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1Active Maths 2 (Strands 1–5): Ch 3 Solutions

Chapter 3 Exercise 3.1

Q. 1. 5 × 8 × 4 = 160

Q. 2. (i) Impossible Christmas day is on the 10th

November (ii) Certain The sun rises in the East (iii) Likely Rolling a number greater than 1 on a die (iv) Unlikely Winning lotto (v) Even Rolling an odd number on a die

Q. 3. (i) 2 × 1 = 2 (ii) 3 × 2 × 1 = 6 (iii) 4 × 3 × 2 × 1 = 24 (iv) 5 × 4 × 3 × 2 × 1 = 120

Q. 4. (i) 6 × 5 × 4 = 120 (ii) 6 × 5 × 4 × 3 = 360 (iii) 6 × 5 × 4 × 3 × 2 × 1 = 720 (iv) 6 × 5 = 30

Q. 5. (i) 10 × 9 × 8 × 7 = 5,040 (ii) 10 × 10 × 10 × 10 = 10,000 (iii) 9 × 8 × 7 × 4 = 2,016

Q. 6. 26 = 64

Q. 7. 26 × 10 × 26 = 6,760

Q. 8. 5 × 5 × 5 × 2 × 2 = 500 e.g. 1, 3, 3, H, H

Q. 9. (i) 1 {1, 2, 3, 4, 5, 6}

(ii) 1 __ 6 1 out of 6 faces

(iii) 1 __ 2 {2, 4, 6}

(iv) 1 __ 6 {6}

0 1

(i)(iii)(ii)

(iv)

1–6

1–2

Q. 10. (i) 35 ____ 100 = 35%

(ii) 100% − 35% = 65%

Q. 11. 15,000

_______ 55,000 = 3 ___ 11

Q. 12. 6 ___ 50 ≈ 0.12

The die could be unfair as a relative frequency close to 1 __ 6 is expected but more trials are needed.

Q. 13. (i) 430 ______ 1,000 = 43%

(ii) 100% − 43% = 57%

After 1,000 flips, the relative frequency should be close to 50%. The coin appears to be biased.

Q. 14. (i) 12 ___ 18 = 66 2 __ 3 %

(ii) 66 2 __ 3 % of 24 = 16

Q. 15. (i) 26 ____ 167 × 100 ____ 1 = 16%

(ii) 3 ___ 24 × 100 ____ 1 = 13%

Yes, in the short term, Myles’ punctuality has improved. However, to compare properly, we would need to see his attendance over the full year.

Q. 16. (i) No. of rolls 20 40 60 80 100

No. of ones 1 5 14 15 24

R.F. 1 ___ 20 or 5% 5 ___ 40 or 12.5% 7 ___ 30 or 23 1 __ 3 % 3 ___ 16 or 18 3 __ 4 % 6 ___ 25 or 24%

(ii) 100 rolls. The result is more accurate, the more times the test is carried out. (iii) Yes, as the relative frequency should be tending to 16%.

2 Active Maths 2 (Strands 1–5): Ch 3 Solutions

Q. 17. 0.02% of 25,000

= 0.0002 × 25,000 = 5 5 parts should have a fault.

Q. 18. (i) 12 ___ 23 (12 + 11 = 23 students)

(ii) 11 ___ 23

Q. 19. 5 + 4 + 9 = 18 marbles

(i) 9 ___ 18 = 1 __ 2

(ii) 4 ___ 18 = 2 __ 9

(iii) 1 __ 2 + 2 __ 9 = 13 ___ 18

(iv) P(red) = 5 ___ 18

P(not red) = 1 − 5 ___ 18

= 13 ___ 18

(v) 5 ___ 18 + 1 __ 2 = 14 ___ 18 = 7 __ 9

Q. 20. (i) P(2) = 1 __ 6

(ii) P(7) = 0

(iii) P(odd) = 1 __ 2

(iv) P(prime) = 1 __ 2 as 2, 3, 5 are the prime numbers from 1 to 6

(v) P(<4) = 1 __ 2 {1, 2, 3}

(vi) P(even and prime) = 1 __ 6 {2}

Q. 21. 13 letters in total, 5 vowels

(i) P(vowel) = 5 ___ 13

(ii) P(E) = 2 ___ 13

(iii) P(P) = 2 ___ 13

(iv) P(not P) = 1 − 2 ___ 13

= 11 ___ 13

Q. 22. 52 playing cards in a pack

(i) P(red) = 26 ___ 52 = 1 __ 2

(ii) P(black) = 26 ___ 52 = 1 __ 2

(iii) P(red or black) = 1 __ 2 + 1 __ 2 = 1

(iv) P(queen and black)

Queen of spades and queen of clubs are black.

∴ 2 ___ 52 = 1 ___ 26

(v) P(queen or black)

2 red queens and 26 black cards

∴ 28 ___ 52 = 7 ___ 13

(vi) P(picture)

Jack, queen, king are picture cards.

∴ 3 × 4 ______ 52 = 12 ___ 52 = 3 ___ 13

Q. 23. (i) {4, 5, 6}

3 __ 6 = 1 __ 2

(ii) P(5 or 6) = 1 __ 3

1 __ 3 × 60 =20 times

Q. 24. P + P + 1.5P + 0.20 + 0.17 = 1

3.5P + 0.37 = 1

3.5P = 0.63

P = 0.18

∴ Q = 0.18

R = 1.5(0.18)

= 0.27

(i) Sector R.F.P 0.18

Q 0.18

R 0.27

S 0.20

T 0.17

(ii) The spinner is NOT fair as the relative frequencies are not equal.

(iii) 0.18 × 1,000 =180 times

Q. 25. (i) 1 __ 2 × 520 = 260

(ii) 12 ___ 52 × 520 = 120


(iii) 4 ___ 52 × 520 = 40

(iv) 2 ___ 52 × 520 = 20

Q. 26. 1 − (0.06 + 6 __ 7 )

= 29 ____ 350

= 8%

Q. 27. (a) (i) 20 ___ 30 = 2 __ 3

(ii) 10 ___ 30 = 1 __ 3

(iii) (10 + 20) − (4 + 5) = 21

21 ___ 30 = 7 ___ 10

(iv) 5 ___ 30 = 1 __ 6

(b) 4 ___ 30 = 2 ___ 15

(c) 15 ___ 30 × 180 = 90 days

Q. 28. (i) P(6) = 1 ___ 20

(ii) P(odd) = 10 ___ 20 = 1 __ 2

(iii) P(single digit) = 9 ___ 20

(iv) P(2, 4, 6, 8) = 4 ___ 20 = 1 __ 5

(v) P(17, 19) = 2 ___ 20 = 1 ___ 10

(vi) P(2, 3, 5, 7, 11, 13, 17, 18, 19, 20)

= 10 ___ 20 = 0.5

(vii) P(6, 12, 18) = 3 ___ 20 = 0.15

Q. 29.

(i) 9 ___ 36 = 1 __ 4 (1, 1) (3, 1) (5, 1) etc

(ii) 6 ___ 36 = 1 __ 6 (1, 1) (2, 2) etc

(iii) 1 − 1 __ 6 = 5 __ 6

(iv) odd × odd = odd

∴ P(odd, odd) = 9 ___ 36 = 1 __ 4

The answers to parts (i) and (iv) are the same because the product of two odd numbers is always odd, whereas the product of two even numbers or the product of an odd number and an even number is always even.

