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

TO AVOID ‘DEATH BY POWERPOINT’ ... Don’t copy everything down, as this presentation is on the HSC Online website Instead, make notes and copy the key steps, then fill in the gaps when you get home Aim to understand the theory TODAY, not later The handout contains all the important facts, while this PowerPoint contain the solutions to the examples If you already know this topic well, then work ahead and check your answers against mine 2

INTRODUCTION 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 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 4

nCk, A FORMULA FOR PASCAL’S TRIANGLE 1 0C0 1 1 1C0 1C1 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 8C78C8 nCk gives the value of row n, term k, if we start numbering the rows and terms from 0 5

C stands for coefficient as well as combination nCk, A FORMULA FOR PASCAL’S TRIANGLE C stands for coefficient as well as combination 6

CALCULATING Mentally

CALCULATING Mentally Formula

CALCULATING Mentally Formula because ...

The sum of terms from k = 0 to n 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 = 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

2. nC1 = nCn-1 = n 2nd and 2nd-last 1 0C0 1 1 1C0 1C1 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 8C78C8 11

4. n+1Ck = nCk-1 + nCk Pascal’s triangle result: each 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 8C78C8 12

n+1Ck = nCk-1 + nCk Pascal’s triangle result 1 0C0 1 1 1C0 1C1 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 8C78C8 15 = 10 + 5 6C4 = 5C3 + 5C4 13

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 =

(a) (a + 3)5 = 5C0 a5 + 5C1 a4 31 + 5C2 a3 32 + 5C3 a2 33 + 5C4 a1 34 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 16

Example 2 (2008 HSC, Question 1(d), 2 marks) Find an expression for the coefficient of x8y4 in the expansion of (2x + 3y)12. 17

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 = ? 18

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 x8 y4 Coefficient is 12C4 28 34 or 10 264 320.  Find an expression for the coefficient of x8y4 in the expansion of (2x + 3y)12. It’s OK to leave the coefficient unevaluated, especially if the question asks for ‘an expression’. 19

Tk is the term that contains xk 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 calling it ‘the kth term’: too confusing HSC questions ask for ‘the term that contains x8’ rather than ‘the 9th term’ 20

Find an expression for the coefficient of x2 Example 3 (2011 HSC, Question 2(c), 2 marks) Find an expression for the coefficient of x2 in the expansion of 21

Example 3 (2011 HSC, Question 2(c), 2 marks) 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  22

FINDING THE GREATEST COEFFICIENT (1 + 2x)8 = 1 + 16x + 112x2 + 448x3 + 1120x4 + 1792x5 + 1792x6 + 1024x7 + 256x8 23

(1 + 2x)8 = 1 + 16x + 112x2 + 448x3 + 1120x4 + 1792x5 + 1792x6 + 1024x7 + 256x8 Example 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. 24

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 Leave out xk as we are only interested in the coefficient 25

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 (b) Show that Leave out xk as we are only interested in the coefficient 26

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 (b) Show that Leave out xk as we are only interested in the coefficient 27

Ratio of consecutive factorials Example 4 (b) (b) Show that Ratio of consecutive factorials 28

Ratio of consecutive factorials Example 4 (b) (b) Show that Ratio of consecutive factorials 29

for the largest possible integer value of k Example 4 (c) Show that the greatest coefficient is 1792. For the greatest coefficient tk+1, we want: tk+1 > tk 16 – 2k > k + 1 -3k > -15 k < 5 k = 4 for the largest possible integer value of k k must be a whole number 30

(c) Show that the greatest coefficient is 1792. Example 4 (c) Show that the greatest coefficient is 1792. Greatest coefficient tk+1 = t5 = 8C5 25 = 56  32 = 1792 tk = 8Ck 2k 31

(1 + x)n = nC0 + nC1 x + nC2 x2 + nC3 x3 + nC4 x4 + ... + nCn xn THE BINOMIAL THEOREM FOR (1 + x)n (1 + x)n = nC0 + nC1 x + nC2 x2 + nC3 x3 + nC4 x4 + ... + nCn xn = 32

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 33

(1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn Sub x = ? Example 5 (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn Sub x = ? [Aiming to prove: ] 34

This will make the x’s disappear and make the LHS become 2n Example 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 35

By considering that (1 + x)2n = (1 + x)n(1 + x)n Example 6 By considering that (1 + x)2n = (1 + x)n(1 + x)n and examining the coefficient of xn on each side, prove that 36

Example 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 ] 37

Example 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 (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 ] 38

