and Variance of a Binomial RV
First note that a Binomial RV can be written as the
sum of n independent Bernoulli RV’s:
X = X1 + X2 + ….+ Xn.
E{X}  E{X1  X2Xn}
 np
Var{X}  Var{X1  X2  Xn}
 Var{X1}  Var{X2} Var{Xn}
 np(1  p)
Counting alpha-particles (1910)
In this classic set of data Rutherford and Geiger counted the number of
scintillations in 72 second intervals caused by radioactive decay of a
quantity of the element polonium. Altogether there were 10097
scintillations during 2608 such intervals. Can we model these data??
Count
Frequency
0
57
1
203
2
383
3
525
4
532
5
408
6
573
Count
Frequency
7
139
8
45
9
27
10
10
11
4
12
1
13
0
5
10
Frequency
500
400
300
200
100
0
0
Number of Observed Alpha-Particles
15
14
1
Characteristics of a Poisson Random Variable
The experiment consists of counting the number of times a certain
event occurs during a given unit of time or in a given area or volume
(or weight, or distance, or any other unit of measurement).
The probability that an event occurs in a given unit of time, area,
volume, is the same for all the units.
The number of events that occur in one unit of time, area, or volume is
independent of the number that occur in other units.
The mean (or expected) number of events in each unit (of time, area
etc..) is denoted by the Greek letter lambda, l.
The probability mass function for a Poisson RV is given by
lx e  l
P x 
x!
 x  0,1,2....
Here x is the number of observed events (alpha particles) that occur
in a given time (or space) unit.
P(x) is the probability of observing x events in any given unit of time.
l remains unchanged regardless of which time unit is being
inspected.
l is often referred to as the rate at which the events occur as itself has
units:
# of particles per 72 second interval
Is is legitimate to describe the Rutherford-Geiger data by the Poisson
Probability Function? As scientists we should
(1) estimate l
(2) calculate P(x) using our estimate l
(3) visually compare the observed counts with expected counts from
the Poisson probability model!!
10097
l 
 3.87  particals / 72 sec. interval
2608
Evaluate
387
. x e  3.87
P x 
x!
 x  0,1,2....
387
. 0 e  3.87
P0 
 e  3.87  0.02085
0!
P1  387
. e  3.87  0.08072
387
. 2 e  3.87
P 2  
 01562
.
2!
387
. 3 e  3.87
P3 
 0.20149
3!
387
. 4 e  3.87
P 4  
 019495
.
4!
Xi
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Observed
57
203
383
525
532
408
273
139
45
27
10
4
0
1
1
P(x)
0.020858
0.080722
0.156197
0.201494
0.194945
0.150888
0.097323
0.053805
0.026028
0.011192
0.004331
0.001524
0.000491
0.000146
0.000040
Expected = P(x)*2608
54.399
210.523
407.361
525.496
508.418
393.515
253.817
140.325
67.882
29.189
11.296
3.974
1.282
0.382
0.105
Obs
500
Exp
Frequency
400
300
200
100
0
0
5
10
Xi
15
How Many Beds??
The mean number of patients admitted per day to the emergency room
of a small hospital is 2.5. If, on a given day, there are only four beds
available for new patients, what is the probability that the hospital will
not have enough beds to accommodate its newly admitted patients??
Let X be the number of patients admitted. We require
P(X > 5) = 1 - [P(X=0) + P(X=2) + P(X=3) + P(X=4)]
using
2.5x e  2.5
P x  
x!
P(0)  e  2.5
 0.0821
P(1)  2.5  P(0)
 0.2052
2.5  P(1)
P ( 2) 
 0.2565
2
2.5  P(2)
P(3) 
 0.2138
3
2.5  P(3)
P ( 4) 
 01336
.
4
P(X<4) = 0.0821+0.2052+0.2565+0.2138+0.1336 = 0.8912
P(X>5) = 1 - 0.8912 = 0.1088
For a Poisson RV we have that

E{X}   xP( x)  l
x0
and

Var{X}    x  l  P( x)  l
2
x0
The Poisson probability distribution also provides a good
approximation to a binomial probability distribution with mean
l = np
when n is large, p is small, and np <7.
The probability mass function for a Poisson RV is given by
lx e  l
P x 
x!
 x  0,1,2....
Now


x
l
P x   e  l 

x0
x 0 x!
 el el
1
so that the sum of the probabilities equal 1, as is the case for any
mass function.
lx e  l
E{X}   xP( x)   x
x!
x0
x0




lx
lx 
lx 1 
l 
l 
 e  x  e 0   x   e  l 

x
!
x
!
x

1
!




x0
x 1
 x 1

l

  lJ 
 e l    e  l le l  l
 J 0 J ! 
l
 
so that E{X} = l for a Poisson RV.