Q. 30.

B

Y

B

B

Y

Y

B

Y

B

Y

B

Y

B

Y

BBB

YBB

YYB

YYY

YBY

BYB

BBY

BYY

1–2

1–2

1–2

1–2

1–2

1–2

1–2

1–21

–2

1–2

1–2

1–2

1–2

1–2

(i) P(BBB) = 1 __ 2 × 1 __ 2 × 1 __ 2 = 1 __ 8

(ii) P(BYY) = 1 __ 8

(iii) P(BYY or YBY or YYB) = 3 __ 8

(iv) P(BBB or BBY or BYB or YBB) = 1 __ 2

Q. 31.

(i) P(6) = 5 ___ 36

(ii) P(even) = 18 ___ 36 = 1 __ 2

(iii) P(2, 3, 5, 7, 11) = 15 ___ 36 = 5 ___ 12

(iv) P(÷2 or ÷5) = 22 ___ 36 = 11 ___ 18

(v) P(10) = 3 ___ 36 = 1 ___ 12

(vi) P(2, 3, 6) = 8 ___ 36 = 2 __ 9

1 2 3 4 5 6

1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6

2 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6

3 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6

4 4, 1 4, 2 4, 3 4, 4 4, 5 4, 6

5 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6

6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6

Red

Bla

ck

1 2 3 4 5 61 2 3 4 5 6 72 3 4 5 6 7 83 4 5 6 7 8 94 5 6 7 8 9 105 6 7 8 9 10 116 7 8 9 10 11 12


Q. 32. (i)

Ford Honda Nissan Volvo Total

Petrol 11 7 11 13 42Diesel 17 20 12 9 58Total 28 27 23 22 100

(ii) Diesel 58 versus 42

(iii) P(Honda) = 27 ____ 100 or 27%

(iv) P(Volvo) = 22 ____ 100 = 11 ___ 50 or 22%

(v) P(Diesel) = 58 ____ 100 = 29 ___ 50 or 58%

(vi) P(Nissan Petrol) = 11 ____ 100 or 11%

(vii) P(Ford out of Diesel) = 17 ___ 58 or 29%

(viii) P(not Nissan) = 1 − 12 ___ 58 = 40 ___ 58

= 20 ___ 29 or 79%

Q. 33. 1 2 3 4 5 6

1 1 2 3 4 5 6

2 2 4 6 8 10 12

3 3 6 9 12 15 18

4 4 8 12 16 20 24

5 5 10 15 20 25 30

6 6 12 18 24 30 36

(i) P(15) = 2 ___ 36 = 1 ___ 18

(ii) P(< 9) = 16 ___ 36 = 4 __ 9

(iii) P(prime) = 6 ___ 36 = 1 __ 6

(iv) P(even or ÷3) = 32 ___ 36 = 8 __ 9

(v) P(even and ÷3) = 15 ___ 36 = 5 ___ 12

(vi) P(factor of 28) = P(1, 2, 4, 7, 14, 28)

= 6 ___ 36

= 1 __ 6

Q. 34.

B

G

R R

R

BB

RG

RR

GR

GG

GB

BG

BR

RB

B

G

R

B

G

B

G

(i) P(GG) = 1 __ 9

(ii) P(one G) = 4 __ 9

(iii) P(R at least once) = 5 __ 9

(iv) P(RR or BB or GG) = 3 __ 9 = 1 __ 3

(v) P(diff colour) = 1 − 1 __ 3 = 2 __ 3

Q. 35. Spain France Italy UK USA Total

Boy 16 8 5 14 17 60

Girl 12 18 10 10 15 65

Total 28 26 15 24 32 125

(i) P(Boy) = 60 ____ 125 = 12 ___ 25

(ii) P(France) = 26 ____ 125

(iii) P(Girl and UK) = 10 ____ 125 = 2 ___ 25

(iv) P (not Italy) = 1 − 15 ____ 125 = 22 ___ 25


(v) P(Italy out of Boys) = 5 ___ 60 = 1 ___ 12

(vi) P(not France from Boys) = 1 − 8 ___ 60

= 13 ___ 15

(vii) P(Europe from Girls)

= 1 − P(USA Girls)

= 1 − 15 ___ 65

= 10 ___ 13

Q. 36. (a) 8 0 2 3 4 7 89 0 2 3 5 5 6 9

10 0 1 1 1 2 3 3 5 5 6 9 911 1 4 5 8 8

(b) (i) P(>90 g) = 23 ___ 30

(ii) P(<100 g) = 13 ___ 30

(iii) P(85 g < x < 105 g) = 16 ___ 30 = 8 ___ 15

(c) P(>110) = 5 ___ 30

5 ___ 30 × 30 × 100 = 500 apples

Q. 37. (i) Combined score 2 3 4 5 6 7 8 9 10 11 12

Frequency 4 6 4 1 5 3 4 2 3 1 3

(ii) P(6) = 5 ___ 36

(iii)

1

2 3 4 5

Combined scores

Freq

uen

cy

6 7 8 9 10 11 12

2

3

5

6

4

(iv) 1 2 3 4 5 61 2 3 4 5 6 72 3 4 5 6 7 83 4 5 6 7 8 94 5 6 7 8 9 105 6 7 8 9 10 116 7 8 9 10 11 12

(v) P(6) = 5 ___ 36

(vi)

1

2 3 4 5

Theoretical combined scores

Freq

uen

cy

6 7 8 9 10 11 12

2

3

5

6

4

(vii) Experimental probability involves the use of relative frequencies to determine the probability of an outcome; theoretical probability does not.

(viii) Do many more rolls of the dice.