Example 6 (1 + x)2n = (1 + x)n(1 + x)n (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 ] nCk = nCn-k 39

Example 6 (1 + x)2n = (1 + x)n(1 + x)n  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 40

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 41

Example 7 (2006 HSC, Question 2(b), 2 marks) [Need to prove ] (i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn Differentiating both sides: 42

Example 7 (2006 HSC, Question 2(b), 2 marks) [Need to prove ] (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  The general term 43

Example 7 (2006 HSC, Question 2(b), 2 marks) [Need to prove ] (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 = ? to prove 44

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 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  45

Example 8 (2008 HSC, Question 6(c), 5 marks) 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 46

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 independent of x will be 47

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 independent of x will be  48

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 independent of x will be  (ii) Given that apply the binomial theorem and the result of part (i) to find a simpler expression for 49

Example 8 (2008 HSC, Question 6(c), 5 marks) (ii) (1 + x)p = pC0 + pC1 x + pC2 x2 + ... + pCp xp  50

Example 8 (2008 HSC, Question 6(c), 5 marks) (ii) If we expanded the RHS, there would be many terms Terms independent of x 51

Example 8 (2008 HSC, Question 6(c), 5 marks) (ii) If we expanded the RHS, there would be many terms Terms independent of x = pC0 qC0 + pC1 x qC1 + pC2 x2 qC2 + ... + pCp xp qCp = 1 + pC1 qC1 + pC2 qC2 + ... + pCp qCp  p ≤ q 52

Example 8 (2008 HSC, Question 6(c), 5 marks) 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  From (i) 53

A very hard identity to prove: And now ... A very hard identity to prove: Example 9 from 2002 HSC Question 7(b), 6 marks 54

Example 9 (2002 HSC, Question 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 55

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 ... [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 56

[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 ] 57

[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] To prove result, we must add c0 + c1 + c2 + ... + cn To get this, sub x = ? into [1] above: [Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 ] 58

Now add [2] and [3] to prove result 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 get this, 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 ] Now add [2] and [3] to prove result 59

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 = c0 + c1 + c1 + 2c2 + c2 + ... + ncn + cn 2n-1 (n + 2) = c0 + 2c1 + 3c2 + ... + (n + 1)cn  [Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 ] Result proved: each coefficient of ck increased by 1 as required 60

Example 9 (2002 HSC, Question 7(b), 6 marks) (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] 61

Don’t forget the constant of integration Example 9 (2002 HSC, Question 7(b), 6 marks) (ii) (1 + x)n = c0 + c1 x + c2 x2 + ... + cn xn [1] Integrate both sides: To find k, sub x = 0: [Aiming to find ] Don’t forget the constant of integration  62

Example 9 (2002 HSC, Question 7(b), 6 marks) (ii) Integrate again to get 1.2, 2.3, 3.4 denominators: To find d, sub x = 0: [Aiming to find ]  63

Example 9 (2002 HSC, Question 7(b), 6 marks) (ii) Sub x = ? for –/+ pattern: [Aiming to find ]  64

Example 9 (2002 HSC, Question 7(b), 6 marks) (ii) Sub x = -1 for –/+ pattern: [Aiming to find ] 65

Example 9 (2002 HSC, Question 7(b), 6 marks) (ii)  [Aiming to find ] 66

Example 10 (2007 HSC, Question 4(a), 4 marks) In a large city, 10% of the population has green eyes. (i) What is the probability that two randomly chosen people have green eyes? (ii) What is the probability that exactly two of a group of 20 randomly chosen people have green eyes? Give your answer to three decimal places. (iii) What is the probability that more than two of a group of 20 randomly chosen people have green eyes? Give your answer to two decimal places. 67

Example 10 (2007 HSC, Question 4(a), 4 marks) (i) P(both green) = 0.1  0.1 = 0.01  68

Example 10 (2007 HSC, Question 4(a), 4 marks) (i) P(both green) = 0.1  0.1 = 0.01  (ii) What is the probability that exactly two of a group of 20 randomly chosen people have green eyes? Give your answer to three decimal places. 69

Example 10 (2007 HSC, Question 4(a), 4 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) = 70

Example 10 (2007 HSC, Question 4(a), 4 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  71

Example 10 (2007 HSC, Question 4(a), 4 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) What is the probability that more than two of a group of 20 randomly chosen people have green eyes? Give your answer to two decimal places. 72

Example 10 (2007 HSC, Question 4(a), 4 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  73

WORK HARD AND BEST OF LUCK 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! robert.yen@cengage.com 74