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Matakuliah
Tahun
Versi
: I0134 – Metoda Statistika
: 2005
: Revisi
Pertemuan 10
Sebaran Binomial dan Poisson
1
Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• Mahasiswa dapat menghitungpeluang,
rataan dan varians peubah acak Binomial
dan Poisson.
2
Outline Materi
• Sebaran Peluang Binomial
• Nilai harapan dan varians sebaran
Binomial
• Sebaran peluang Poisson
• Nilai harapan dan varians sebaran
Poisson
3
Binomial and Poisson Probability Distributions
Binomial Probability Distribution
l
Consider a situation where there are only two possible outcomes (a “Bernoulli trial”)
Example:
u flipping a coin
head or tail
u rolling a dice
6 or not 6 (i.e. 1, 2, 3, 4, 5)
Label the possible outcomes by the variable k
We want to find the probability P(k) for event k to occur
Since k can take on only 2 values we define those values as:
k = 0 or k = 1
u let P(k = 0) = q (remember 0 ≤ q ≤ 1)
u something must happen so
P(k = 0) + P(k = 1) = 1
(mutually exclusive events)
P(k = 1) = p = 1 - q
u We can write the probability distribution P(k) as:
P(k) = pkq1-k (Bernoulli distribution)
u coin toss: define probability for a head as P(1)
P(1) = 0.5 and P(0=tail) = 0.5 too!
u dice rolling: define probability for a six to be rolled from a six sided dice as P(1)
P(1) = 1/6 and P(0=not a six) = 5/6.
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l
What is the mean () of P(k)?
1
 kP(k)
  k0
1
 P(k)

0  q 1 p
p
q p
discrete distribution
k0
l
What is the Variance (2) of P(k)?
1
2
 k P(k )

 2  k  01
  2  02 P(0)  12 P(1)   2  p  p 2  p(1  p)  pq
 P(k )
k 0
l
l
l
Suppose we have N trials (e.g. we flip a coin N times)
what is the probability to get m successes (= heads)?
Consider tossing a coin twice. The possible outcomes are:
no heads: P(m = 0) = q2
one head: P(m = 1) = qp + pq (toss 1 is a tail, toss 2 is a head or toss 1 is head, toss 2 is a tail)
= 2pq
we don't care which of the tosses is a head so
2
two heads: P(m = 2) = p
there are two outcomes that give one head
Note: P(m=0)+P(m=1)+P(m=2)=q2+ qp + pq +p2= (p+q)2 = 1 (as it should!)
We want the probability distribution function P(m, N, p) where:
m = number of success (e.g. number of heads in a coin toss)
N = number of trials (e.g. number of coin tosses)
p = probability for a success (e.g. 0.5 for a head)
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l
l
If we look at the three choices for the coin flip example, each term is of the form:
CmpmqN-m m = 0, 1, 2, N = 2 for our example, q = 1 - p always!
coefficient Cm takes into account the number of ways an outcome can occur without regard to order.
for m = 0 or 2 there is only one way for the outcome (both tosses give heads or tails): C0 = C2 = 1
for m = 1 (one head, two tosses) there are two ways that this can occur: C1 = 2.
Binomial coefficients: number of ways of taking N things m at time
CN , m 

   m!( NN! m)!
N
m
0! = 1! = 1, 2! = 1·2 = 2, 3! = 1·2·3 = 6, m! = 1·2·3···m
Order of occurrence is not important
u e.g. 2 tosses, one head case (m = 1)
n we don't care if toss 1 produced the head or if toss 2 produced the head
Unordered groups such as our example are called combinations
Ordered arrangements are called permutations
For N distinguishable objects, if we want to group them m at a time, the number of permutations:
N!
PN,m 
(N  m)!
u example: If we tossed a coin twice (N = 2), there are two ways for getting one head (m = 1)
u example: Suppose we have 3 balls, one white, one red, and one blue.
n Number of possible pairs we could have, keeping track of order is 6 (rw, wr, rb, br, wb, bw):
3!
P3,2 
6
(3  2)!
n If order is not important (rw = wr), then the binomial formula gives
3!
C3,2 
 3 number of “two color” combinations
2!(3  2)!