(ix) P(A wins) = 12 ___ 36 = 1 __ 3

P(B wins) = 24 ___ 36 = 2 __ 3

No, as B has twice as much chance of winning as A.


Q. 38. (a) 72% of 2,700 = 1,944 (Total: Fiction)

2,700 − (1,944 + 176) = 580 (Total: Non-fiction)

55% of 620 = 341 NF

Hardback 25% of 620 = 155 C

20% of 620 = 124 F

176 − 155 = 21 C

Softback 580 − 341 = 239 NF

1,944 − 124 = 1,820 F

2,080 Total softback

2,700 − 2,080 = 620 Total hardback (used as check)

Version

CategoryHardback Softback Totals

Fiction 124 1,820 1,944

Non-fi ction 341 239 580

Classics 155 21 176

Totals 620 2,080 2,700

(b) (i) P(NF SB) = 239 ______ 2,700

(ii) P(NF or HB) = 239 + 620 __________ 2,700 (NF softback + Total hardbacks)

= 859 ______ 2,700

(iii) P(SB F) = 1,820

______ 2,080 Fiction _____________ Total softbacks

= 7 __ 8

(c) P(HB C) = 155 ______ 2,700 × 200 = 11 13 ___ 27

∴ 11 hardback classics

Q. 39. (i) P(green) = 20 − (3 + 8)

____________ 20

= 9 ___ 20

(ii) P(not green) = 1 − 9 ___ 20

= 11 ___ 20

(iii) P(yellows) increased by 50%

8 ___ 20 × 1.5 = 0.6

8 + x ______ 20 + x = 0.6

8 + x = 0.6 (20 + x)

8 + x = 12 + 0.6x

0.4x = 4 x = 10 10 marbles

20 − (3 + 8)


Q. 40. (i), (iii), (v) Male and female have the same probability of occurring (50% or 0.5)

Outcomes

M

F

MM {M, M, M}

F

F

(iii) 2nd (v) 3rd(i) 1st

{M, M, F}{M, F, M}

{M, F, F}{F, M, M}

{F, M, F}{F, F, M}

{F, F, F}

M

F

MM

F

F

{M, M}

{M, F}

{F, M}

{F, F}M

F

(ii) 1 __ 2 = 0.5

(iv) 2 __ 4 = 1 __ 2 = 0.5

(vi) 4 __ 8 = 1 __ 2 = 0.5

(vii) 1 __ 8 = 0.125

(viii) 1 __ 8 = 0.125

(ix) 3 __ 8 = 0.375

(x) 3 __ 8 = 0.375

Exercise 3.2

Q. 1. P(B, W) = 3 __ 4 × 1 __ 4 = 3 ___ 16

Q. 2. (i) P(late) = 1 − 0.7 = 0.3

(ii) P(late, late) = 0.3 × 0.3 = 0.09

Q. 3. P(20, 20) = 3 __ 8 × 3 __ 8 = 9 ___ 64

Q. 4. P(black, red) = 3 __ 7 × 4 __ 7 = 12 ___ 49

Q. 5. P(T, T) = 0.2 × 0.2= 0.04

Q. 6. P(red, black) = 1 __ 5 × 2 ___ 10 = 1 ___ 25

Q. 7. P(lose, win) = (1 − 0.6) × 0.4 = 0.16

Q. 8. 3, 6, 9 are divisible by 3.

∴ P(both ÷ 3) = 3 __ 9 × 3 __ 9 = 1 __ 9

Q. 9.

B

G

B

G

B

G

B, B

B, G

G, B

G, G

7—10

3—10

7—10

3—10

7—10

3—10

(i) P(B, B) = 7 ___ 10 × 7 ___ 10 = 49 ____ 100

(ii) P(G, B) = 3 ___ 10 × 7 ___ 10 = 21 ____ 100

(iii) P(G, B) + P(B, G)

= 3 ___ 10 × 7 ___ 10 + 7 ___ 10 × 3 ___ 10

= 42 ____ 100 = 21 ___ 50

Q. 10. (i) P(G, B) = 15 ___ 25 × 10 ___ 25 = 6 ___ 25

(ii) P(B, G) = 10 ___ 25 × 15 ___ 25 = 6 ___ 25

(iii) P(B, B) = 10 ___ 25 × 10 ___ 25 = 4 ___ 25

(iv) P(B, B or G, G)

= 4 ___ 25 + 15 ___ 25 × 15 ___ 25

= 4 ___ 25 + 9 ___ 25 = 13 ___ 25

(v) P(B, G or G, B)

= 6 ___ 25 + 6 ___ 25 = 12 ___ 25

Q. 11. (i) P(10, 10) = 8 ___ 22 × 8 ___ 22 = 16 ____ 121

(ii) P(50, 50) = 14 ___ 22 × 14 ___ 22 = 49 ____ 121

(iii) P(10, 10 or 50, 50)

= 16 ____ 121 + 49 ____ 121 = 65 ____ 121


(iv) P(10, 50 or 50, 10) = 1 − P(same)

= 1 − 65 ____ 121 = 56 ____ 121

or 8 ___ 22 × 14 ___ 22 + 14 ___ 22 × 8 ___ 22

= 28 ____ 121 + 28 ____ 121 = 56 ____ 121

(v) P(≥60c worth)

= 1 − P(10,10)

= 1 − 16 ____ 121 = 105 ____ 121

Q. 12.

R

N

R

N

R

N

R,R

0.40.4

0.6

0.6

0.4

0.6

R,N

N,R

N,N

0.24

0.16

0.36

0.24

1+

Mon Tue Outcomes Prob.

(i) P(N, N) = 0.24

(ii) P(at least 1 rain) = 1 − P(N,N)

= 1 − 0.24

= 0.76

(iii) P(N,R or R,N) = 0.36 + 0.16

= 0.52

Q. 13. (i) P(W) = 1 __ 2

(ii) P(L,W ) = 1 __ 2 × 1 __ 2 = 1 __ 4

(iii) P(L,L,W ) = 1 __ 2 × 1 __ 2 × 1 __ 2 = 1 __ 8

P(L,L,L) = 1 __ 2 × 1 __ 2 × 1 __ 2 = 1 __ 8

Q. 14. G – Good F – Faulty

G

F

G

F

G

F

G,G

G,F

F,G

F,F

0.95

0.95

0.050.95

0.05

0.05

(i) P(F, F) = 0.05 × 0.05 = 0.0025

(ii) P(G, G) = 0.95 × 0.95 = 0.9025

(iii) P(at least 1 F) = 1 − P(G, G)

= 1 − 0.9025

= 0.0975

Q. 15. (i) P(W) = 1 __ 6

(ii) P(L,W) = 5 __ 6 × 1 __ 6 = 5 ___ 36

(iii) P(L,L,W) = 5 __ 6 × 5 __ 6 × 1 __ 6 = 25 ____ 216

P(L,L,L) = 5 __ 6 × 5 __ 6 × 5 __ 6 = 125 ____ 216

Q. 16.

R

B

R

G

R

G

R,R

R,G

B,R

B,G

1+

Outcomes Prob.

