WS Fall Semester Final Review 2018 [198 marks]

1a.

Markscheme

(A1)(A1)(A1)(A1)(C4)

[4 marks]

1b.

MarkschemeIncorrect (A1)

Natural numbers are positive integers. Integers can also be negative. (or equivalent) (R1) (C2)

Note: Accept a correct justification. Do not award (R0)(A1).Accept: a statement with an example of an integer which is not natural.

[2 marks]

2.

Markscheme

(A1)(A1)(A1)(A1)(A2) (C6)

Note: Row 1 has been given in the question.Row 2 to row 5: Award (A1) for each correct row.Row 6: Award (A1) for both not selected and selected; award (A1) for both and selected. Do not penalize if crosses (orsimilar) appear in the empty cells.

N Z Q R

[4 marks]

[2 marks]

[6 marks]

3a.

Markscheme

(A1)(A1)(A1)(A1)(A1)(A1) (C6)

Note: Award (A1) for each number correctly placed.

Award (A0) for any entry in more than one region.

[1 mark]

3b.

Markscheme

(A1)(A1)(A1)(A1)(A1)(A1) (C6)

Note: Award (A1) for each number correctly placed.

Award (A0) for any entry in more than one region.

[1 mark]

[1 mark]

[1 mark]

3c.

Markscheme

(A1)(A1)(A1)(A1)(A1)(A1) (C6)

Note: Award (A1) for each number correctly placed.

Award (A0) for any entry in more than one region.

[1 mark]

3d.

Markscheme

(A1)(A1)(A1)(A1)(A1)(A1) (C6)

Note: Award (A1) for each number correctly placed.

Award (A0) for any entry in more than one region.

[1 mark]

[1 mark]

[1 mark]

3e.

Markscheme

(A1)(A1)(A1)(A1)(A1)(A1) (C6)

Note: Award (A1) for each number correctly placed.

Award (A0) for any entry in more than one region.

[1 mark]

3f.

Markscheme

(A1)(A1)(A1)(A1)(A1)(A1) (C6)

Note: Award (A1) for each number correctly placed.

Award (A0) for any entry in more than one region.

[1 mark]

4a.

Markschemer = 0.01924 (A1) (C1)

Note: Accept 0.0192 and 0.019

[1 mark]

[1 mark]

[1 mark]

[1 mark]

4b.

Markschemer = 1.924 × 10 (A1)(ft)(A1)(ft) (C2)

Notes: Award (A1) for 1.924, (A1) for 10 . Accept 1.92 and 1.9. Follow through from their part (a).

[2 marks]

−2

−2

4c.

Markscheme

(A1)(A1)(A1) (C3)

Notes: Award (A1) for each true statement circled. Do not follow through from part (a). Award (A1)(A1)(A0) if 1 extra term seen. Award(A1)(A0)(A0) if 2 extra terms seen. Award (A0)(A0)(A0) if all terms circled. Accept other indications of the correct statements i.e.highlighted or ticks.

[3 marks]

5a.

Markscheme24500 × 25 (M1)

Note: Award (M1) for multiplying 24500 by 25.

= 613000 (612500) (A1) (C2)

[2 marks]

5b.

Markscheme (M1)

Note: Award (M1) for correct substitution into the percentage error formula.

= 0.842 (0.841832) (A1)(ft) (C2)

Note: Follow through from part (a).

[2 marks]

∣∣ ∣∣ × 100612500−617700617700

[2 marks]

[3 marks]

[2 marks]

[2 marks]

5c.

Markscheme8.42 × 10 (A1)(ft)(A1)(ft) (C2)

Note: Award (A0)(A0) for answers of the type 84.2 × 10 . Follow through from part (b). Ignore ‘%’ sign.

[2 marks]

−1

−2

6a.

Markscheme (M1)

Note: Award (M1) for correct substitution into Pythagoras.

Accept correct substitution into cosine rule.

(A1)

(AG)

Note: Both the rounded and unrounded value must be seen for the (A1) to be awarded.

[2 marks]

BD2 = 402 + 842

BD = 93.0376…

= 93

6b.

Markscheme (M1)(A1)

Note: Award (M1) for substitution into cosine formula, (A1) for correct substitutions.