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Binomial distribution: the probability of m success out of N trials:
l
P(m, N , p)  CN , m p m q N  m 
 p
N
m
m N m
q

N!
p mq N  m
m!( N  m)!
p is probability of a success and q = 1 - p is probability of a failure
0.14
0.40
0.10
P (k , 50, 1/3)
P (k , 7, 1/3)
0.12
Expectation Value
= np = 7 * 1/3 = 2.333...
0.30
Expectation Value
= np = 50 * 1/3 = 16.667...
0.20
0.08
0.06
0.04
0.10
0.02
0.00
0.00
0
l
2
4
k
6
8
10
0
5
10
15
k
20
25
30
To show that the binomial distribution is properly normalized, use Binomial Theorem:
k
( a  b) k  
 a
k
l
l 0
N
N
 P(m, N , p)  
m0
k l l
 p
m0
N
m
b
m N m
q
 ( p  q) N  1
binomial distribution is properly normalized
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Mean of binomial distribution:
N

 mP (m, N , p)
m0
N
 P(m, N , p)
N
N
m0
m0
  mP (m, N , p )   m
 p
N
m
m N m
q
m0
A cute way of evaluating the above sum is to take the derivative:
 
  N N m N m 
0
 m p q


p  m  0

N
m
m0
p
1
N
m
m0
 p
N
m
m N m
q
 p
N
m
m 1 N  m
q
 N (1  p )
N
 
 p
m0
N
1

N
m
 p
m0
N
m
m
m
( N  m)(1  p ) N  m 1  0
(1  p )
N m
 (1  p )
1
N
m
m0
 p
N
m
m
(1  p ) N  m
p 1  N (1  p ) 1  1  (1  p ) 1 
  Np
Variance of binomial distribution (obtained using similar trick):
N
 (m   )2 P(m, N, p)
 2  m0
N
 Npq
 P(m, N, p)
m0
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Example: Suppose you observed m special events (success) in a sample of N events
u The measured probability (“efficiency”) for a special event to occur is:
m

N
What is the error on the probability ("error on the efficiency"):
Npq

N (1  )
 (1  )
we will derive this later in the course
  m 


N
N
N
The sample sizeN (N) should
be as large as possible to reduce certainty in the probability measurement
Let’s relate the above result to Lab 2 where we throw darts to measure the value of p.
If we inscribe a circle inside a square with side=s then the ratio of the area of the circle

to the rectangle is:
area of circle pr 2 p ( s / 2) 2 p
 2 

2
area of square s
4
s
So, if we throw darts at random at our rectangle then the probability () of a dart landing inside the
circle is just the ratio of the two areas, p/4. The we can determine p using:
The error in p is related to the error in  by:
p4.
p  4
 (1   )
N
We can estimate how well we can measure p by this method by assuming that p/4= (3.14159…)/4:
p  4
 (1   )
N

1.6
using   p / 4
N
This formula “says” that to improve our estimate of p by a factor of 10 we have to throw 100 (N) times as
many darts! Clearly, this is an inefficient way to determine p.
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Example: Suppose a baseball player's batting average is 0.300 (3 for 10 on average).
u
Consider the case where the player either gets a hit or makes an out (forget about walks here!).
prob. for a hit:
p = 0.30
prob. for "no hit”: q = 1 - p = 0.7
u
On average how many hits does the player get in 100 at bats?
= Np = 100·0.30 = 30 hits
u
What's the standard deviation for the number of hits in 100 at bats?
 = (Npq)1/2 = (100·0.30·0.7)1/2 ≈ 4.6 hits
we expect ≈ 30 ± 5 hits per 100 at bats
u
Consider a game where the player bats 4 times:
probability of 0/4 = (0.7)4 = 24%
probability of 1/4 = [4!/(3!1!)](0.3)1(0.7)3 = 41%
Pete Rose’s lifetime
batting average: 0.303
probability of 2/4 = [4!/(2!2!)](0.3)2(0.7)2 = 26%
probability of 3/4 = [4!/(1!3!)](0.3)3(0.7)1 = 8%
probability of 4/4 = [4!/(0!4!)](0.3)4(0.7)0 = 1%
probability of getting at least one hit = 1 - P(0) = 1-0.24=76%
10
Poisson Probability Distribution
l
l
l
l
The Poisson distribution is a widely used discrete probability distribution.
Consider the following conditions:
p is very small and approaches 0
u example: a 100 sided dice instead of a 6 sided dice, p = 1/100 instead of 1/6
u example: a 1000 sided dice, p = 1/1000
u radioactive decay
N is very large and approaches ∞
u number of Prussian soldiers kicked
u example: throwing 100 or 1000 dice instead of 2 dice
to death by horses per year per army corps!
The product Np is finite
u quality control, failure rate predictions
Example: radioactive decay
Suppose we have 25 mg of an element
very large number of atoms: N ≈ 1020
Suppose the lifetime of this element t = 1012 years ≈ 5x1019 seconds
probability of a given nucleus to decay in one second is very small: p = 1/t = 2x10-20/sec
Np = 2/sec finite!
The number of decays in a time interval is a Poisson process.
Poisson distribution can be derived by taking the appropriate limits of the binomial distribution
N!
P(m, N, p) 
p m q Nm
m!(N  m)!
N!
N(N 1)  (N  m 1)(N  m)!

