THE BINOMIAL THEOREM Robert Yen Please take a handout This PowerPoint can be downloaded from HSC Online from HSC Online.

THE BINOMIAL THEOREMTHE BINOMIAL THEOREM

Robert YenRobert Yen

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INTRODUCTIONINTRODUCTION

Binomial theorem = expanding (a + x)n

Difficult topic at end of Ext 1 course: high-level algebra

Targeted at better Extension 1 students aiming for Band E4

Master this topic to get ahead in the exam

No shortcuts for this topic: must learn the theory

BINOMIAL EXPANSIONS AND PASCAL’S TRIANGLEBINOMIAL EXPANSIONS AND PASCAL’S TRIANGLE

Coefficients

(a + x)1 = a + x 1 1

(a + x)2 = a2 + 2ax + x2 1 2 1

(a + x)3 = a3 + 3a2x + 3ax2 + x3 1 3 3 1

(a + x)4 = a4 + 4a3x + 6a2x2 + 4ax3 + x4 1 4 6 4 1

(a + x)5 = a5 + 5a4x + 10a3x2 + 10a2x3 + 5ax4 + x5 1 5 10 10 5 1

For (a + x)n, the powers of a are ↓ and the powers of x are ↑ The sum of the powers in each term is always n

nnCCkk, A FORMULA FOR PASCAL’S TRIANGLE, A FORMULA FOR PASCAL’S TRIANGLE

1 0C0

1 1 1C0 1C1

1 2 1 2C0 2C1

2C2

1 3 3 1 3C0 3C1

3C2 3C3

1 4 6 4 1 4C0 4C1

4C2 4C3

4C4

1 5 10 10 5 1 5C0 5C1

5C2 5C3

5C4 5C5

1 6 15 20 15 6 1 6C0 6C1

6C2 6C3

6C4 6C5

6C6

1 7 21 35 35 21 7 1 7C0 7C1

7C2 7C3

7C4 7C5

7C6

7C7

1 8 28 56 70 56 28 8 1 8C0 8C1

8C2 8C3

8C4 8C5

8C6 8C7

8C8

nCk gives the value of row n, term k,

if we start numbering the rows and terms from 0

nnCCkk, A FORMULA FOR PASCAL’S TRIANGLE, A FORMULA FOR PASCAL’S TRIANGLE

n

kCk

n

C stands for coefficient as well as combination

CALCULATINGCALCULATING

MentallyMentally

102

20

123

3453

5

C

35C

1026

120

!2!3

!53

5

C


MentallyMentally

FormulaFormula

35C

102

20

123

3453

5

C

!!

!

knk

nCk

n

1026

120

!2!3

!53

5

C


MentallyMentally

FormulaFormula

because ...because ...

102

20

123

3453

5

C

!2!3

!5

12

12

123

345

123

3453

5

C

!!

!

knk

nCk

n

35C

THE BINOMIAL THEOREMTHE BINOMIAL THEOREM

(a + x)n = nC0 an + nC1 an-1 x + nC2 an-2 x2 + nC3 an-3 x3

+ nC4 an-4 x4 + ... + nCn xn

=

kknn

kk

n xaC

0

Don’t worry too much about writing in notation: just have a good idea of the general term

