THE BINOMIAL THEOREM
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Transcript THE BINOMIAL THEOREM
THE BINOMIAL THEOREM
Robert Yen
Hurlstone Agricultural High School
This presentation is accompanied by workshop notes
INTRODUCTION
Expanding (a + x)n
Difficult topic: high-level algebra
Targeted at better Extension 1 students
Master this topic to get ahead in the HSC exam
No shortcuts for this topic
BINOMIAL EXPANSIONS AND PASCAL’S TRIANGLE
(a + x)1 = a + x
(a + x)2 = a2 + 2ax + x2
(a + x)3 = a3 + 3a2x + 3ax2 + x3
(a + x)4 = a4 + 4a3x + 6a2x2 + 4ax3 + x4
Coefficients
11
121
1331
14641
(a + x)5 = a5 + 5a4x + 10a3x2 + 10a2x3 + 5ax4 + x5 1 5 10 10 5 1
nC ,
k
A FORMULA FOR PASCAL’S TRIANGLE
1
11
121
1331
14641
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
nC
k
0C
0
1C 1C
0
1
2C 2C 2C
0
1
2
3C 3C 3C 3C
0
1
2
3
4C 4C 4C 4C 4C
0
1
2
3
4
5C 5C 5C 5C 5C 5C
0
1
2
3
4
5
6C 6C 6C 6C 6C 6 C 6C
0
1
2
3
4
5
6
7 C 7C 7C 7C 7C 7C 7C 7C
0
1
2
3
4
5
6
7
8C 8C 8C 8C 8C 8C 8C 8C 8C
0
1
2
3
4
5
6
7
8
gives the value of row n, term k,
if we start numbering the rows and terms from 0
nC ,
k
A FORMULA FOR PASCAL’S TRIANGLE
n
n
Ck
k
CALCULATING
Mentally
5
C3
5 4 3 20
C3
10
3 2 1 2
5
CALCULATING
5
C3
Mentally
5 4 3 20
C3
10
3 2 1 2
5
Formula
5!
120
C3
10
3!2! 6 2
5
n
n!
Ck
k!n k !
CALCULATING
5
C3
Mentally
5 4 3 20
C3
10
3 2 1 2
5
Formula
5!
120
C3
10
3!2! 6 2
5
n
n!
Ck
k!n k !
because ...
5 4 3 5 4 3 2 1
5!
C3
3 2 1 3 2 1 2 1 3! 2!
5
THE 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
n
=
Ck a
n
k 0
The sum of terms
from k = 0 to n
nk k
x
Don’t worry too much about
writing in notation: just have a
good idea of the general term
PROPERTIES OF nCk
1. nC0 = nCn = 1
1st and last
2. nC1 = nCn-1 = n
2nd and 2nd-last
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
n+1C
k
= nCk-1 + nCk
1
11
121
1331
14641
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
Pascal’s triangle result
0C
0
1C 1C
0
1
2C 2C 2C
0
1
2
3C 3C 3C 3C
0
1
2
3
4C 4C 4C 4C 4C
0
1
2
3
4
5C 5C 5C 5C 5C 5C
0
1
2
3
4
5
6C 6C 6C 6C 6C 6 C 6C
0
1
2
3
4
5
6
7 C 7C 7C 7C 7C 7C 7C 7C
0
1
2
3
4
5
6
7
8C 8C 8C 8C 8C 8C 8C 8C 8C
0
1
2
3
4
5
6
7
8
15 = 10 + 5
6C = 5C + 5C
4
3
4
Example 1
(a) (a + 3)5 =
Example 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 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)
(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
For coefficient of x8y4, substitute k = ?
= 1 mark
Example 2 (2008 HSC, Question 1(d), 2 marks)
(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) x8y4
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’.
Tk is not the kth term
Tk is the term that contains xk
Simpler to write out the first few terms rather than memorise
the notation
Better to avoid referring to ‘the kth term’: too confusing
HSC questions ask for ‘the term that contains x8’ rather than
‘the 9th term’
Example 3 (2005 HSC, Question 2(b), 3 marks)
12
1
2
1
1 12
1
12
11
10
12
12
2 x 2 C0 2 x C1 2 x 2 C2 2 x 2 ...
x
x
x
General term Tk =
1
12C (2x)12-k
2
k
x
k
= 12Ck 212-k x12-k (-x-2)k
= 12Ck 212-k x12-k (-1)k x-2k
= 12Ck 212-k x12-3k (-1)k
For term independent of x:
Example 3 (2005 HSC, Question 2(b), 3 marks)
12
1
2
1
1 12
1
12
11
10
12
12
2
x
C
2
x
C
2
x
C
2
x
2
2 ...
