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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

Sadie

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...

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Keywords

/10 0 0.0032 0.0064 0.033 0.0512 0.0574 0.064 0.1 0.128 0.1445 0.15 0.2 0.2048 0.211 0.3 0.3277 0.3487 0.3633 0.4 0.4096 0.5 0.6 0.6080 0.655 0.7361 0.8 0.822 0.8555 0.9298 0.9872 0.99 0.9984 0.9999 1 1.0000 1.024 1.5 10 100 101 102 1024 103 104 105 106 107 108 109 10th 11 110 111 1110 112 113 114 116 12 120 126 12th 13 135 1461 15 1512 16 161 166 18 2 2.1 2.347 20 20th 21 210 216 22 25 252 253 256 26 28 3 3.2 30 3032 3367 34 343 35 36 37 4 4.096 4.2 40 4096 42 45 5 5.0 5.1 5.2 5.3 56 5a 5b 6 6.554 611 616 62 66 661 666 7 70 729 75 77 777 8 8.192 84 9 93 99 abl accept accord achiev activ adopt aeb ag alreadi also altern alway analysi anoth answer anyon appendix appli applic appropri archer argu arrang arriv arrow ashok ask associ assum attempt attend avail averag away b babi back bag bar batch battl bay begin benefit bi bias binomi black bolt book born bottom bowl bowler boy branch button c cage calcul cambford card case centr certain cezann chanc chapter check children choic choos close cn co co-oper coeffici coffe coin colour combin come compar complet complex comput consid consider consign consist constant contain correct correspond count cours cover crossword cumul d d.p day decid decim decor defect defin denot describ design determin develop deviat diagram diamet die differ digit direct discard discuss distribut divid dock done doubl downward draw drawn drunk drunkard e e.s.p easier easili eater ed eight eighteen either els end entri equal equatori es evalu even everi evid exact examin exampl except exercis exercise1b expect experi explain extend f face facil fact factori fail fall fals famili famous faulti feel fellow figur find finish first five flash follow food formal formula fought found four fourth fraction frequenc function game general get give given go goe graph guess h head heet henc henri hit identifi ii inaccur inclus ind independ inspect instanc instead instruct interest introduct investig involv jame k key know known label ladi larg last least lectur leisur less let lift light like list local look machin made major make man mani mankoya marbl mark match math may mean meet mercuri met method might mile miscellan miss monday monkey morn move much mug multipl multipli must n nd need next nn notat note number nw object observ obtain occas occur often old one onlin oo oper oppon opposit option otherwis outcom over p packag paint pair paper particip particular pascal pass past pattern pen peopl per perhap period person pick pin place plan play point politician portrait possibl potato potteri pq practis predict present press prob probabilitri probabl problem produc professor proof proport psalm publish pure put puzzl q quantiti question quiz r random read realli reason recognis record red regular reject repeat replac repli requir rest result retail return reward rewritten right roll round row sadi sampl sarah save say score seat second section see seen select set seven sever shatter sheet shock shoot shot show shown signific simpli sinc singl sit six sixteen solut sowton spread standard start state statement statist stay step student studi succeed success sunday surviv swim tabl tail take tallest target tell ten tenni term test testament tetrahedron tg thatcher theo theorem thermal thirti thoma three throw thrown time toss toward tree tri trial triangl true tube turn twelv twenti twice two type unbias uncondit understand univers upward use usual v vacuum valu variabl varianc varieti visit w walk want way wednesday week whether whilst white whole wholesal wicket win without word work would write written x y yvett z zambia