The sum of terms from k = 0 to n

PROPERTIES OF nCk

1. nC0 = nCn = 1 1st and last

2. nC1 = nCn-1 = n 2nd and 2nd-last

1 0C0

1 1 1C0 1C1

1 2 1 2C0 2C1

2C2

1 3 3 1 3C0 3C1

3C2 3C3

1 4 6 4 1 4C0 4C1

4C2 4C3

4C4

1 5 10 10 5 1 5C0 5C1

5C2 5C3

5C4 5C5

1 6 15 20 15 6 1 6C0 6C1

6C2 6C3

6C4 6C5

6C6

1 7 21 35 35 21 7 1 7C0 7C1

7C2 7C3

7C4 7C5

7C6

7C7

1 8 28 56 70 56 28 8 1 8C0 8C1

8C2 8C3

8C4 8C5

8C6 8C7

8C8

PROPERTIES OF nCk

3. nCk = nCn-k Symmetry

4. n+1Ck = nCk-1 + nCk Pascal’s triangle result: each

coefficient is the sum of the two

coefficients in the row above it

1 0C0

1 1 1C0 1C1

1 2 1 2C0 2C1

2C2

1 3 3 1 3C0 3C1

3C2 3C3

1 4 6 4 1 4C0 4C1

4C2 4C3

4C4

1 5 10 10 5 1 5C0 5C1

5C2 5C3

5C4 5C5

1 6 15 20 15 6 1 6C0 6C1

6C2 6C3

6C4 6C5

6C6

1 7 21 35 35 21 7 1 7C0 7C1

7C2 7C3

7C4 7C5

7C6 7C7

1 8 28 56 70 56 28 8 1 8C0 8C1

8C2 8C3

8C4 8C5

8C6 8C7

8C8

n+1Ck = nCk-1 + nCk Pascal’s triangle result

1 0C0

1 1 1C0 1C1

1 2 1 2C0 2C1

2C2

1 3 3 1 3C0 3C1

3C2 3C3

1 4 6 4 1 4C0 4C1

4C2 4C3

4C4

1 5 10 10 5 1 5C0 5C1

5C2 5C3

5C4 5C5

1 6 15 20 15 6 1 6C0 6C1

6C2 6C3

6C4 6C5

6C6

1 7 21 35 35 21 7 1 7C0 7C1

7C2 7C3

7C4 7C5

7C6 7C7

1 8 28 56 70 56 28 8 1 8C0 8C1

8C2 8C3

8C4 8C5

8C6 8C7

8C8

15 = 10 + 56C4 = 5C3 + 5C4

Example 1Example 1

(a) (a + 3)5 =

Example 1Example 1

(a) (a + 3)5 = 5C0 a5 + 5C1 a4 31 + 5C2 a3 32 + 5C3 a2 33 + 5C4 a1 34

+ 5C5 35

= a5 + 5a4.3 + 10a3.9 + 10a2.27 + 5a.81 + 243

= a5 + 15a4 + 90a3 + 270a2 + 405a + 243

(b) (2x – y)4 =

Example 1Example 1

(a) (a + 3)5 = 5C0 a5 + 5C1 a4 31 + 5C2 a3 32 + 5C3 a2 33 + 5C4 a1 34

+ 5C5 35

= a5 + 5a4.3 + 10a3.9 + 10a2.27 + 5a.81 + 243

= a5 + 15a4 + 90a3 + 270a2 + 405a + 243

(b) (2x – y)4 = 4C0 (2x)4 + 4C1 (2x)3(-y)1 + 4C2 (2x)2(-y)2

+ 4C3 (2x)1(-y)3 + 4C4 (-y)4

= 16x4 + 4(8x3)(-y) + 6(4x2)(y2) + 4(2x)(-y3) + y4

= 16x4 – 32x3y + 24x2y2 – 8xy3 + y4

Example 2 (2008 HSC, Question 1(d), 2 marks)Example 2 (2008 HSC, Question 1(d), 2 marks)

Find an expression for the coefficient of Find an expression for the coefficient of xx88yy44

in the expansion of (2in the expansion of (2xx + 3 + 3yy))1212..


(2x + 3y)12 = 12C0 (2x)12 + 12C1 (2x)11(3y)1 + 12C2 (2x)10(3y)2 + ...

General term Tk = 12Ck (2x)12-k(3y)k = 1 mark

For coefficient of x8y4, substitute k = ?


(2x + 3y)12 = 12C0 (2x)12 + 12C1 (2x)11(3y)1 + 12C2 (2x)10(3y)2 + ...

General term Tk = 12Ck (2x)12-k(3y)k = 1 mark

For coefficient of x8y4, substitute k = 4:

T4 = 12C4 (2x)8(3y)4

= 12C4 28 34 x8 y4

Coefficient is 12C4 28 34 or 10 264 320.

It’s OK to leave the coefficient unevaluated, especially if the

question asks for ‘an expression’.

Find Find an expression an expression for the for the coefficient of coefficient of xx88yy44

in the expansion of (2in the expansion of (2xx + 3 + 3yy))1212..