(A1)(G2)

[3 marks]

cosC = (932 = 1152 + 602 − 2 × 115 × 60 × cosC)1152+602−932

2×115×60

= 53.7∘ (53.6679…∘)

6c.

Markscheme (M1)(M1)(A1)(ft)

Note: Award (M1) for correct substitution into right-angle triangle area. Award (M1) for substitution into area of triangle formula and(A1)(ft) for correct substitution.

(A1)(ft)(G3)

Notes: Follow through from part (b).

[4 marks]

(40)(84) + (115)(60)sin(53.6679…)12

12

= 4460 m2 (4459.30… m2)

[2 marks]

[2 marks]

[3 marks]

[4 marks]

6d.

Markscheme (M1)

Note: Award (M1) for correct substitution in the area formula used by ‘Ancient Egyptians’.

(A1)(G2)

[2 marks]

(40+60)(84+115)

4

= 4980 m2 (4975 m2)

6e.

Markscheme (M1)

Notes: Award (M1) for correct substitution into percentage error formula.

(A1)(ft)(G2)

Notes: Follow through from parts (c) and (d)(i).

[2 marks]

∣∣ ∣∣ × 1004975−4459.30…4459.30…

= 11.6 (%) (11.5645…)

[2 marks]

[2 marks]

7a.

Markscheme (M1)(A1)

Note: Award (M1) for substitution in compound interest formula, (A1) for correct substitution.

OR

N = 5

I = 3.5

PV = 1000

P/Y = 1

C/Y = 4

Note: Award (A1) for C/Y = 4 seen, (M1) for other correct entries.

OR

N = 5 × 4

I = 3.5

PV = 1000

P/Y = 1

C/Y = 4

Note: Award (A1) for C/Y = 4 seen, (M1) for other correct entries.

= 1190.34 (USD) (A1)

Note: Award (M1) for substitution in compound interest formula, (A1) for correct substitution.

[3 marks]

1000(1 + )4×53.5

4×100

7b.

Markscheme190.34 (USD) (A1)(ft) (C4)

Note: Award (A1)(ft) for subtraction of 1000 from their part (a)(i). Follow through from (a)(i).

[1 mark]

7c.

Markscheme (M1)

Note: Award (M1) for division of 170 by their part (a)(ii).

= 0.89 (A1)(ft) (C2)

Note: Follow through from their part (a)(ii).

[2 marks]

170190.34

[3 marks]

[1 mark]

[2 marks]

8a.

Markscheme (M1)(A1)

Note: Award (M1) for substitution into the compound interest formula, (A1) for correct substitution. Award at most (M1)(A0) if notequated to 17500.

OR

N = 5

PV = ±14000

FV = 17500

P/Y = 1

C/Y = 1 (A1)(M1)

Note: Award (A1) for C/Y = 1 seen, (M1) for all other correct entries. FV and PV must have opposite signs.

= 4.56 (%) (4.56395… (%)) (A1) (G3)

[3 marks]

17500 = 14000(1 + )5r

100

∓

8b.

Markscheme14000 × 66.91 (M1)

Note: Award (M1) for multiplying 14000 by 66.91.

936740 (INR) (A1) (G2)

Note: Answer must be given to the nearest whole number.

[2 marks]

[3 marks]

[2 marks]

8c.

Markscheme (M1)(A1)(ft)

Note: Award (M1) for substitution into the compound interest formula, (A1)(ft) for their correct substitution.

OR

N = 60

I% = 5.2

PV = ±936740

P/Y= 12

C/Y= 12 (A1)(M1)

Note: Award (A1) for C/Y = 12 seen, (M1) for all other correct entries.

OR

N = 5

I% = 5.2

PV = ±936740

P/Y= 1

C/Y= 12 (A1)(M1)

Note: Award (A1) for C/Y = 12 seen, (M1) for all other correct entries

= 1214204 (INR) (A1)(ft) (G3)

Note: Follow through from part (b). Answer must be given to the nearest whole number.

[3 marks]

936740 × (1 + )12×55.2

12×100

8d.

Markscheme (M1)

Note: Award (M1) for dividing their (c) by 67.16.

(USD) (M1)(A1)(ft) (G3)

Note: Award (M1) for finding the difference between their conversion and 17500. Answer must be given to the nearest whole number.Follow through from part (c).

