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Chapter 5 Binomial Distribution

5 BINOMIAL
DISTRIBUTION
Objectives
After studying this chapter you should

be able to recognise when to use the binomial distribution;

understand how to find the mean and variance of the
distribution;

be able to apply the binomial distribution to a variety of
problems.

Note: Statistical tables can be found in many books and are also available online.

5.0

Introduction

'Bi' at the beginning of a word generally denotes the fact that the meaning involves 'two' and binomial is no exception. A random variable follows a binomial distribution when each trial has exactly two possible outcomes. For example, when Sarah, a

practised archer, shoots an arrow at a target she either hits or misses each time. If X is 'the number of hits Sarah scores in ten shots', then the probabilities associated with 0, 1, 2, ..., 10 hits can be expected to follow a particular pattern, known as the binomial distribution.

5.1

Finding the distribution

You have already met this type of distribution in Chapter 4, as can be seen in the following example.

Example
Ashoke, Theo and Sadie will each visit the local leisure centre to swim on one evening next week but have made no arrangement
between themselves to meet or go on any particular day. The
random variable X is 'the number of the three who go to the
leisure centre on Wednesday'. Find the probability distribution for X.

99

Chapter 5 Binomial Distribution

Solution
The probabilities of 0, 1, 2 or 3 people going on Wednesday can be found by using the tree diagram method covered in Section 1.5. The following tree diagram shows probabilities for how many go on Wednesday.

Prob.
X

6
7

1
7

s

e
go

on

ed
W

do

es

6
7

tg

6
7

1
7

1
1
7
1
7

oo

Ashoke

2

nW
ed

2
6
7

616
××
777
661
××
777

0

666
××
777

6
7

You can see that

and

P( X = 2 ) =

66
×
77
11
×
77

1

1
7

Theo

3

1
7
6
7
1
7

1

6
7

P ( X = 3) =  1  ,
 7

1
×
7
1
×
7
6
×
7

1
×
7
6
×
7

2

1
7

6
7

no

1
×
7
1
×
7
1
×
7

3

1
7

6
P ( X = 0) =  
 7

3

116
161
611
××
+
××
+
××
777
777
777
2

 1  6
= 3×   
 7  7

P ( X = 1) =

166
616
661
××
+
××
+
××
777
777
777
2

 1  6
=3×   .
 7  7

This gives the table below:
x

0

P( X = x )

 6

 7

=

100

1
3

216
343

 1 6
3   
 7 7

=

108
343

2
2

3

2

 1  6
3   
 7  7

=

18
343

 1

 7

=

3

1
343

Chapter 5 Binomial Distribution

The method used in the example above can be extended to a
fourth person so that there will be sixteen branches to cover all the possibilities as shown in the diagram below, with X now
'the number of the four people who go on Wednesday'.
W
X

.

Prob.

4

()

3

()()

3

()()

2

()()

3

()()

2

()()

2

2

()()

2

1

( )( )

3

()()

2

()()

2

2

()()

2

1

( )( )

2

()()

1

( )( )

3

1

( )( )

3

0

()

1
7

4

3

1
7

3

1
7

6
7

2

1
7

6
7

3

1
7

6
7

2

1
7

6
7

6
7

1
7

3

1
7

6
7

2

1
7

6
7

6
7

1
7

2

1
7

6
7

1
7

3

6
7

6
7

1
7

3

6
7

2

1
7

2

6
7

2

1
7

6
7

6
7

2

4

The resulting probability distribution is
x

0

P( X = x )

 6
 7

1
4

16
4   
 7  7

2
3

2

6
1
6   
 7  7

3
2

4

3

 1
 7

6
1
4   
 7  7

4

The fractions are as you might expect. For instance, looking at 12 62
P ( X = 2) = 6     ,
 7  7
since you are interested in having two present and the

( 1 )2 is explained, and as two are
7
2
not to attend this produces the ( 6 ) .
7
probability for each is 1 , the
7

101

Chapter 5 Binomial Distribution

Explain the reason for the coefficient 6.
This has come from the second of the tree diagrams: there are six branches corresponding to two being present. This is

(1)s
7

because there are six ways of writing down two

(6)s
7

and two

in a row and each produces a branch.

The six ways are shown below:

1

×

7
6
7

1

×

7
×

1
7

×

6

6
×,
77

1

1

6

×

7

6
×,
77

7

6

×

7
×

1
7

×

1

6
×,
77

1

6

6

×

7

1
×,
77

6

×

6

×

1

7
×

7

6
7

1
×,
77

1
×.
77

Pascal's Triangle
1

There are four fractions to write down, two of each type, and the number of different ways of combining these is six. You don't want to draw a tree diagram every time so another method can 1
be d...