TTkk is not the is not the kkthth term term

Tk is the term that contains xk

Simpler to write out the first few terms rather than memorise the notation

Better to avoid calling it ‘the kth term’: too confusing

HSC questions ask for ‘the term that contains x8’ rather than ‘the 9th term’

Example 3 (2011 HSC, Question 2(c), 2 marks)Example 3 (2011 HSC, Question 2(c), 2 marks)

Find an expression for the coefficient of Find an expression for the coefficient of xx22

in the expansion of in the expansion of 8

43 .x

x


General term Tk = 8Ck (3x)8-k

= 8Ck 38-k x8-k (-4)k x-k

= 8Ck 38-k x8-2k (-4)k

For the coefficient of x2:8 – 2k = 2 -2k = -6 k = 3

T3 = 8C3 38-3 x8-2×3 (-4)3

= 8C3 35 (-4)3 x2

= -870 192 x2

Coefficient is -870 192 or 8C3 35 (-4)3

8 1 2

8 7 68 8 80 1 2

4 4 43 3 3 3 ...x C x C x C x

x x x

4k

x

FINDING THE GREATEST COEFFICIENT

(1 + 2x)8 = 1 + 16x + 112x2 + 448x3 + 1120x4

+ 1792x5 + 1792x6 + 1024x7 + 256x8

(1 + 2x)8 = 1 + 16x + 112x2 + 448x3 + 1120x4

+ 1792x5 + 1792x6 + 1024x7 + 256x8

Example 4Example 4

Suppose (1 + 2x)8 =

(a) Find an expression for tk, the coefficient of xk.

(b) Show that

(c) Show that the greatest coefficient is 1792.

8

0

.kk

k

t x

12 8

.1

k

k

kt

t k

Example 4Example 4

(a) (1 + 2x)8 = 8C0 18 + 8C1 17 (2x)1 + 8C2 16 (2x)2 + 8C3 15 (2x)3 +...

General term Tk = 8Ck 18-k (2x)k

= 8Ck 1 (2k) xk

= 8Ck 2k xk

tk = 8Ck 2k Leave out xk as we are only interested in the coefficient

Example 4Example 4

(a) (1 + 2x)8 = 8C0 18 + 8C1 17 (2x)1 + 8C2 16 (2x)2 + 8C3 15 (2x)3 +...


= 8Ck 1 (2k) xk

= 8Ck 2k xk

tk = 8Ck 2k

(b) Show that

Leave out xk as we are only interested in the coefficient

12 8

.1

k

k

kt

t k

Example 4Example 4

(a) (1 + 2x)8 = 8C0 18 + 8C1 17 (2x)1 + 8C2 16 (2x)2 + 8C3 15 (2x)3 +...


= 8Ck 1 (2k) xk

= 8Ck 2k xk

tk = 8Ck 2k

(b) Show that

Leave out xk as we are only interested in the coefficient

12 8

.1

k

k

kt

t k

kk

kk

k

k

C

C

t

t

2

28

11

81

Example 4Example 4

(b) (b) Show that

Ratio of consecutive factorials

kk

kk

k

k

C

C

t

t

2

28

11

81

!

!1

n

n

12 8

.1

k

k

kt

t k

Example 4Example 4

(b) (b) Show that

Ratio of consecutive factorials

kk

kk

k

k

C

C

t

t

2

28

11

81

18! 8!2

! 8 !1 ! 8 1 !

8 !8! !. . .2

8! 1 ! 7 !

1 8. .2

1 12 8

1

k kk k

kk

k k

k

kk

k

81...567

1...5678

7!

8! eg

1!

!1

nn

n

12 8

.1

k

k

kt

t k

Example 4Example 4


For the greatest coefficient tk+1, we want:

tk+1 > tk

16 – 2k > k + 1

-3k > -15

k < 5

k = 4

k must be

a whole number

1

1

82

11

k

k

t

t

k

k

for the largest possible integer

value of k

Example 4Example 4


Greatest coefficient tk+1 = t5

= 8C5 25

= 56 32

= 1792

tk = 8Ck 2k

THE BINOMIAL THEOREM FOR (1 + THE BINOMIAL THEOREM FOR (1 + xx))nn

(1 + x)n = nC0 + nC1 x + nC2 x2 + nC3 x3 + nC4 x4 +

... + nCn xn

=

kn

kk

n xC0

Example 5 (similar to 2010 HSC, Question 7(b)(i), 1 mark)Example 5 (similar to 2010 HSC, Question 7(b)(i), 1 mark)

Expand (1 + x)n and substitute

an appropriate value of x to prove that

nk

n

k

nC 20

Example 5Example 5

(1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn

Sub x = ?