0
1
2
2
x
x
x
General term Tk =
12C (2x)12-k 1
2
k
x
k
= 12Ck 212-k x12-k (-x-2)k
= 12Ck 212-k x12-k (-1)k x-2k
= 12Ck 212-k x12-3k (-1)k
For term independent of x:
12 – 3k = 0
k=4
T4 = 12C4 28 x0 (-1)4
= 126 720
FINDING THE GREATEST COEFFICIENT
(1 + 2x)8 = 1 + 16x + 112x2 + 448x3 + 1120x4
+ 1792x5 + 1792x6 + 1024x7 + 256x8
Example 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 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
tk 1 8 Ck 1 2k 1
8
(b)
tk
Ck 2k
Leave out xk as we are only
interested in the coefficient
Example 4
8
k 1
t
C
2
k 1
(b) k 1
8
tk
Ck 2k
8!
8!
21
k 1!8 k 1! k!8 k !
k! 8 k !
Ratio of consecutive factorials
.
.2
n 1!
k 1! 7 k !
n!
1 8k
.
.2
k 1 1
28 k
k 1
Example 4
8
k 1
t
C
2
k 1
(b) k 1
8
tk
Ck 2k
8!
8!
21
k 1!8 k 1! k!8 k !
k! 8 k !
Ratio of consecutive factorials
.2
.
n 1! n 1
k 1! 7 k !
n!
1 8k
.2
.
8! 8 7 6 5 ... 1
eg
8
k 1 1
7!
7 6 5 ... 1
28 k
k 1
Example 4
(c) For the greatest coefficient tk+1, we want:
tk+1 > tk
tk 1
1
tk
28 k
1
k 1
16 – 2k > k + 1
-3k > -15
k<5
k=4
for the largest
possible integer
value of k
k must be a whole
number
Example 4
(c) Greatest coefficient tk+1 = t5
= 8C5 25
= 56 32
= 1792
tk = 8Ck 2k
THE BINOMIAL THEOREM FOR (1 + x)n
(1 + x)n = nC0 + nC1 x + nC2 x2 + nC3 x3 + nC4 x4 +
... + nCn xn
n
=
Ck x
k 0
n
k
Example 5
(1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
Sub x = ?
n
[Aiming to prove:
k 0
n
Ck 2 n ]
Example 5
(1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
This will make the x’s disappear
and make the LHS become 2n
Sub x = 1:
(1 + 1)n = nC0 + nC1 (1) + nC2 (12) + ... + nCn (1n)
2n = nC0 + nC1 + nC2 + ... + nCn
n
2 n Ck
n
k 0
Example 6
(1 + x)2n = 2nC0 + 2nC1 x + 2nC2 x2 + ... + 2nC2n x2n
Term with xn = 2nCn xn
Coefficient of xn = 2nCn
C
n
[Aiming to prove
2
n
k 0
k
2 nCn ]
Example 6
(1 + x)2n = 2nC0 + 2nC1 x + 2nC2 x2 + ... + 2nC2n x2n
Term with xn = 2nCn xn
Coefficient of xn = 2nCn
(1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
(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
C
n
[Aiming to prove
2
n
k 0
k
2 nCn ]
Example 6
(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
nC
C
n
[Aiming to prove
2
n
k 0
k
2 nCn ]
k=
nC
n-k
Example 6
By equating coefficients of xn on both sides of
(1 + x)2n = (1 + x)n.(1 + x)n
2nC
n
= (nC0)2 + (nC1)2 + (nC2)2 + ... + (nCn)2
n
2n
Cn Ck
n
k 0
2
Example 7 (2006 HSC, Question 2(b), 2 marks)
(i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
The general term
Differentiating both sides:
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
(ii) Substitute x = ? to prove result:
[Aiming to prove: n 3n-1 = nC1 + ... + r nCr 2r-1 + ... + n nCn 2n-1 ]
Example 7 (2006 HSC, Question 2(b), 2 marks)
(i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
Differentiating both sides:
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
(ii) Substitute x = 2 to prove result:
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
Example 8 (2008 HSC, Question 6(c), 5 marks)
(i) (1 + x)p+q = p+qC0 + p+qC1 x + p+qC2 x2 + ... + p+qCp+q xp+q
The term of
1 x p q
xq
independent of x will be
