3.2 BINOMIAL DISTRIBUTION
CONTENTS
3.2 Binomial Distribution ................................................................................................... 2
3.2.1 Characteristics of a Binomial Experiment ...........................................................2
3.2.2 Binomial Probability Distribution Function ........................................................2
3.2.3 Mean and Variance of Binomial Distribution ......................................................5
3.2.4 Application of Binomial Distribution in Computer Science................................6
3.2 EXERCISES ................................................................................................................. 6
EXAMPLES
Example 1: Binomial Distribution: Tossing a coin 6 times problem ................................................... 2
Example 2: Binomial Distribution: Telemarketer’s problem .............................................................. 3
Example 3: Binomial Distribution: Mean and standard deviation of die problem .............................. 5
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3.1 Binomial Distribution
3.2.1 Characteristics of a Binomial Experiment
1.
2.
3.
4.
The experiment consists of n identical and independent trials.
There are only two possible outcomes on each trial. We will denote one outcome by S (for
success) and the other by F (for failure).
The probability of S or F remains the same from trial to trial. The probability of S is denoted
by p, and the probability of F is denoted by q. Note that q = 1 – p.
Random Variable X, is the number of successes in the n trials is said to follow Binomial
Distribution with parameters n and p (x=0,1,2,3,…,n)
Remark: Notation: X~b(x; n, p)
3.2.2 Binomial Probability Distribution Function
Definition: The probability of exactly x successes in n repeated trials is given by
n!
b(x; n, p) = p( x )  nCx p x qn  x 
p x (1  p)n  x
x ! (n  x )!
Where:
n
Cx
=
the binomial coefficient
p
=
Probability of a ‘Success’ on a single trial
q
=
(1 – p): Probability of a ‘Failure’ on a single trial
n
=
Number of identical and independent trials
x
=
Number of ‘Successes’ in n trials (x = 0, 1, 2, ..., n)
n–x =
Number of failures in n trials
Example 1: Binomial Distribution: Tossing a coin 6 times problem
A coin is tossed 6 times in a row. Note the number of heads as success. Find the probability of
i.
ii.
iii.
iv.
Getting exactly 2 heads
Getting at least 4 heads
No heads
At least one head
i. The probability of getting exactly 2 heads
The formula for probability of exactly x successes in n repeated trials is given by
n!
p ( x )  nC x p x q n  x 
p x (1  p)n  x
x ! (n  x )!
Given, n=6; p=1/2 and x=2 i.e find p(2).
2
6 2
2
4
6!
6!
1 1
 1   1  15
Thus, p(2) 

   
    
2!(6  2)!  2   2 
2!(6  2)!  2   2  64
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ii. The probability of getting at least 4 heads
The formula for probability of exactly x successes in n repeated trials is given by
n!
p x (1  p)n  x
x ! (n  x )!
Given, n=6; p=1/2 ; x=4,5,6
p ( x )  nC x p x q n  x 
To find, p(at least 4 heads) =p(x≥4) = p(4)+p(5)+p(6)
4
64
4
2
6!
1 1

   
4!(6  4)!  2   2 
5
6!
1 1

   
5!(6  5)!  2   2 
5
6 5
6
6!
1 1

   
6!(6  6)!  2   2 
1
6
6!
6!
6!
1 1
1 1
1 1

    
    
   
4!(6  4)!  2   2  5!(6  5)!  2   2  6!(6  6)!  2   2 
=15/64 + 6/64 +1/64 = 22/64 =11/32
6 0
0
iii. The probability of no heads
The formula for probability of exactly x successes in n repeated trials is given by
n!
p ( x )  nC x p x q n  x 
p x (1  p)n  x
x ! (n  x )!
Given, n=6; p=1/2 ; x=0
0
6
6
6!
1
1 1 1
p(0) 
      
0!(6  0)!  2   2   2  64
iv. The probability of at least one head
The formula for probability of exactly x successes in n repeated trials is given by
n!
p ( x )  nC x p x q n  x 
p x (1  p)n  x
x ! (n  x )!
Given, n=6; p=1/2 ; x=1,2,3,4,5,6
To find, p(at least 1 heads) = p(x≥1) =p(1)+p(2)+p(3)+p(4)+p(5)+p(6) =1-p(0)
0
6
6