[Aiming to prove: [Aiming to prove: ] ]n

k

n

k

nC 20

Example 5Example 5

(1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn

Sub x = 1:

(1 + 1)n = nC0 + nC1 (1) + nC2 (12) + ... + nCn (1n)

2n = nC0 + nC1 + nC2 + ... + nCn

This will make the x’s disappear and make the LHS become 2n

k

n

k

nn C

0

2

Example 6Example 6

By considering that (1 + x)2n = (1 + x)n(1 + x)n

and examining the coefficient of xn on each side, prove that

2 2

0

.n

n nk n

k

C C

Example 6Example 6 (1 + x)2n = (1 + x)n(1 + x)n

(1 + x)2n = 2nC0 + 2nC1 x + 2nC2 x2 + ... + 2nC2n x2n

Term with xn = 2nCn xn

Coefficient of xn = 2nCn

[Aiming to prove ][Aiming to prove ] n

nn

kk

n CC 2

0

2


(1 + x)2n = 2nC0 + 2nC1 x + 2nC2 x2 + ... + 2nC2n x2n

Term with xn = 2nCn xn

Coefficient of xn = 2nCn

(1 + x)n.(1 + x)n = (nC0 + nC1 x + nC2 x2 + ... + nCn xn)

(nC0 + nC1 x + nC2 x2 + ... + nCn xn)

If we expanded the RHS, there would be many terms

[Aiming to prove ][Aiming to prove ]

nn

n

kk

n CC 2

0

2


(1 + x)n.(1 + x)n = (nC0 + nC1 x + nC2 x2 + ... + nCn xn)

(nC0 + nC1 x + nC2 x2 + ... + nCn xn)

Terms with xn

= nC0(nCn xn) + nC1 x (nCn-1 xn-1) + nC2 x2 (nCn-2 xn-2) + ...

+ nCn xn (nC0)

Coefficient of xn

= nC0 nCn + nC1

nCn-1 + nC2 nCn-2 + ... + nCn

nC0

= (nC0)2 + (nC1)2 + (nC2)2 + ... + (nCn)2

by symmetry of Pascal’s triangle

[Aiming to prove ][Aiming to prove ]

nCk = nCn-k

nn

n

kk

n CC 2

0

2


By equating coefficients of xn on both sides of

(1 + x)2n = (1 + x)n.(1 + x)n

2nCn = (nC0)2 + (nC1)2 + (nC2)2 + ... + (nCn)2

n

kk

nn

n CC0

22

Example 7 (2006 HSC, Question 2(b), 2 marks)Example 7 (2006 HSC, Question 2(b), 2 marks)

(i) By applying the binomial theorem to (1 + x)n and differentiating, show that

(ii) Hence deduce that

1 1 11 2 ... ... .1 2

n r nn n n nn x x r x n x

r n

1 1 13 ... 2 ... 2 .1

n r nn n nn r n

r n


[Need to prove ]

(i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn

Differentiating both sides:

1 1 11 2 ... ...1 2


r n


[Need to prove ]



n(1 + x)n-1 = 0 + nC1 + 2 nC2 x + ... + r nCr xr-1 + ... + n nCn xn-1

= nC1 + 2 nC2 x + ... + r nCr xr-1 + ... + n nCn xn-1

The general term

1 1 11 2 ... ...1 2


r n


[Need to prove ]





(ii) Substitute x = ? to prove

1 1 13 ... 2 ... 2 .1

n r nn n nn r n

r n

1 1 11 2 ... ...1 2


r n






(ii) Substitute x = 2 to prove

n(1 + 2)n-1 = nC1 + 2 nC2 2 + ... + r nCr 2r-1 + ... + n nCn 2n-1

n 3n-1 = nC1 + 4 nC2 + ... + r nCr 2r-1 + ... + n nCn 2n-1

1 1 13 ... 2 ... 2 .1

n r nn n nn r n

r n


Let p and q be positive integers with p ≤ q.

(i) Use the binomial theorem to expand (1 + x)p+q , and hence

write down the term of which is independent of x.