1—3

5—7

10—21

2—21

4—21

5—7

5—21

1—3

× =

2—7

2—3

1—3

2—3

(i) P(two same colour) = P(R,R)

= 5 ___ 21

(ii) P(two diff. colour) = 1 − P(R,R)

= 1 − 5 ___ 21

= 16 ___ 21

The game is not fair as the probabilities are not equal as seen in the above questions.

Q. 17. (i) WRU – Watches R Us

BW – Best Watches

WRU

BW

F

G

F

G

WRU,F

0.70.98

0.3

0.02

0.95

0.05

WRU,G

BW,F

BW,G

0.7 × 0.02 = 0.014

0.686

0.015

0.285

1+

Outcomes Prob.


(ii) P(not faulty) = P(WRU,G and BW, G) = 0.686 + 0.285 = 0.971Q. 18. (i)

C

W

L

NL

L

NL

C,L

0.150.7

0.85

0.3

0.89

0.11

C,NL

W,L

W,NL

0.045

0.105

0.0935

0.7565

1+

Outcomes Prob.

C–car, W–walk, L–late, NL–not late

(ii) P(C,L) = 0.15 × 0.3 = 0.045

(iii) P(not late) = P(C,NL) + P(W,NL)

= 0.105 + 0.7565 = 0.8615

(iv) On time

0.105 0.09350.7565

Walks

0.7565 _______ 0.8615 = 0.8781

Q. 19. (i)

R

D

T

N

T

N

R,T

R,N

D,T

D,N

1

+

Outcomes Prob.

1—5

2—3

2—3

2—15

1—5

× =

2—3

8—15

4—5

× =

1—3

1—4

3—4

× =

1—3

1–—12

1—4

× =

1—3

4—5

3—4

1—4

———

R - Rain

D - Dry

T - Top of mountain

N - Not reached top of mountain

(ii) P(top) = P(R,T) + P(D,T)

= 2 ___ 15 + 1 __ 4

= 23 ___ 60

(iii) Dry

1—12

2—15

1—4

Summiting

1 __ 4 ___

23 ___ 60 = 15 ___ 23


Q. 20.

K

M

K

M

K

M

K,K

K,M

M,K

M,M

Outcomes Prob.

1—2

1—3

1—3

1—6

1—2

× =

1—3

1—6

1—2

× =

2—3

2—15

1—5

× =

2—3

4—5

× =

2—3

1—2

1—5

4—5

1

+

8–—15

———

K - Knocks down coconut

M - Misses coconut

(i) P(one coconut) = P(K,M) + P(M,K)

= 1 __ 6 + 2 ___ 15

= 3 ___ 10

(ii) P(both coconuts) = P(K,K) = 1 __ 6

(iii) P(atleast one coconut) = 1 − P(missing both times)

= 1 − 8 ___ 15

= 7 ___ 15

(iv) P(no coconuts) = P(M,M) = 8 ___ 15

Exercise 3.3

Q. 1. #U = 5 + 7 + 8 + 10 = 30

∴ 30 teenagers in the group

(i) P(tea) = 5 + 7 ______ 30 = 12 ___ 30 = 2 __ 5

(ii) P(coffee) = 7 + 8 ______ 30 = 15 ___ 30 = 1 __ 2

(iii) P(tea + coffee) = 7 ___ 30

(iv) P(neither tea nor coffee) = 10 ___ 30

= 1 __ 3

Q. 2. #U = 225 + 50 + 100 + 0 = 375

∴ 375 students in the college

(i) P(zoology) = 225 + 50 _________ 375

= 275 ____ 375 = 11 ___ 15

(ii) P(botany) = 50 + 100 _________ 375

= 150 ____ 375 = 2 __ 5

(iii) P(zoology + botany) = 50 ____ 375 = 2 ___ 15

Q. 3. (a) #U = 4 + 3 + 5 + 1 = 13

(i) P(R ∪ Q) = 4 + 3 + 5 _________ 13 = 12 ___ 13

(ii) P(R ∩ Q) = 3 ___ 13

(iii) P(R \ Q) = 4 ___ 13

(iv) P(R ∩ Q)’ = 1 ___ 13

(b) #U = 1 + 5 + 7 = 13

(i) P(R ∪ Q) = 1 + 7 ______ 13 = 8 ___ 13

(ii) P(R ∩ Q) = 0 ___ 13

(iii) P(R \ Q) = 1 ___ 13

(iv) P(R ∪ Q)’ = 5 ___ 13

Q. 4. #U = 14 + 3 + 23 + 5

= 45 (45 customers)

(i) P(brand X) = 14 + 3 _______ 45 = 17 ___ 45

(ii) P(did not buy brand X)

= 23 + 5 _______ 45 = 28 ___ 45

( also 1 − 17 ___ 45 = 28 ___ 45 )


(iii) P(bought neither brand)

= 5 ___ 45 = 1 __ 9

Q. 5. #U = 10 + 30 + 12 + 8 = 60

60 students asked

(i) P(Strand 1) = 10 + 30 ________ 60 = 40 ___ 60 = 2 __ 3

(ii) P(Strand 1 and 2) = 30 ___ 60 = 1 __ 2

(iii) P(Strand 1 or 2) = 10 + 30 + 12 _____________ 60

= 52 ___ 60

P(Strand 1 or 2) = 13 ___ 15

(iv) P(neither strand) = 8 ___ 50 = 2 ___ 25

Q. 6. (a)

5

U(12)

A(6) B(5)

42 1

Total: 6 + 5 + 5 = 16

but #U = 12

Liked A and B = 16 − 12 = 4

(4 people double counted)

Liked A only = 6 − 4 = 2

Liked B only = 5 − 4 = 1

(i) Number who like A or B

= 2 + 4 + 1 = 7

(ii) Number who like A only = 2

(iii) Number who liked one brand only = 2 + 1 = 3

(iv) P(like A or B) = 7 ___ 12

(v) P(brand A only) = 2 ___ 12 = 1 __ 6

(vi) P(one brand only) = 3 ___ 12 = 1 __ 4

Q. 7. (i) U(1)

B(0.4) W(0.7)