Example 8 (2008 HSC, Question 6(c), 5 marks)
(i) (1 + x)p+q = p+qC0 + p+qC1 x + p+qC2 x2 + ... + p+qCp+q xp+q
The term of
1 x p q
xq
pq
Cq x q
x
q
independent of x will be
, that is,
pq
Cq
Example 8 (2008 HSC, Question 6(c), 5 marks)
(i) (1 + x)p+q = p+qC0 + p+qC1 x + p+qC2 x2 + ... + p+qCp+q xp+q
The term of
1 x p q
xq
pq
Cq x q
x
q
independent of x will be
, that is,
pq
Cq
(ii) (1 + x)p = pC0 + pC1 x + pC2 x2 + ... + pCp xp
q
2
q
1
1
1
q
q 1
q
q
1 C0 C1 C2 ... Cq
x
x
x
x
1
1
q
q 1
q
q
C0 C1 C2 2 ... Cq q
x
x
x
q
1
p
1 x 1 pC0 pC1 x pC2 x 2 ... pC p x p
x
1 q 1
1
q
C0 C1 C2 2 ... Cq q
Example 8 (2008 HSC,
x Question
x 6(c), 5 marks)
x
q
q
p
(ii) 1 x 1 pC0 pC1 x pC2 x 2 ... pC p x p
1
x
q
1
1
1
qC0 qC1 qC2 2 ... qCq q
x
x
x
If we expanded the RHS, there would be many terms
Terms independent of x
=
1 q 1
1
q
C0 C1 C2 2 ... Cq q
Example 8 (2008 HSC,
x Question
x 6(c), 5 marks)
x
q
q
p
(ii) 1 x 1 pC0 pC1 x pC2 x 2 ... pC p x p
1
x
q
1
1
1
qC0 qC1 qC2 2 ... qCq q
x
x
x
If we expanded the RHS, there would be many terms
Terms independent of x
=
=
1
+ ... + pCp xp qCp 1
p
x
pC qC + pC x qC 1 + pC x2 qC
0
0
1
1
2
2 x2
x
p
q
p
q
p
q
1 + C1 C1 + C2 C2 + ... + Cp Cp
p≤q
1 q 1
1
q
C0 C1 C2 2 ... Cq q
Example 8 (2008 HSC,
x Question
x 6(c), 5 marks)
x
q
q
(ii) 1 x p 1 1 pC0 pC1 x pC2 x 2 ... pC p x p
q
x
1
1
1
qC0 qC1 qC2 2 ... qCq q
x
x
x
Terms independent of x
=
pC qC
0
0
+
p C x q C 1 + p C x2 q C
1
1
2
2
x
1 + ... + pC xp qC 1
2
p
p
p
x
x
= 1 + pC1 qC1 + pC2 qC2 + ... + pCp qCp
By equating the terms independent of x on both sides of
From
(i)
1 x p q 1 x p 1 1 q
q
x
x
p+qC = 1 + pC qC + pC qC + ... + pC qC
q
1
1
2
2
p
p
1 + pC1 qC1 + pC2 qC2 + ... + pCp qCp = p+qCq
And now ...
A fairly hard identity to prove:
Example 9 from 2002 HSC
Question 7(b), 6 marks
Example 9 (2002 HSC, Question 7(b), 6 marks)
(i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
= c0 + c1 x + c2 x2 + ... + cn xn [1]
To save time, this
question asks us to
abbreviate nCk to ck
Identity to be proved involves (n + 2) 2n-1 so try ...
[Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 ]
Example 9 (2002 HSC, Question 7(b), 6 marks)
(i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
= c0 + c1 x + c2 x2 + ... + cn xn [1]
Identity to be proved involves (n + 2) 2n-1 so try ...
Differentiating both sides:
n(1 + x)n-1 = c1 + 2c2 x + ... + ncn xn-1
Substituting x = ? to give 2n-1 on the LHS:
[Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 ]
Example 9 (2002 HSC, Question 7(b), 6 marks)
(i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
= c0 + c1 x + c2 x2 + ... + cn xn [1]
Identity to be proved involves (n + 2) 2n-1 so try ...
Differentiating both sides:
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]
How do we make the LHS say (n + 2) 2n-1?
[Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 ]
Example 9 (2002 HSC, Question 7(b), 6 marks)
(i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
= c0 + c1 x + c2 x2 + ... + cn xn [1]
Differentiating both sides:
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]
Add 2 (2n-1) to both sides. But that’s 2n.
[Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 ]
Example 9 (2002 HSC, Question 7(b), 6 marks)
(i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
= c0 + c1 x + c2 x2 + ... + cn xn [1]
Differentiating both sides:
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 prove the 2n identity, sub x = 1 into [1] above:
(1 + 1)n = c0 + c1 1 + c2 12 + ... + cn 1n
2n = c0 + c1 + c2 + ... + cn [3]
[Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 ]
Example 9 (2002 HSC, Question 7(b), 6 marks)
(i)
n 2n-1 = c1 + 2c2 + ... + ncn [2]
2n = c0 + c1 + c2 + ... + cn [3]
[2] + [3]:
n 2n-1 + 2n = c1 + 2c2 + ... + ncn + c0 + c1 + c2 + ... + cn
2n-1 (n + 2) = c0 + 2c1 + 3c2 + ... + (n + 1)cn
c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1
Each coefficient ck
increases by 1 as
required
[Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 ]
Example 9 (2002 HSC, Question 7(b), 6 marks)
(ii) 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]
c0
cn
c1
c2
n
1
[Aiming to find
]
n 1n 2
1.2 2.3 3.4
Example 9 (2002 HSC, Question 7(b), 6 marks)
(ii) (1 + x)n = c0 + c1 x + c2 x2 + ... + cn xn [1]
Integrate both sides:
1
c1 2 c2 3
cn n 1
n 1
1 x c0 x x x
x k
n 1
2
3
n 1
To find k, sub x = 0:
Don’t forget the
1 n 1
1 0 0 0 0 k
constant of integration
n 1
1
k
n 1
1
c1 2 c2 3
cn n 1
1
n 1
1 x c0 x x x
x
n 1
2
3
n 1
n 1
c0
cn
c1
c2
n
1
[Aiming to find
]
n 1n 2
1.2 2.3 3.4
1
k 9 (2002 HSC, Question 7(b), 6 marks)
Example
n 1
1
c1 2 c2 3
cn n 1
1
n 1
(ii)
1 x c0 x x x
x
n 1
2
3
n 1
n 1
Integrate again to get 1.2, 2.3, 3.4 denominators:
1
c0 x 2 c1x3 c2 x 4
cn x n 1
x
n2
1 x
d
n 1n 2
n 1n 2 n 1
2
2.3
3.4
To find d, sub x = 0:
1
1n 2 0 0 0 0 0 d
n 1n 2
1
d
n 1n 2
2
3
4
n2
1
c
x
c
x
c
x
c
x
n
2
n
c0 x c1 c02 1 n 2 c
1
n
[Aiming
n 1ton find
2 1.2 2.3 31..42 2.3 1 3.n4 1n 2n] 1n 2
x
1
1
d 9 (2002 HSC, Question 7(b), 6 marks)
Example
n 1n 2
(ii)
cn x n 2
c0 x 2 c1 x3 c2 x 4
1
n2
1 x
n 1n 2
n 1n 2
3.4
2.3
1.2
1
x
n 1 n 1n 2
Sub x = ? for –/+ pattern:
c0
cn
c1
c2
n
1
[Aiming to find
]
n 1n 2
1.2 2.3 3.4
1
d 9 (2002 HSC, Question 7(b), 6 marks)
Example
n 1n 2
cn x n 2
c0 x 2 c1 x3 c2 x 4
1
n2
1 x
(ii)
n 1n 2
n 1n 2
3.4
2.3
1.2
1
x
n 1 n 1n 2
Sub x = -1 for –/+ pattern:
2
3
4
n2
1
c
(
1
)
c
(
1
)
c
(
1
)
c
(
1
)
0n 2 0
1
2
n
n 1n 2
n 1n 2
1.2
2.3
3.4
1
1
n 1 n 1n 2
c0 c1 c2
cn (1) n 12
1
1
0
n 1n 2 n 1 n 1n 2
1.2 2.3 3.4
c0
cn 1
cc1 (1c)2 n
n
c0[Aiming
c1 tocfind
1
n
1
2
n 1n 2 ]
1.2 2.3 3.4
n 1n 2 n 1 n 1n 2
1.2 2.3 3.4
Example 9 (2002 HSC, Question 7(b), 6 marks)
c0 c1 c2
cn (1) n 1
1
1
(ii) 0
n 1n 2 n 1 n 1n 2
1.2 2.3 3.4
c
c
c
cn
1
1
0 1 2 (1) n
n 1n 2 n 1 n 1n 2
1.2 2.3 3.4
n 2 1
n 1n 2
n 1
n 1n 2
1
n 2
2
c0
cn
c1
c2
n
1
[Aiming to find
]
n 1n 2
1.2 2.3 3.4
Example 10 (2007 HSC, Question 4(a), 6 marks)
(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) =
Example 10 (2007 HSC, Question 4(a), 6 marks)
(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
Example 10 (2007 HSC, Question 4(a), 6 marks)
(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
HOW TO STUDY FOR MATHS (P-R-A-C)
1. Practise your maths
2. Rewrite your maths
3. Attack your maths
4. Check your maths
WORK HARD AND BEST OF LUCK
FOR YOUR HSC EXAMS!