(ii) Given that apply the binomial

theorem and the result of part (i) to find a simpler expression

for

1p q

q

x

x

1 11 1 ,

p q qp

q

xx

x x

1 ... .1 1 2 2

p q p q p q

p p


(i) (1 + x)p+q = p+qC0 + p+qC1 x + p+qC2 x2 + ... + p+qCp+q xp+q

The term of independent of x will be

q

qp

x

x 1




qqp

q

qq

qp

Cx

xC

is,that ,

q

qp

x

x 1




(ii) Given that apply the binomial

theorem and the result of part (i) to find a simpler expression

for

q

qp

x

x 1

qqp

q

qq

qp

Cx

xC

is,that ,

1 11 1 ,

p q qp

q

xx

x x

1 ... .1 1 2 2

p q p q p q

p p


(ii) (1 + x)p = pC0 + pC1 x + pC2 x2 + ... + pCp xp

1 1Find 1 ... using 1 1

1 1 2 2

p q qp

q

p q p q p q xx

p p x x


(ii)


Terms independent of x

1 1Find 1 ... using 1 1

1 1 2 2

p q qp

q

p q p q p q xx

p p x x


(ii)


Terms independent of x

= pC0 qC0 + pC1 x qC1 + pC2 x2 qC2 + ... + pCp xp qCp

= 1 + pC1 qC1 + pC2 qC2 + ... + pCp

qCp

px

1

x

1

2

1

x

p ≤ q

1 1Find 1 ... using 1 1

1 1 2 2

p q qp

q

p q p q p q xx

p p x x


By equating the terms independent of x on both sides of

p+qCq = 1 + pC1 qC1 + pC2 qC2 + ... + pCp

qCp

1 + pC1 qC1 + pC2 qC2 + ... + pCp

qCp = p+qCq

q

pq

qp

xx

x

x

111

1From (i)

1 1Find 1 ... using 1 1

1 1 2 2

p q qp

q

p q p q p q xx

p p x x

And now ...And now ...

A very hard identity to prove:A very hard identity to prove:

Example 9 from 2002 HSCExample 9 from 2002 HSC

Question 7(b), 6 marksQuestion 7(b), 6 marks


The coefficient of xk in (1 + x)n, where n is a positive integer, is denoted by ck (so nck = ck).

(i) Show that c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1.

(ii) Find the sum

Write your answer as a simple expression in terms of n.

To save time, this question

asks us to abbreviate nCk to ck

0 1 2 1 .

1.2 2.3 3.4 1 2n nc cc c

n n



= c0 + c1 x + c2 x2 + ... + cn xn [1]

Identity to be proved involves (n + 2) 2n-1 so try ...

[Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 ]

To save time, this question asks us to abbreviate nCk to ck



= c0 + c1 x + c2 x2 + ... + cn xn [1]



n(1 + x)n-1 = c1 + 2c2 x + ... + ncn xn-1

Substituting x = ? to give 2n-1 on the LHS:




= c0 + c1 x + c2 x2 + ... + cn xn [1]



n(1 + x)n-1 = c1 + 2c2 x + ... + ncn xn-1

Substituting x = 1 to give 2n-1 on the LHS:

n(1 + 1)n-1 = c1 + 2c2 1 + ... + ncn 1n-1

n 2n-1 = c1 + 2c2 + ... + ncn [2]

To prove result, we must add c0 + c1 + c2 + ... + cn

To get this, sub x = ? into [1] above:




= c0 + c1 x + c2 x2 + ... + cn xn [1]


n(1 + x)n-1 = c1 + 2c2 x + ... + ncn xn-1

Substituting x = 1:

n(1 + 1)n-1 = c1 + 2c2 1 + ... + ncn 1n-1

n 2n-1 = c1 + 2c2 + ... + ncn [2]

To get this, sub x = 1 into [1] above:

(1 + 1)n = c0 + c1 1 + c2 12 + ... + cn 1n

2n = c0 + c1 + c2 + ... + cn [3]


Now add [2] and [3] to prove result


(i) n 2n-1 = c1 + 2c2 + ... + ncn [2]

2n = c0 + c1 + c2 + ... + cn [3]

[2] + [3]:

n 2n-1 + 2n = c0 + c1 + c1 + 2c2 + c2 + ... + ncn + cn

2n-1 (n + 2) = c0 + 2c1 + 3c2 + ... + (n + 1)cn


Result proved: each coefficient of ck increased

by 1 as required


(ii) Find the sum

Write your answer as a simple expression in terms of n.