0.10.3 0.6

Probabilities must add up to 1

P(black) = 0.4 given

P(black + white) = 0.1 given

∴ P(black only) = 0.4 − 0.1 = 0.3

∴ P(white only) = 1 − 0.3 − 0.1

= 0.6

(ii) P(white only) = 0.6

Q. 8. (i)

0.2

U(1)

S(0.4) R(0.5)

0.10.4 – 0.1

= 0.3

0.5 – 0.1

= 0.4


0.3 + 0.1 + 0.4 = 0.8

∴ 1 − 0.8 = 0.2 missing

(ii) P(not picked for either team)

= 0.2

Q. 9. (i) Q1(0.5)

0.5 – x= 0.5 – 0.3

= 0.2

0.6 – x= 0.6 – 0.3

= 0.3

0.2

x= 0.3

Q2(0.6)

U(1)


∴ 0.5 − x + x + 0.6 − x + 0.2 = 1

1.3 − x = 1

0.3 = x

(ii) P(both Q correct) = 0.3

(iii) P(1st Q only) = 0.2

x x


Q. 10.

C(35)

35 – x= 35 – 5

= 30

50 – x= 50 – 5

= 45

20

x= 5

R(50)

U(100)

35 − x + x + 50 − x + 20 = 100

105 − x = 100

∴ 5 = x

(i) P(rock music) = 50 ____ 100 = 1 __ 2

(ii) P(classical only) = 30 ____ 100 = 3 ___ 10

(iii) P(rock and classical) = 5 ____ 100 = 1 ___ 20

(iv) P(rock or classical) = 80 ____ 100 = 4 __ 5

Q. 11.

13 20

3

14

H(27) F(34)

U(50)

(i) P(H) = 27 ___ 50 = 0.54

(ii) P(F) = 34 ___ 50 = 17 ___ 25 = 0.68

(iii) P(both football + hurling) = 14 ___ 50

= 7 ___ 25

(iv) P(hurling only) = 13 ___ 50

Q. 12. #U = 10 + 1 + 1 + 5 + 6 + 3 + 2 + 2 = 30

(i) P(comedy) = 10 + 1 + 1 + 3 ______________ 30

= 15 ___ 30 = 1 __ 2

(ii) P(drama only) = 5 ___ 30 = 1 __ 6

(iii) P(all 3 types) = 1 ___ 30

(iv) P(comedy + horror) = 3 + 1 ______ 30

= 4 ___ 30 = 2 ___ 15

(v) P(comedy + horror only) = 3 ___ 30

= 1 ___ 10

(vi) P(none of film types) = 2 ___ 30

= 1 ___ 15

Q. 13. #U = 11 + 8 + 6 + 2 + 1 + 5 = 33

(i) P(paper P) = 11 + 8 + 2 + 1 ______________ 33

= 22 ___ 33 = 2 __ 3

(ii) P(Q only) = 6 ___ 33 = 2 ___ 11

(iii) P(Q and R) = 0 + 2 ______ 33 = 2 ___ 33

(iv) P(Q or R) = 8 + 6 + 2 + 1 + 5 _________________ 33

= 22 ___ 33 = 2 __ 3

Q. 14. (i) B(22)

U(85)

C(19)

4

19 – 5 – 7 – 3

= 4

7

72 – 10

– 7 – 5

= 50

22 – 10

– 7 – 3

= 2

10 – 7

= 3

17 – 7

= 10

12 – 7

= 5

A(72)

A = Amazing Value B = Best Bargains C = Crazy Value

(ii) P(A only) = 50 ___ 85 = 10 ___ 17

(iii) P(A and B only) = 10 ___ 85 = 2 ___ 17

(iv) P(any of these stores)

= 50 + 10 + 7 + 5 + 2 + 3 + 4 ___________________________ 85

= 81 ___ 85

(also found by 1 − 4 ___ 85 = 81 ___ 85 , i.e. 1 − probability of not shopping at any of these stores)

(v) P(none of these stores) = 4 ___ 85


Q. 15. (i) F(19)

U(45)

G(13)

13

116

13 – 1 – x

= 12 – x

= 12 – 8

= 4

13 – 4 – 1

= 8

19 – 4 – x

= 15 – x

= 15 – 8

= 7

x = 8

4

1

E(13)

8 + 4 + 15 − x + 1 + x + 12 − x + 13 = 45

∴ 53 − x = 45

8 = x

(ii) P(at least one country)

= 8 + 4 + 7 + 8 + 4 + 1 _____________________ 45

= 32 ___ 45

(iii) P(only one country) = 8 + 7 + 4 _________ 45

= 19 ___ 45

(iv) P(two countries) = 4 + 8 + 1 _________ 45

= 13 ___ 45

Q. 16. (a) T(20)

U(100)

V(60)

x = 12

116

60 – 15 – 10 – 2

= 33

50 – 5

– 10 – 15

= 20

20 – 5

– 10 – 2

= 3

2

5

10

15

C(50)

(b) #U = 20 + 5 + 10 + 15 + 3 + 2 + 33 + x

= 88 + x but given #U = 100 ∴ x + 88 = 100 x = 12 ∴ P(no instrument) = 12 ____ 100 = 3 ___ 25

(c) P(only one instrument)

= 20 + 3 + 33 ____________ 100 = 56 ____ 100 = 14 ___ 25

Q. 17. #(A ∪ B ∪ C) = 28

#(A) = 10, #(B) = 18, #(C) = 18

#(A ∩ C) = 5, #(B ∩ C) = 9

and #(A ∩ B ∩ C) = 2

B(18)

U(28)

C(18)

116

18 – 3 – 2 – 7

= 6

10 – 3 – 2 – x

= 5 – x

= 5 – 4

= 1

18 – 7 – 2 – x

= 9 – x

= 9 – 4

= 5

9 – 2

= 7

x = 5

25 – 2

= 3

A(10)

5 − x + x + 9 − x + 3 + 2 + 7 + 6 = 28

32 − x = 28 4 = x (i) P[(A ∩ B) \ C] = 4 ___ 28 = 1 __ 7

(ii) P[(A ∪ B) \ C] = 1 + 4 + 5 _________ 28

= 10 ___ 28 = 5 ___ 14

(iii) P(B’) = 1 + 3 + 6 _________ 28

= 10 ___ 28 = 5 ___ 14

Revision ExercisesQ. 1. (a) (i)

Whi

te d

ie 1

st

Black die 2nd1 2 3 4 5 6

1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 62 2, 1 2, 2 2, 3 2, 4 2, 5 2, 63 3, 1 3, 2 3, 3 3, 4 3, 5 3, 64 4, 1 4, 2 4, 3 4, 4 4, 5 4, 65 5, 1 5, 2 5, 3 5, 4 5, 5 5, 66 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6

x x

14 Active Maths 2 (Strands 1–5): Ch 3 SolutionsAAAAAAAAAAActive Maths 2 ((Strands 1–5):) Ch 3 Solutions

(ii) There are 36 possible outcomes.

