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7.1. The Binomial Probability Distribution
Example:
X2 :
representing the number of heads as flipping a fair coin twice.
H : head
X2  0
X2 1
X2  2
□
□
T
T
□
□
H
T
T
H
□
□
H
H
T : tail .
 P( X 2  0) 
1 1  2 1 1
   
2 2  0  2 2
(1 combination)
1 1  2 1 1
(2 combinations)
 P( X 2  1)  2      
2 2  1 2 2
 P( X 2  2) 
1 1  2 1 1
   
2 2  2  2 2
(1 combination)
 2  1 
 P( X 2  i)  f 2 (i)    
 i  2 
 (number of combinatio ns)  ( the probabilit y of every combinatio n)
2
,
i  0, 1, 2.
X3 :
representing the number of heads as flipping a fair coin 3 times.
X3  0
□
□
□
T
T
T  P( X 3  0)        (1 combination)
2 2 2  0  2 
1 1 1
1
 3 1
3
X3 1
X3  2
X3  3
□
□
□
H
T
T
T
H
1 1 1  3 1
T  P( X 3  1)  3       
2 2 2  1  2 
T
T
H
□
□
□
H
H
T
H
T
1 1 1  3 1
H  P( X 3  2)  3       
2 2 2  2  2 
T
H
H
□
□
□
H
H
H  P( X 3  3)        (1 combination)
2 2 2  3 2 
1 1 1
 3 1
3
(3 combinations)
3
(3 combinations)
3
 3 1 
 P( X 3  i)  f 3 (i)    
 i  2 
 (number of combinatio ns)  ( the probabilit y of every combinatio n)
3
,
i  0, 1, 2, 3.
Xn :
representing the number of heads as flipping a fair coin n times.
Then,
n
□ □ …………… □
 n 1
1
T…………….T  P( X n  0)       
 2
 0  2 
n
Xn  0
T
2
n
(1 combination)
n
□ □ …………… □
H T……..……..T
 n 1
1
T  P( X n  1)  n       
 2
 1  2 
n
n
Xn 1
T H………




n
(n combinations)
T T …………..H


 n  1 
P( X n  i)  f n (i)    
 i  2 
 (number of combinatio ns)  (the probabilit y of every combinatio n)
n
Note: the number of combinations is equivalent to the number of ways as
drawing i balls (heads) from n balls (n flips).
Example:
Z3 :
representing the number of successes over 3 trials.
S : Success
F : Failure
3
Suppose the probability of the success is
1
2
while the probability of failure is
.
3
3
Then,
□
□
□
F
 3 1
1
2
2
F  P(Z 3  0)           
 3   3   0  3   3 
0
Z3  0
F
3
0
3
(1 combination)
□
□
□
S
F
F
F
S
 3 1 2
1 2
F  P(Z 3  1)  3         
3  3
 1  3  3 
2
Z3  1
2
(3 combinations)
F
F
S
□
□
□
S
S
F
F
1
2  3 1
2
S  P(Z 3  2)  3          
 3  3  2  3   3 
2
Z3  2
S
2
(3 combinations)
F
S
S
□
□
□
S
 3 1
1
2
2
S  P(Z 3  3)           
 3  3 
 3 3   3 
3
Z3  3
S
0
3
0
(1 combination)
3i
 3 1   2 
 P( Z 3  i)  f 3 (i)      
 i  3   3 
 (number of combinatio ns)  ( the probabilit y of every combinatio n)
i
4
,
i  0, 1, 2, 3.
Zn :
representing the number of successes over n trials.
Then,
n
□ □ …………… □
 n 1
2
F F…………….F  P(Z n  0)       
 3
 0  3 
n
Zn  0
0
 2
 
 3
n
(1 combination)
n
□ □ ……… □
S F…….. .F
Zn  1
n
F S……… F




 1  2 
 P( Z n  1)  n    
 3  3 
n 1
 n  1  2 
    
 1  3  3 
n 1
(n combinations)
F F … .S


n i
 n  1   2 
P( Z n  i)  f n (i)      
 i  3   3 
 (number of combinatio ns)  (the probabilit y of every combinatio n)
i
5
From the above example, we readily describe the binomial experiment.
Properties of Binomial Experiment
 X: representing the number of successes over n independent
identical trials.
 The probability of a success in a trial is p while the probability of
a failure is (1-p).
Binomail Probability Distribution:
Let X be the random variable representing the number of successes
of a Binomial experiment. Then, the probability distribution function
for X is
 n
n! i
n i
n i
P( X  i)  f x (i)    p i 1  p  
p 1  p  , i  0,1, 2,, n .
i!n  i !
i
Properties of Binomial Probability Distribution:
A random variable X has the binomial probability distribution f (x)
with parameter p , then
E ( X )  np
and
Var ( X )  np(1  p) .
[Derivation:]
6
n
 n i
n!
n i
n i
E ( X )   i  f (i )   i    p 1  p    i 
p i 1  p 
i!n  i !
i 0
i 0
i 0
i
n
n
n!
n!
n i
n i
i
 i 
p 1  p   
p i 1  p 
i!n  i !
i 1
i 1 i  1! n  i !
n
n
n  1!
( n 1) ( i 1)
 p i 1 1  p 
i  1!n  1  i  1!
i 1
n 1
n  1! p j 1  p n1 j ( j  i  1)
 np 
j 0 j!n  1  j !
n  1! p j 1  p n1 j is the probabilit y
 np (since
j!n  1  j !
n
  np  
distributi on of a binomial random variable over n  1
trials)
The derivation of
Var ( X )  np(1  p)
is left as exercise.
How to obtain the binomail probability distribution:
(a) Using table of Binomail distribution.
(b) Using computer
 by some software, for example, Excel or Minitab.
Online Exercise:
Exercise 7.1.1
Exercise 7.1.2
7
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