 Nm
(N  m)!
(N  m)!
q
Nm
 (1 p)
Nm
p2 (N  m)(N  m 1)
( pN)2
 1 p(N  m) 
   1 pN 
   e pN
2!
2!
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N m m  pN
P(m, N, p) 
p e
m!
Let   Np
e   m
P(m,  ) 
m!
m
m  e   m
m0
m0
note :  P(m,  )  
m!
e

m  m

m0
m!
 e  e   1
m is always an integer ≥ 0
The mean and variance of
u  does not have to be an integer
a Poisson distribution are the
It is easy to show that:
 = Np = mean of a Poisson distribution
same number!
2
 = Np =  = variance of a Poisson distribution
l Radioactivity example with an average of 2 decays/sec:
i) What’s the probability of zero decays in one second?
e2 20 e2 1 2
p(0,2) 

 e  0.135 13.5%
0!
1
ii) What’s the probability of more than one decay in one second?
e2 20 e2 21
p( 1,2) 1 p(0,2)  p(1,2) 1

1 e2  2e2  0.594  59.4%
0!
1!

iii) Estimate the most probable number of decays/sec?

P(m, )  0
m
m*

u To solve this problem its convenient to maximize lnP(m, ) instead of P(m, ).
e  m 
ln P(m,  )  ln
   m ln   ln m!

m!


u
12
u
In order to handle the factorial when take the derivative we use Stirling's Approximation:
ln m! m ln m  m


ln P(m,  ) 
(  m ln   ln m!)
m
m


(  m ln   m ln m  m)
m
 ln   ln m  m
ln10!=15.10
ln50!=148.48
10ln10-10=13.03 14%
50ln50-50=145.601.9%
1
1
m
0
m*  
The most probable value for m is just the average of the distribution
u This is only approximate since Stirlings Approximation is only valid for large m.
u Strictly speaking m can only take on integer values while  is not restricted to be an integer.

If you observed m events in a “counting” experiment, the error on m is
  m

13
Comparison of Binomial and Poisson distributions with mean = 1
0.5
0.4
0.35
poisson 
binomial N=3, p=1 /3
0.3
Probability
Probability
0.4
0.3
0.2
binomial N=10,p =0.1
po iss on 
0.25
0.2
Not much
difference
between them!
0.15
0.1
0.1
0.05
0
0
0
1
2 m
N
3
4
5
0.0
1.0
2.0
3.0
m
N
4.0
5.0
6.0
7.0
For N large and  fixed: Binomial Poisson
14
Uniform distribution and Random Numbers
What is a uniform probability distribution: p(x)?
p(x)=constant (c) for a x b
p(x)=zero everywhere else
Therefore p(x1)dx1= p(x2)dx2 if dx1=dx2  equal intervals give equal probabilities
For a uniform distribution with a=0, b=1 we have p(x)=1
1
1
1
0
0
0
 p( x)dx  1   cdx  c  dx  c  1
What is a random number generator ?
A number picked at random from a uniform distribution with limits [0,1]
All major computer languages (FORTRAN, C) come with a random number generator.
FORTRAN: RAN(iseed)
The following FORTRAN program generates 5 random numbers:
iseed=12345
do I=1,5
y=ran(iseed)
type *, y
enddo
end
0.1985246
0.8978736
0.2382888
0.3679854
0.3817045
If we generate “a lot” of random numbers
all equal intervals should contain the same
amount of numbers. For example:
generate: 106 random numbers
expect: 105 numbers [0.0, 0.1]
105 numbers [0.45, 0.55]
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