Answer involves dividing by (k + 1)(k + 2) and alternating

–/+ pattern so try integrating (1 + x)n from (i) twice and substituting x = -1.

(1 + x)n = c0 + c1 x + c2 x2 + ... + cn xn [1]

0 1 2 1 .

1.2 2.3 3.4 1 2n nc cc c

n n


(ii) (1 + x)n = c0 + c1 x + c2 x2 + ... + cn xn [1]

Integrate both sides:

To find k, sub x = 0:

[Aiming to find ]

kxn

cx

cx

cxcx

nnnn

13221

01

1321

1

1

Don’t forget the constant of integration

1

1

1321

1

11

1

000011

1

132210

1

1

nx

n

cx

cx

cxcx

n

nk

kn

nnn

n

211

4.33.22.1210

nn

cccc nn


(ii)

Integrate again to get 1.2, 2.3, 3.4 denominators:

To find d, sub x = 0:

[Aiming to find ]

dn

x

nn

xcxcxcxcx

nn

nnn

1214.33.221

21

1 142

31

202

211

4.33.22.1210

nn

cccc nn


(ii)

Sub x = ? for –/+ pattern:

[Aiming to find ] 211

4.33.22.1210

nn

cccc nn


(ii)

Sub x = -1 for –/+ pattern:


4.33.22.1210

nn

cccc nn


(ii)


4.33.22.1210

nn

cccc nn

2

1

21

1

21

12

21

1

1

1

21)1(

4.33.22.1

21

1

1

1

21

1)1(

4.33.22.10

210

2210

n

nn

n

nn

n

nnnnn

cccc

nnnnn

cccc

nn

nn

Example 10 (2007 HSC, Question 4(a), 4 marks)Example 10 (2007 HSC, Question 4(a), 4 marks)

In a large city, 10% of the population has green eyes.In a large city, 10% of the population has green eyes.

(i) What is the probability that two randomly chosen people have (i) What is the probability that two randomly chosen people have green eyes?green eyes?

(ii) What is the probability that exactly two of a group of 20 (ii) What is the probability that exactly two of a group of 20 randomly chosen people have green eyes? Give your answer randomly chosen people have green eyes? Give your answer to three decimal places.to three decimal places.

(iii) What is the probability that more than two of a group of 20 (iii) What is the probability that more than two of a group of 20 randomly chosen people have green eyes? Give your answer randomly chosen people have green eyes? Give your answer to two decimal places.to two decimal places.


(i) P(both green) = 0.1 0.1

= 0.01



= 0.01

(ii) What is the probability that exactly two of a group of 20 (ii) What is the probability that exactly two of a group of 20 randomly chosen people have green eyes? Give your answer randomly chosen people have green eyes? Give your answer to three decimal places.to three decimal places.



= 0.01

(ii) p = 0.1, q = 1 – 0.1 = 0.9, n = 20

P(X = 2) =



= 0.01

(ii) p = 0.1, q = 1 – 0.1 = 0.9, n = 20

P(X = 2) = 20C2 0.12 0.918

= 0.28517 ...

0.285



= 0.01

(ii) p = 0.1, q = 1 – 0.1 = 0.9, n = 20

P(X = 2) = 20C2 0.12 0.918

= 0.28517 ...

0.285

(iii) What is the probability that more than two of a group of 20 (iii) What is the probability that more than two of a group of 20 randomly chosen people have green eyes? Give your answer randomly chosen people have green eyes? Give your answer to two decimal places.to two decimal places.


(i) P(both green) = 0.1 0.1 = 0.01

(ii) p = 0.1, q = 1 – 0.1 = 0.9, n = 20

P(X = 2) = 20C2 0.12 0.918

= 0.28517 ... 0.285

(iii) P(X > 2) = 1 – P(X = 0) – P(X = 1) – P(X = 2)

= 1 – 20C0 0.10 0.920 – 20C1 0.110.919 – 0.285 from (ii)

= 1 – 0.920 – 20(0.1)0.919 – 0.285 = 0.32325 ... 0.32

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