[3 marks]

121420467.16

( ) − 17500 = 579121420467.16

9a.

MarkschemeNote: In this question, the first time an answer is not to 2 dp the final (A1) is not awarded.

(M1)

Note: Award (M1) for dividing 350 by 0.1559.

(A1) (C2)

[2 marks]

3500.1559

= 2245.03 (ARS)

[3 marks]

[3 marks]

[2 marks]

9b.

Markscheme (M1)

Note: Award (M1) for multiplying their answer to part (a) by 1.02.

(A1)(ft) (C2)

OR

(M1)

Note: Award (M1) for multiplying their answer to part (a) by 0.02.

(A1)(ft) (C2)

Note: Follow through from part (a).

[2 marks]

2245.03 × 1.02

= 2289.93 (ARS)

2245.03 × 0.02

= 44.9006

2245.03 + 44.90

= 2289.93 (ARS)

9c.

Markscheme (M1)

Note: Award (M1) for dividing 4228.38 by 585.

(A1) (C2)

[2 marks]

4228.38585

= 7.23

10a.

Markscheme5d = 46 − 21 OR u + 2d = 21 and u + 7d = 46 (M1)

Note: Award (M1) for a correct equation in d or for two correct equations in u and d.

(d =) 5 (kg) (A1) (C2)

[2 marks]

1 1

1

[2 marks]

[2 marks]

[2 marks]

10b.

Markschemeu + 2 × 5 = 21 (M1)

OR

u + 7 × 5 = 46 (M1)

Note: Award (M1) for substitution of their d into either of the two equations.

(u =) 11 (kg) (A1)(ft) (C2)

Note: Follow through from part (a)(i).

[2 marks]

1

1

1

10c.

Markscheme (M1)

Note: Award (M1) for correct substitution into arithmetic series formula.

= 462 (kg) (A1)(ft) (C2)

Note: Follow through from parts (a) and (b).

[2 marks]

(2 × 11 + (12 − 1) × 5)122

11a.

Markscheme60 + 10 × 10 (M1)(A1)

Note: Award (M1) for substitution into the arithmetic sequence formula, (A1) for correct substitution.

= ($) 160 (A1)(G3)

[3 marks]

11b.

Markscheme (M1)(A1)(ft)

Note: Award (M1) for substituting the arithmetic series formula, (A1)(ft) for correct substitution. Follow through from their first term andcommon difference in part (a).

= ($) 1380 (A1)(ft)(G2)

[3 marks]

(2 × 60 + 11 × 10)122

[2 marks]

[2 marks]

[3 marks]

[3 marks]

11c.

Markscheme60 × 1.1 (M1)(A1)

Note: Award (M1) for substituting the geometric progression nth term formula, (A1) for correct substitution.

= ($) 156 (155.624…) (A1)(G3)

Note: Accept the answer if it rounds correctly to 3 sf, as per the accuracy instructions.

[3 marks]

10

11d.

Markscheme (M1)(A1)(ft)

Note: Award (M1) for substituting the geometric series formula, (A1)(ft) for correct substitution. Follow through from part (c) for their firstterm and common ratio.

= ($)1280 (1283.05…) (A1)(ft)(G2)

[3 marks]

60(1.112−1)

1.1−1

11e.

Markscheme (M1)(M1)

Note: Award (M1) for correctly substituted geometric and arithmetic series formula with n (accept other variable for “n”), (M1) forcomparing their expressions consistent with their part (b) and part (d).

OR

(M1)(M1)

Note: Award (M1) for two curves with approximately correct shape drawn in the first quadrant, (M1) for one point of intersection withapproximate correct position.

Accept alternative correct sketches, such as

Award (M1) for a curve with approximate correct shape drawn in the 1 (or 4 ) quadrant and all above (or below) the x-axis, (M1) forone point of intersection with the x-axis with approximate correct position.

17 (A2)(ft)(G3)

Note: Follow through from parts (b) and (d).An answer of 16 is incorrect. Award at most (M1)(M1)(A0)(A0) with working seen. Award (G0) if final answer is 16 without working seen.

[4 marks]

> (2 × 60 + (n − 1) × 10)60(1.1n−1)

1.1−1n

2

st th

[3 marks]

[3 marks]

[4 marks]

12a.

