Cambridge International Mathematics

ANSWERS 733

11 fHHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH,

THTH, TTHH, THHT, TTTH, TTHT, THTT, HTTT, TTTTg

a 161 b^38 c 165 d^1516 e^14

12 a

bi 131 ii 521 iii^14 iv 133 v 261 vi^12

vii 134 viii 0

13 a^1114 b 285 c 281

EXERCISE 25G.1

1a 101 b^15 2a 143 b 214

3a 218 b^17 c^27 4a 0 : 0441 b 0 : 6241

5a 152 b^25 c^15 d 154

6a i 0 : 405 ii 0 : 595 b 0 : 164 c 0 : 354

EXERCISE 25G.2

1a i 145 ii^1556 iii^1556 iv 283

bBecause these 4 events are the only possible outcomes. One

of them must occur.

2a i^13 ii 154 iii 158

bThe possibilities are: WW, WY, YW, YY.

The 3 events do not cover all these possibilities.

So, the probability sum should not be 1.

3a 141 b 561 c 281

EXERCISE 25H

1a b 254

c 251

d^1625

e^1625

2a i^59 ii^49

bci^2581

ii^1681

iii^2081

iv^4081

3a b^1180

40 : 034 5 2360 6a^2342 b^1942

EXERCISE 25I

1a^49 b^19

c^29 d^49

2a^27 b^17 c^27 d^47

3a^16 b 185 c 185 d 185

These cases cover all possibilities, so their probabilities must add

up to 1.

4a b 335

c 227

5a b 17128

c 17155

6a 251 b^2425 c 49757 ¼ 0 :001 41

d^45844975 ¼ 0 : 921 e 1998

EXERCISE 25J

1aAandB,AandD,AandE,AandF,BandD,

BandF,CandD

bi^23 ii^56 iii^23 iv^12 v 1 vi^56

2a 175 b 178 c 176 3 154

4a b i 121

ii 127

cP(H)+P(5)¡P(H and5) = 126 + 122 ¡ 121

= 127 =P(Hor5)

5a b i 181

ii^59

cP(3) +P(4)¡P(3and4) =^1136 +^1136 ¡ 362

=^59 =P(3or4)

suit

A

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2 34 56 78910 JQK

card value

1st spin 2nd spin

B Y G B Y G B Y G

B

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win

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muddy

not

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lose

lose

Qr_

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tile 1 tile 2

G

G

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B

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1st ticket 2nd ticket

R

R

R

Y

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1st egg 2nd egg

S

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D

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qJ_w_

qG_w_

qH_q_

qG_q_

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1st chocolate 2nd chocolate

H

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3 465

coin

die

die 1

die 2

1

1

2

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3

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4

4

5

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6

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IB MYP_3 ANS

cyan magenta yellow black

(^05255075950525507595)
100 100
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Y:\HAESE\IGCSE01\IG01_an\733IB_IGC1_an.CDR Thursday, 20 November 2008 10:42:15 AM PETER