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Maths Extension 1 – Binomial Theorem & Binomial Probability

Binomial Theorem & Binomial Probability ! ! ! ! ! ! ! ! ! !

Pascal’s Triangle General Expansion Special Expansion Finding Terms Equidistant Coefficients Pascal’s Relation Pairing Off Method Sums of Coefficients Greatest Coefficient Extra stuff

! Success and Failure

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Maths Extension 1 – Binomial Theorem & Binomial Probability

Pascal’s Triangle 1 1 1 1 1 1

3 4

6

1 3

6

5

1

1 2

10 15

1 4

10 20

1 5

15

1 6

1

1 2 4 8 16 32 64

20 21 2 2 23 24 25 26

2n-1 n = line/ row

n

Cr

=

n! r!( n r )!

Where Tr+1 term is important to identify a particular term Tn = Tr+1

n

General Expansion a x n n n n n r r a xn = n C0 a n x 0 + C1a 1 x 1 + … + Cr a x + … + n C n a 0 x ___1st _____________________Term r + 1__________last

n Special Expansion 1 x n = nC0 x 0 + nC1 x1 + … + nCr xr + … + nC n x n 1 x

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Maths Extension 1 – Binomial Theorem & Binomial Probability

Finding Terms Example 1 Find Term 4 7 = 7C0 x0 1 x = x0 7

C4

+ 7C1 x1

+ 7C 2 x 2

+ 7C 3 x 3

+ 7C4 x4

+ 7C5 x5

+ 7C 6 x 6

+ 7C 7 x 7

+ 7x1

+ 21x 2

+ 35x 3

+ 35x4

+ 21x 5

+ 7x 6

+ x7

7! 4! 7 4 ! 7! = 4!.3! 5040 = 144 = 35

=

Example 2 Find the term independent of x in 3 x 2 n Tr+1 = C r a n r x r

1 2x

3 2 = Cr 3 x

3

3 r

1 r 2x

6 2 0 = x r .xr = x 6 – 3r = 0 r 2

Equidistant Coefficients Equidistant coefficients are equal Proof LHS = nC r

n! r! n r ! n! = r! n r ! =

n

Cr

n

Cn

r

RHS = nC n

r

n! n r ! n (n r) ! n! = n r!n n r! n! = r! n r ! =

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Maths Extension 1 – Binomial Theorem & Binomial Probability

Pascal’s Relation n C r 1 nC r n 1Cr 1 1 1 1

1 2

1

3

Example n Cr 1 + nCr 2 +1

3

nth row n 1 row

1

1

= n Cr =3

Proof 1 LHS = nC r nC r 1 n! n! = r! n r ! r 1 ! n r 1 ! n! n r 1 r .n! = r! n r 1 ! n r 1! >>> n r! 1.2.3... n r n r 1 >>> 1.2.3... n r >>> n r 1

RHS

=

n 1

Cr n 1! = r! n r 1 !

n! n r 1 r r! n r 1 ! n! n 1 = r! n r 1 ! n 1! = r! n r 1 ! = RHS =

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Maths Extension 1 – Binomial Theorem & Binomial Probability

Proof 2 Expand 1 x n 1 in two ways. Compare the coefficients of xr . 1st n 1 xn1 = 1 x .1 x n n r n r n n 0 n 1 = 1 x . C0 x C 1x ... Cr 1 x 1 Cr x ... Cn x Coefficients of xr = 1 n Cr nCr 1 xr = nCr n Cr 1 xr 2nd 1 x

n 1

=

n 1

C0 x 0

n 1

C1 x1 ...

n 1

Cr xr

...

n 1

n1

Cn 1 x

Coefficients of xr 1 = n Cr n

Cr

n

Cr

n 1

Cr

1

Pairing Off Method 1 x

Find the coefficient of x in the expansion of x **Do Not Expand = 4C x 4 1 0 + 4C x3 1 0 x

C 0 x3

3

1 0 x

1 1 x

+ 3C x 2 1

1 2 x

+ 4 C2 x 2

1 1 x

+ 3C x 1 2

+ 4C 3 x 1

1 2 x

4

x

1 3 x

3

x

+ 4C 4 x 0

+ 3C x 0 3

1

1 4 x

1 3 x

Coefficients of x are: = 4C 0. 3C 3 + 4C1.3 C2 + 4C 2 . 3C1 + 4C 3.3 C0 = –1 + 12 – 18 + 4 = –3