MarkschemeOR (M1)

Note: Award (M1) for dividing any by .

(A1) (C2)

[2 marks]

162486

54162

un+1 un

= (0.333, 0.333333…)13

12b.

Markscheme (M1)

Note: Award (M1) for their correct substitution into geometric sequence formula.

(A1)(ft) (C2)

Note: Follow through from part (a).

Award (A1)(A0) for or with or without working.

[2 marks]

486( )n−1= 21

3

n = 6

u6 = 2 u6

12c.

Markscheme (M1)

Note: Award (M1) for correct substitution into geometric series formula.

(A1)(ft) (C2)

[2 marks]

S30 =486(1−

30)13

1− 13

= 729

13a.

Markscheme (A1) (C1)

[1 mark]

(0.5)12

[2 marks]

[2 marks]

[2 marks]

[1 mark]

13b.

Markscheme (M1)

Note: Award (M1) for their correct substitution into the geometric sequence formula. Accept a list of their five correct terms.

(A1)(ft) (C2)

Note: Follow through from their common ratio from part (a).

[2 marks]

18 × ( )412

1.125 (1.13, )98

13c.

Markscheme (M1)(M1)

Notes: Award (M1) for their correct substitution into the geometric sequence formula with a variable in the exponent, (M1) for

comparing their expression with .

Accept an equation.

(A1)(ft) (C3)

Note: Follow through from their common ratio from part (a). “ ” must be a positive integer for the (A1) to be awarded.

[3 marks]

18 × ( )n−1< 10−31

2

10−3 ( )11000

n = 16

n

14a.

Markscheme

(A2) (C2)

Note: Award (A2) for all correct entries, (A1) for 3 correct entries.

[2 marks]

[2 marks]

[3 marks]

[2 marks]

14b.

Markschemecontinuous (A1) (C1)

[1 mark]

14c.

Markscheme (A1) (C1)

Note: Accept equivalent notation such as or .

Award (A0) for “60-70” (incorrect notation).

[1 mark]

60 (cm) < trout length ⩽ 70 (cm)

(60, 70] ]60, 70]

14d.

Markscheme (M1)

Note: Award (M1) for their 4 divided by their 22.

(A1)(ft) (C2)

Note: Follow through from their part (a). Do not accept 0.181818….

[2 marks]

× 100422

= 18.2 (18.1818…)

15a.

Markscheme180 (A1) (C1)[1 mark]

15b.

Markscheme36, 24 (A1)(A1) (C2)

Note: Award (A0)(A1) for two incorrect values that add up to 60.

[2 marks]

15c.

Markscheme125 (accept 125.5) (A1)

[1 mark]

[1 mark]

[2 marks]

[1 mark]

[2 marks]

[1 mark]

15d.

Markscheme (M1)

Note: Award (M1) for correct substitution of their mid-interval values, multiplied by their frequencies, into mean formula.

=156 (155.625) (A1)(ft) (C3)

Note: Follow through from parts (b) and (c)(i).

[3 marks]

4×25+36×75+34×125+46×175+24×225+16×275160

16a.

Markscheme (A1)

[1 mark]

60

16b.

Markscheme (A1)

[1 mark]

12.5

16c.

Markscheme (M1)

Note: Award (M1) for an attempt to substitute their mid-interval values (consistent with their answer to part (b)) into the formula for themean.

Award (M1) where a table is constructed with their (consistent) mid-interval values listed along with the frequencies.

(A1)(ft)(G2)

Note: Follow through from their answer to part (b).

[2 marks]

3×2.5+5×7.5+…+10×27.560

= ( , 17.9, 17.9166…)107560

21512

16d.

Markscheme (A1)(A1)

[2 marks]

a = 34, b = 60

[2 marks]

[1 mark]

[1 mark]

[2 marks]

[2 marks]

16e.

Markscheme(i)

(A1)

Note: Accept.

Accept any answer between and

.

(Accept 21.5, but do not accept 21.)

(ii) (A1)

Note: Accept. Do not accept

.

Answer must be an integer.

(iii) (M1)

(A1)(G2)

Notes: Award (M1) for subtraction from. Accept

.

Answer must be an integer.