(iii) P(W2, B4) = 1 ___ 36

(iv) P(W6, B6) = 1 ___ 36

(v) P(W1, B odd) = 3 ___ 36 = 1 ___ 12

(vi) P(W even, B odd)

= 9 ___ 36 = 1 __ 4

(vii) P(W ≥ 4 and B ≤ 3) = 9 ___ 36 = 1 __ 4

(b) (i)

B

Y

B

Y

B

Y

B,B

B,Y

Y,B

Y,Y

Outcomes

2—7

4—7

3—7

5—7

2—7

5—7

1st pick 2nd pick

B = blue, Y = yellow

(ii) P(B, B) = 4 __ 7 × 2 __ 7 = 8 ___ 49

P(Y, Y) = 3 __ 7 × 5 __ 7 = 15 ___ 49

P(B, Y) = 4 __ 7 × 5 __ 7 = 20 ___ 49

P(B, B or Y, Y) = 8 ___ 49 + 15 ___ 49 = 23 ___ 49

(c) 3 yellow, 2 blue, 1 orange so 6 balls in total

P(Y) = 3 __ 6 , P(B) = 2 __ 6 , P(O) = 1 __ 6

∴ P(Y, B, O) = 3 __ 6 × 2 __ 6 × 1 __ 6 = 1 ___ 36

Q. 2. (a) Event Probability

A 13 ___ 52 = 1 __ 4

B 1 __ 2

C 1

D 0

E 1 __ 7

(b)

10

D E A B C

1—7

1—4

1—2

Q. 3. (a) (i) 180 different lunches

5 × 9 × d = 180

d = 4

∴ 4 desserts needed

(ii) 5 − 1 = 4 starters, 9 − 1 = 8 main courses

∴ 4 × 8 × 4 = 128 different 3 course lunches possible

(b) (i) P(AB–) = 1 ____ 100 or 0.01

(ii) P(Mary safely receives blood)

= 8 + 2 ______ 100 = 1 ___ 10 or 0.1

(iii) P(Aaron’s blood can be given)

= 47 + 2 + 26 + 9 _______________ 100 = 84 ____ 100

= 21 ___ 25 or 0.84

(iv) O– can be given to the entire population and thus is a very flexible source of blood for emergencies etc.

Q. 4. (a) Colour Frequency Relative frequency Daily frequency (v)Red 70 0.14 336Blue 100 0.2 480

Yellow 45 0.09 216White 55 0.11 264Black 90 0.18 432Silver 140 0.28 672Total 500 1.0 2,400

(i) Black cars: 500 − (70 + 100 + 45 + 55 + 140) = 500 − 410 = 90


(iii) Add all the relative frequencies together – if correct they will total to 1.

70 ____ 500 + 100 ____ 500 + 45 ____ 500 + 55 ____ 500 + 90 ____ 500 + 140 ____ 500 = 500 ____ 500 = 1

(iv) P(Red) = 70 ____ 500 = 7 ___ 50 or 0.14 (v) See table above.

(vi) No. A test is reliable if repeated runs of the test would give the same results. There is no reason to say that if this test was run again it would be different because of the sample not being random. The colour of a vehicle is random and running the test at different times of the day or on different days would not necessarily make the test more reliable.

(b) Number on die 1 2 3 4 5 6

Frequency 92 101 115 98 89 105

(i) Die tossed 600 times: 600 − (92 + 101 + 115 + 98 + 105) = 600 − 511 = 89

∴ 5 appeared 89 times.

(ii) If the die was fair (unbiased) each number would be expected to occur 1 out of 6 times overall

∴ In 600 tosses each number would be expected approximately 100 times.

In most cases the results are close enough to 100 to suggest that it is a fair die.

(iii) P(even) = 101 + 98 + 105 _______________ 600 = 304 ____ 600

∴ After 300 tosses 304 ____ 600 × 300 = 152 times

∴ Expect an even number 152 times out of 300

Q. 5. (a) Male Female TotalMarried 0.029 0.048 − 0.029 = 0.019 0.048

Unmarried 0.515 − 0.029 = 0.486 0.952 − 0.486 = 0.466 1 − 0.048 = 0.952

Total 0.515 0.485 1

From the two way table P(unmarried female) = 0.466

(b) No. of throws 10 20 30 40 50 60 70 80 90 100 110 120

No. of 4s 1 5 8 10 10 12 13 13 15 15 16 18

Relative freq. 1 ___ 10 4 ___ 10 3 ___ 10 2 ___ 10 0 ___ 10 2 ___ 10 1 ___ 10 0 ___ 10 2 ___ 10 0 ___ 10 1 ___ 10 2 ___ 10

= 0.1 = 0.4 0.3 = 0.2 = 0 = 0.2 = 0.1 = 0 = 0.2 = 0 = 0.1 = 0.2

(ii)

010 20 30 40 50 60 70 80 90 100 110 120

0.1

0.2

0.3

0.4

0.5

Rela

tive f

req

uen

cy

No. of throws


(iii) The graph shows that as the number of throws increases the relative frequency approaches its theoretical probability of 0.16.