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Maths Extension 1 – Binomial Theorem & Binomial Probability

Sums of Coefficients n n r

n

Cr

2n

0

OR C 0 nC 1 nC 2 ... nCn

2n

Greatest Coefficient a bn Tr 1 Tr

n

Cr an r br 1 1 n Cr 1 an r br b n! r 1! n r 1! . = . a r! n r ! n! b n r 1 = . 1 a r

=

Example 1 Find the greatest coefficient in 5 2 x 12 . = 5 6 13 = . r 5 78 6r = 5r 78 6 r 78

r r Tr

1

T8

2 x 12 when x

3

r 1 r r r is an integer

1

5r 11r 78 11 =7 12 = C7 a12 7 b7 = 6.93 x 1011

Example 2 12 Find the greatest coefficient in 7 3 x when x 6 13 r = . 1 7 r 78 6 r 7r 78 13r r =7

2

= 12C6 76 66 = 5.07 x 1012

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Maths Extension 1 – Binomial Theorem & Binomial Probability

Extra Examples Example 1 8 Find the middle term of a 2b 4 4 8 C 4 a 4 2b 4 = 70a 16b

There are 9 terms. Middle term is T5.

= 1120 a 4b 4 Example 2 11

C5

1 6 x

x2

5

x2

1 x

Fine the middle term of

11

There are 12 terms. Middle term is T6, T7.

= 462x 4

11

C6

1 5 x

x2

6

= 462x 7

Example 3 1 x 8 1 x 8 1 x 16 16 Prove C 3 2 8C 0.8C 3 8C 1. 8C 2 RHS =

8

... 8 C8 x8

8

... 8 C8 x8

C0 x 0 8C1 x1 C0 x 0 8C1 x1

Coefficients of x3

LHS =

16

=

16

16 Prove C8

= 8C0 .8 C3 + 8C 1. 8C 2 + 8C 2 .8C1 + 8C 3.8 C0 8 = 2 C0 .8 C3 8C1 .8 C2

C3 x3 C3

C0

8

2

...

C8

8

2

= 8C 0. 8C 8 + 8C1.8 C7 + … + 8C0 .8 C8 2

= 8C0 + = Proven

8

C1

2

+ … + 8C8

2

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Maths Extension 1 – Binomial Theorem & Binomial Probability

Success and Failure n q p nCr q n r p r Let p be the one you want Let q be the one you don’t want There are only 2 options – Binomial

“Success” “Failure”

Example 1 A die is tossed 3 times. “6” outcome S

SSS

1 3 5 0 6 6

p 3q 0

F

SSF

1 2 5 1 6 6

p 2q1

S

SFS

1 2 5 1 6 6

p 2q1

F

SFF

1 1 5 2 6 6

p1q 2

S

FSS

1 2 5 1 6 6

p 2q1

F

FSF

1 1 5 2 6 6

p1q 2

S

FFS

1 1 5 2 6 6

p1q 2

F

FFF

1 0 5 3 6 6

p 0 q3

S S F ?

S F F

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Maths Extension 1 – Binomial Theorem & Binomial Probability

Example 2 An archer has a record of hitting the target 3 out of 4 occasions. He tries 5 times. Find the probability a) Exactly 3 hits b) Exactly 4 hits c) Bull’s eye only in 2nd round d) 1 bull’s eye A 5C3 q 2 p 3 B 5C 4q 1 p 4 5C 5q 0 p 5 1 1 3 4 1 35 = 10 1 2 3 3 = 5 4

4

=

C

4

135 512

4

4

405 243 1024 1024 81 = 128 5 C1q 4 p 1 4 3 = 5 1

=

1 3 4 4 2

1 1 4 4 3

D

1 4

4

= 1 4 4 3 = 1024

4

15 = 1024

Example 3 For a certain species of bird, there is a 3 in 5 chance that a fledgling will survive the 1st month. From a breed of 10 chicks, find the probability that: a) P (0 survive) b) P (more than 1 survive) c) P (3 survive) 10 0 10 10 A C0 3 2 =1 2 5

5

5

1024 9765625 =1 10 C0 3 0

= B

1

P0

P1

5

2 10 5

16384 9762625 9748881 = 9765625 7 3 = 120 25 5 414720 = 9762625 = 0.0425

10

C1

3 1 2 9 5 5

=1

C

10

C3 q 7 p 3

3

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