[4 marks]

⩽ 21.25 minutes

21.25

2121.5

5

< 66

60 − 45

= 15

6015 ± 1

17a.

Markscheme (M1)

Note: Award (M1) for correct substitutions into mean formula.

(A1) (C2)

[2 marks]

14+2×15+7×16+17+4×18+19+20+3×2220

(=) 17.5

17b.

Markscheme16.5 (A1) (C1)

[1 mark]

[4 marks]

[2 marks]

[1 mark]

17c.

Markscheme

(A1)(A1)(A1)(ft) (C3)

Note: Award (A1) for correct endpoints, (A1) for correct quartiles, (A1)(ft) for their median. Follow through from part (a)(ii), but only ifmedian is between 16 and 18.5. If a horizontal line goes through the box, award at most (A1)(A1)(A0). Award at most (A0)(A1)(A1) if aruler has not been used.

[3 marks]

18a.

Markscheme48 (A1) (C1)

18b.

Markscheme (M1)

Note: Award (M1) for 58 and 40 seen.

(A1) (C2)

58 − 40

= 18

18c.

Markscheme

(A1)(A1)(ft)(A1)(ft) (C3)

Note: Award (A1) for the correct maximum and minimum, (A1)(ft) for their correct median and (A1)(ft) for 40 and their upper quartile.

Follow through from parts (a) and (b).

Award a maximum of (A1)(A1)(ft)(A0) if the horizontal line goes through the box or if a ruler has clearly not been used.

[3 marks]

[1 mark]

[2 marks]

[3 marks]

19a.

Markscheme (M1)

Note: Award (M1) for correct substitution into median formula or for arranging all 9 values into ascending/descending order.

(A1) (C2)

[2 marks]

= 10x+112

(x =) 9

19b.

Markscheme2.69 (2.69072…) (A2)(ft)

Note: Follow through from part (a).

[2 marks]

19c.

Markscheme13 − 8 (M1)Note: Award (M1) for 13 and 8 seen.

= 5 (A1)(ft) (C4)Note: Follow through from part (a).

[2 marks]

20a.

MarkschemeA' (A1)

Note: Accept alternative set notation for complement such as U − A.

[1 mark]

20b.

Markscheme OR (A1)

Note: Accept alternative set notation for complement.

[1 mark]

C ∩ D′ D′ ∩ C

20c.

Markscheme OR (A2) (C4)

Note: Accept equivalent answers, for example .

[2 marks]

(E ∩ F) ∪ G G ∪ (E ∩ F)

(E ∪ G) ∩ (F ∪ G)

[2 marks]

[2 marks]

[2 marks]

[1 mark]

[1 mark]

[2 marks]

20d.

Markscheme

(A1)

[1 mark]

20e.

Markscheme

(A1) (C2)

[1 mark]

21a.

Markschemea = 0.2 (A1)

[1 mark]

21b.

Markschemeb = 0.85 (A1)

[1 mark]

21c.

Markscheme0.25 × 0.8 (M1)

Note: Award (M1) for a correct product.

(A1)(G2)

[2 marks]

= 0.2 ( , 20%)15

[1 mark]

[1 mark]

[1 mark]

[1 mark]

[2 marks]

21d.

Markscheme0.25 × 0.8 + 0.75 × 0.15 (A1)(ft)(M1)

Note: Award (A1)(ft) for their (0.25 × 0.8) and (0.75 × 0.15), (M1) for adding two products.

(A1)(ft)(G3)

Note: Award the final (A1)(ft) only if answer does not exceed 1. Follow through from part (b)(i).

[3 marks]

= 0.313 (0.3125, , 31.3%)516

21e.

Markscheme (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for a correct numerator (their part (b)(i)), (A1)(ft) for a correct denominator (their part (b)(ii)). Follow through fromparts (b)(i) and (b)(ii).

(A1)(ft)(G3)

Note: Award final (A1)(ft) only if answer does not exceed 1.

[3 marks]

0.25×0.80.25×0.8+0.75×0.15

= 0.64 ( , 64% )1625

21f.

Markscheme(x =) 3 (A1)

[1 Mark]

21g.

Markscheme(y =) 10 (A1)(ft)

Note: Following through from part (c)(i) but only if their x is less than or equal to 13.