Q. 6. (a)

R

W

G

R

R

W

R

W

R

W

R

W

R

G

R,G,R

W,R,R

W,G,R

W,G,W

W,R,W

R,R,R

R,G,W

R,R,W

1–2

1–21–

2

1–2

1–2

1–2

1–2

1–2

1–2

1–2

1–2

1–2

1–2

1–2

Outcomes3rd pick2nd pick1st pick

(i) P(3 balls same colour)

= 1 __ 2 R

× 1 __ 2 R

× 1 __ 2 R

= 1 __ 8

(ii) P(3 balls different colour)

= 1 __ 2 R

× 1 __ 2 G

× 1 ___ 2 W

+ 1 ___ 2 W

× 1 __ 2 G

× 1 __ 2 R

= 1 __ 8 + 1 __ 8 = 1 __ 4

(iii) P(1st 2 balls same colour) = 1 __ 2 R × 1 __

2 R

= 1 __ 4

(iv) P(at least one red ball chosen) = 7 __ 8

(b) (i) P

F

P

F

P

F,P

F,F

Outcomes

0.4

0.40.6

0.6

1st test 2nd test

(ii) P(pass on 1st attempt) = 0.4

P(fail on both attempts) = 0.6 × 0.6 = 0.36

P(pass on second attempt) = 0.6 × 0.4 = 0.24

(c) (i) P(red) = 2 __ 5

(ii) P(red) = 2 ______ 5 + 5 = 2 ___ 10 = 1 __ 5

(iii) P(red) = 1 __ 3

Also P(red) = 2 + x _____ 15 after

5 more marbles added

∴ 2 + x _____ 15 = 1 __ 3

3(2 + x) = 15

2 + x = 5

∴ x = 3

3 more red marbles added

Q. 7. (a) Second die

Firs

t die

1 2 3 4 5 61 1 2 3 4 5 62 2 4 6 8 10 123 3 6 9 12 15 184 4 8 12 16 20 24

(i) P(even) = 18 ___ 24 = 3 __ 4

(ii) P(odd) = 6 ___ 24 = 1 __ 4

(iii) P(prime) = 5 ___ 24

(iv) P(<12) = 16 ___ 24 = 2 __ 3

(v) P(perfect square) = 6 ___ 24 = 1 __ 4

1, 4, 9 and 16 are the only perfect squares present

(vi) P(perfect cube) = 3 ___ 24 = 1 __ 8

1 and 8 perfect cubes present.

(b) (i) P(GGG) = 0.7 × 0.5 × 0.3

= 0.105

(ii) P(RRR) = 0.3 × 0.5 × 0.7

= 0.105


(iii) P(GRG) = 0.7 × 0.5 × 0.3

= 0.105

P(1st set G) = 0.7 then

P(1st set R) = 0.3 as both

probabilities must add to 1.

Similarly P(2nd set G) = 0.5

then P(2nd set R) = 0.5 also.

Then P(3rd set G) = 0.3

gives P(3rd set R) = 0.7.

(c)

17 4

1

3

D(20) C(7)U(25)

(i) #(C/D) = 7 – 3 = 4

(ii) # (C∪D)’ = 25 – (17 + 3 + 4) = 1

(iii) P(cat only) = 4 ___ 25 = 0.16

(iv) P(neither) = 1 ___ 25 = 0.04

Q. 8. (i) P(1) = 1 __ 2 also P(7) = 1 __ 2

∴ P(1, 1, 1) = 1 __ 2 × 1 __ 2 × 1 __ 2 = 1 __ 8

(ii) P(same number each time)

= P(1, 1, 1) OR P(7, 7, 7)

= 1 __ 8 + 1 __ 8 = 1 __ 4

(b) (i) Two possible sets of countries a journalist could have chosen to interview:

(1) Italy, Iceland, Norway

(2) Spain, Cyprus, Turkey (Student answers will vary.)

(ii) 6 in group A × 10 in group B × 5 in group C

= 300 different sets of countries

(iii) Turkey would be chosen more often, as it is one of only

5 countries in group C compared to Russia, which is 1 of 10 countries and thus has a lower probability of being chosen.

(c) (i) The students will need to complete a large number of trials (several hundred if possible) and record how many times each number on the die is shown. Students would then calculate the relative frequency for each number on the die by dividing the number of occurrences by the total number of trials. Students would then compare these relative frequencies to the theoretical probabilities for each outcome on the die.

(ii) Outcome 1 2 3 4 5 6Freq. 83 85 80 71 100 81

(iii) The expected frequency for rolling a 1 on a die is 0.16̇ thus a relative frequency of 0.166 appears fair. Similarly for rolling a 2, 3 and 6.

However rolling the die does seem to be biased in favour of rolling a 5 with a much higher relative frequency after 500 trials. ∴ The die is not fair.

Q. 9. (i) Colour Number Relative freq.

Red 75 75 ____ 131 = 0.573

Amber 24 24 ____ 131 = 0.183

Green 32 32 ____ 131 = 0.244

Total 131 1

(ii) Expected red = 0.573 × 250 = 143.25 Expected green = 0.244 × 250 = 61 ∴ 143 − 61 = 82 more red lights

seen


(b) (i) 2nd die

1st d

ie

1 2 3 4 5 61 2 3 4 5 6 72 3 4 5 6 7 83 4 5 6 7 8 94 5 6 7 8 9 105 6 7 8 9 10 116 7 8 9 10 11 12

(ii) P(total of 11) = 2 ___ 36 = 1 ___ 18

∴ Probability of winning €10

= 1 ___ 18

(iii) Expected to win atleast once

= 18 × 1 ___ 18 = 1

∴ After 18 goes

(iv) Expected win = 40 × 1 ___ 18 = 2.2̇

∴ 2 wins expected.

(v) P(WWW) = 1 ___ 18 × 1 ___ 18 × 1 ___ 18

= 0.00017

to 2 significant figures

(vi)

1—18

1—18

1—18

17—18

17—18

17—18

1—18

17—18

1—18

17—18

1—18

17—18

1—18

17—18

W

L

W

W

L

L

W

L

W

L

W

L

W

L

W,W,W

L,W,W

L,L,W

L,L,L

L,W,L

W,L,W

W,W,L

W,L,L

Outcomes

P(at least 1 win in 3 games)

= 1 − P(no wins)

= 1 − 17 ___ 18 × 17 ___ 18 × 17 ___ 18

= 1 − 0.842

= 0.158 to 3 significant figures

(vii) P(wins at least twice)

= 1 ___ 18 × 1 ___ 18 × 1 ___ 18

+ 1 ___ 18 × 1 ___ 18 × 17 ___ 18

+ 1 ___ 18 × 17 ___ 18 × 1 ___ 18

+ 17 ___ 18 × 1 ___ 18 × 1 ___ 18

= 0.0089 to 2 significant figures

Q. 10. (a) #U = 150

22 + 12 + 25 + 5 + 6 + 10 + 48 = 128

∴ 150 − 128 = 22 people read none of the papers

(i) (1) P(Times)

= 22 + 12 + 6 + 5 _______________ 150

= 45 ____ 150 = 3 ___ 10

(2) P(only one paper)

= 22 + 25 + 48 _____________ 150

= 45 ____ 150 = 19 ___ 30

(3) P(none of papers)

= 22 ____ 150 = 11 ___ 75

(4) P(Ind or Times or both but not Examiner)

= 22 + 12 + 25 _____________ 150 = 59 ____ 150

(ii) P(Gemma reads Times) = 11 ___ 69

(iii) (1) The number of pages in each newspaper is a discrete variable.