6!
1 63
1 1 1 
 1  p(0)  1  

        1
64 64
 0!(6  0)!  2   2   2  
Example 2: Binomial Distribution: Telemarketer’s problem
You are a telemarketer selling service contracts for Macy’s. You have sold 20 in your last 100 calls.
If you call 12 people tonight, what’s the probability of
A.
B.
C.
D.
No sales?
Exactly 2 sales?
At most 2 sales?
At least 2 sales?
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Solution
Given: n = 12, p =20/100=1/5
n!
p ( x )  nC x p x q n  x 
p x (1  p)n  x
x ! (n  x )!
A. Probability of no sale i.e. x=0
0
12
12!
1  4
p(0) 
   
0!(12  0)!  5   5 
p(0) = .0687
B. Probability of exactly two sales
2
12 2
12!
1  4
p(2) 
   
2!(12  2)!  5   5 
p(2) = 0.2835
C. Probability of at most two sales: p(at most 2) = p(0) + p(1) + p(2)
0
120
12!
1  4
p(0) 
     0.0687
0!(12  0)!  5   5 
121
1
p(1) 
12!  1   4 
   
1!(12  1)!  5   5 
2
 0.2062
12 2
12!
1  4
p(2) 
     0.2835
2!(12  2)!  5   5 
Thus, p(at most 2)= p(X≤2) = p(0) + p(1) + p(2) = .0687 + .2062 + .2835 = 0.5584
D. Probability of at least two sales
p(at least 2) = p(x≥2) = p(2) + p(3)...+ p(12)
= 1 – [p(0) + p(1)]
Where,
0
12
12!
1  4
p(0) 
     0.0687
0!(12  0)!  5   5 
1
121
12!  1   4 
p(1) 
     0.2062
1!(12  1)!  5   5 
= 1 – .0687 – .2062= 0.7251
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Theorem (ASIDE): Show that p(x) is the probability distribution i.e. Σ p(x) = 1
Solution:
n
n
Binomial Expansion Formula: ( a  b) n     a i bn i
i 0  i 
n
n
n
  p( x )     p x (1  p ) n  x
x 0
x 0  x 
Here a  p and b  (1  p ),
  p  (1  p )   1
n
 p(x) is a "Legitimate" Probability Distribution
QED
3.2.3 Mean and Variance of Binomial Distribution
Mean
µ= E(x) =np
Standard Deviation
  npq
Example 3: Binomial Distribution: Mean and standard deviation of die problem
(i)
A fair die is tossed 180 times. Find the expected number of times the face 6 occurs and the standard
deviation.
Solution:
A fair die is tossed 180 times, p=1/6; q = (1-p) = 5/6.
The expected number of sixes is µ= np = 180 x 1/6 =30.
The standard deviation is   npq =√180 x (1/6) x(5/6) = 5
(ii)
A recent survey found that 61% of all adults aged 50 years and over wear glasses for driving. In a
random sample of 70 adults over 50 what is the mean and standard deviation of the number who
wear glasses?
Solution:
Probability of wearing glasses for driving of adults over 50 is p=0.61 and q=(1-p) = (1-0.61) =0.39
The number of adults sampled (=n) = 70)
As this follows the Binomial distribution, the
Mean = np = 70 x 0.61 =42.7 Number of adults wearing glasses.
Variance = npq =70 x 0.61 x 0.39 =16.653
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3.2.4 Application of Binomial Distribution in Computer Science
To model the number of successes
1. The number of processors that are up in a multiprocessor system.
2. The number of packets that reach the destination without loss.
3. The number of bits in a packet that are not affected by noise.
4. The number of items in a batch that have certain characteristics.
3.2 EXERCISES
Q1. What are the characteristics of a binomial experiment?
Q 2. Define binomial probability distribution function.
Q 3. A coin is tossed 5 times in a row. Note the number of heads as success. Find the probability of
getting:
1. exactly 3 heads
2. at least 4 heads
3. at most 2 heads
4. No heads
5. At least one head
Q4. A recent survey found that 61% of all adults over 50 wear glasses for driving. In random
sample of 70 adults over 50 what is the mean and standard deviation of the number who wear
glasses? (Answer: Mean 42.7 adults over 50 wear glasses and standard deviation: 4.08)
Q5. A fair die is tossed 7 times, call a toss a success (p) if a 5 or a 6 appears. Find the probability
that a 5 or a 6 occurs exactly 3 times (i.e. X=3).
Ans.: p(X=3) = 560/2187
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