(2) The length of time spent reading each newspaper is a continuous variable.


(b) P(B) = 0.4

∴ x = 0.4 − (0.1 + 0.05 + 0.05)

x = 0.4 − 0.2

x = 0.2

P(C) = 0.35

∴ y = 0.35 − (0.05 + 0.05 + 0.05)

y = 0.35 − 0.15

y = 0.2

Finally U = 1

∴ z = 1 − (0.3 + 0.1 + 0.05 + 0.2 + 0.05 + 0.05 + 0.2)

z = 1 − 0.95

z = 0.05

Q. 11. (a) Dee (D) Elie (E) Fiona (F) Gemma (G) Holly (H) Isabel (I)Andy (A) AD AE AF AG AH AI

Bob (B) BD BE BF BG BH BI

Charles (C) CD CE CF CG CH CI

(i) P(B and G chosen) = 1 ___ 18

(ii) P(A) = 6 ___ 18 = 1 __ 3

(iii) P(C doesn’t get job) = 12 ___ 18 = 2 __ 3

(b)

M(150)

U(200)

T(125)

200 – [5 + 45 + 25 + 10 + 10 + 70 + 35] = 200 – 200

= 0

116

125 – 10 – 10 – 70

= 35

70 – 45

– 10 – 10

= 5

150 – 45

– 10 – 70

= 25

80 – 10

= 70

55 – 10

= 45

10

20 – 10

= 10

I(70)

(i) P(I only) = 5 ____ 200 = 1 ___ 40

(ii) P(any source but Internet) = 25 + 70 + 35 _____________ 200 = 130 ____ 200 = 13 ___ 20

(iii) P(TV and word of mouth) = 80 ____ 200 = 2 __ 5

(iv) P(word of mouth and Internet but not TV) = 45 ____ 200 = 9 ___ 40


(c) (i)

0.4

U(1)

A (0.4) B (0.3)

0.10.3 0.2

P(neither wins) = 1 − (0.3 + 0.1 + 0.2) = 1 − 0.6 = 0.4

(ii) U(1)

A (0.4) B (0.3)

0.4 0.3

0.3

P(neither wins)

= 1 − (0.4 + 0.3)

= 1 − 0.7

= 0.3

Q. 12. (i) Die

Spin

ner 1 2 3 4 5 6

1 2 3 4 5 6 73 4 5 6 7 8 97 8 9 10 11 12 13

Sophie’s idea

Money back: P(13) = 1 ___ 18

Win €1: P(even) = 9 ___ 18

Lose: P(anything else) = 1 − 10 ___ 18 = 8 ___ 18

Under Sophie’s idea

Money back: 1 __ 8 × 180 = 10

10 people get 50 out back = €5

Win €1: 9 ___ 18 × 180 = 90 people

90 people get to win €1 = €90

Amy’s idea

Money back: P(even) = 9 ___ 18

Win €1: P(odd, not prime) = 2 ___ 18

Lose: P(anything else) = 1 − 11 ___ 18 = 7 ___ 18

Under Amy’s idea

Money back: 9 ___ 18 × 180 = 90

90 people get 50 cent back = €45

Win €1: 2 ___ 18 × 180 = 20 people

20 people get to win €1 = €20

In summary:

Sophie’s idea Amy’s idea

Amount generated for charity 180 × 50c = €90 180 × 50c = €90

Win (charity loses money) − €90 − €20

Money back (charity loses money) − €5 − €45

Amount left for charity − €5 + €25

∴ Amy's idea is best, as it has on average a gain of €25 for the charity, while Sophie's idea would on average lose €5.

(ii) Any idea that results in a better return for the charity than Amy’s or Sophie’s.

Students’ answers will vary.


Q. 13. (a) Sample space for coin flipped four times:

{HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}

(i) P(HHHH) = 1 ___ 16

(ii) P(exactly 3 heads) = 4 ___ 16 = 1 __ 4

(iii) P(20 or more H) = 11 ___ 16

(iv) P(less than 2H) = 5 ___ 16

(v) P(no heads) = 1 ___ 16

(b) (i)

3

not

3

3

not

3

3

not

3

3, 3

3, not 3

not 3, 3

not 3, not 3

Outcomes

1—5

1—10

9—10

4—5

1—5

4—5

1st pick 2nd pick

(ii) P(3, 3) = 1 ___ 10 × 1 __ 5 = 1 ___ 50

(iii) No. As the probability of winning is quite low people may not be interested in buying a raffle ticket.

(iv) Under new rules:

P(3, 3) = 5 ___ 10 × 1 __ 5 = 1 ___ 10

(c)

C

S

C

S

C

S

C,C

C,S

S,C

S,S

Outcomes

3—8

4—9

5—9

5—8

4—8

4—8

1st cake 2nd cake

(i) P(CC) = 4 __ 9 × 3 __ 8 = 1 __ 6

(ii) P(SS) = 5 __ 9 × 4 __ 8 = 5 ___ 18

(iii) P(C then S) = 4 __ 9 × 5 __ 8 = 5 ___ 18

(iv) P(C and S) = 4 __ 9 × 5 __ 8 + 5 __ 9 × 4 __ 8

= 5 __ 9

Q. 14. (a) P(B) = 2 __ 3

(b) P(HR) = 1 __ 2 × 1 __ 3 = 1 __ 6

(c) P(TB) = 1 __ 2 × 2 __ 3 = 1 __ 3

Q. 15. (a) The probability of getting a 1 when a fair die is tossed is 1 __ 6 = 0.16̇

(b) Number on die 1 2 3 4 5 6

Frequency 70 82 86 90 91 81

Relative frequency 70 ____ 500 82 ____ 500 86 ____ 500 90 ____ 500 91 ____ 500 81 ____ 500

= 0.14 = 0.16 = 0.17 = 0.18 = 0.18 = 0.16

Die tossed 500 times, so 3 tossed 500 − (70 + 82 + 90 + 91 + 81) = 500 − 414 = 86

(c) Relative frequency = frequency

__________________ total number of trials

(see table above)

(d) The more trials that are done the closer the relative frequency of getting a 1 comes to the theoretical frequency. As a fair die was used for the trials, more trials need to be done to see this happen.

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