[1 Mark]

21h.

Markscheme54 − (10 + 3 + 4 + 2 + 6 + 8 + 13) (M1)

Note: Award (M1) for subtracting their correct sum from 54. Follow through from their part (c).

= 8 (A1)(ft)(G2)

Note: Award (A1)(ft) only if their sum does not exceed 54. Follow through from their part (c).

[2 marks]

[3 marks]

[3 marks]

[1 mark]

[1 mark]

[2 marks]

21i.

Markscheme6 + 8 + 13 (M1)

Note: Award (M1) for summing 6, 8 and 13.

27 (A1)(G2)

[2 marks]

22a.

Markscheme15 (A1) (C1)

[1 mark]

22b.

Markschemeno (A1) (C1)

Note: Accept “it is only offered in Winter and Spring”.

[1 mark]

22c.

Markschemevolleyball, golf, cycling (A1) (C1)

Note: Responses must list all three sports for the (A1) to be awarded.

[1 mark]

22d.

Markscheme4 (A1) (C1)

[1 mark]

22e.

MarkschemeOR (or equivalent) (A2) (C2)

[2 marks]

(F ∪ W ∪ S)′ F ′ ∩ W ′ ∩ S ′

[2 marks]

[1 mark]

[1 mark]

[1 mark]

[1 mark]

[2 marks]

23a.

Markscheme

(A1)(A1) (C2)

Note: Award (A1) for 30 in correct area, (A1) for 60 and 10 in the correct areas.

[2 marks]

23b.

Markscheme (A1)(ft)(A1)(ft) (C2)

Note: Award (A1)(ft) for correct numerator of 30, (A1)(ft) for correct denominator of 90. Follow through from their Venn diagram.

[2 marks]

( , 0.333333… , 33.3333… %)3090

13

23c.

Markscheme (R1)

Note: Award (R1) for multiplying their by .

therefore the events are independent (A1)(ft) (C2)

Note: Award (R1)(A1)(ft) for an answer which is consistent with their Venn diagram.

Do not award (R0)(A1)(ft).

Do not award final (A1) if is not calculated. Follow through from part (a).

[2 marks]

P(S) × P(M) = × =34

13

14

13

(as P(S ∩ M) = )14

P(S) × P(M)

24a.

Markscheme20 (A1) (C1)

[1 mark]

[2 marks]

[2 marks]

[2 marks]

[1 mark]

24b.

Markscheme (A1)(A1) (C2)

Note: Award (A1) for correct numerator, (A1) for correct denominator.

[2 marks]

(0.11627… , 11.6279… % )543

24c.

Markscheme (A1)(M1)

Note: Award (A1) for first or second correct product seen, (M1) for adding their two products or for multiplying their product by two.

(A1) (C3)

[3 marks]

× + ×737

1236

1237

736

= (0.12612… , 12.6126% )14111

25a.

Markscheme

(A1)(A1)(A1) (C3)

Note: Award (A1) for each correct pair of probabilities.

[3 marks]

25b.

Markscheme (A1)(ft)(M1)

Note: Award (A1)(ft) for two correct products of probabilities taken from their diagram, (M1) for the addition of their products.

(A1)(ft) (C3)

Note: Follow through from part (a).

[3 marks]

0.8 × 0.99 + 0.2 × 0.95

= 0.982 (98.2%, )491500

[2 marks]

[3 marks]

[3 marks]

[3 marks]

26a.

Markscheme (A1) (C1)

[1 mark]

x12

26b.

Markscheme

(A1)(A1)(ft) (C2)

Notes: Award (A1) for 15 placed in the correct position, award (A1)(ft) for and their placed in the correct positions of diagram.

Do not penalize the absence of 0 inside the rectangle and award at most (A1)(A0) if any value other than 0 is seen outside the circles.Award at most (A1)(A0) if 35 and 70 are seen instead of and their .

[2 marks]

x x12

x x12

26c.

Markscheme or equivalent (M1)

Note: Award (M1) for adding the values in their Venn and equating to 120 (or equivalent).

(A1)(ft) (C2)

Note: Follow through from their Venn diagram, but only if the answer is a positive integer and is seen in their Venn diagram.

[2 marks]

x + x + 15 = 12012

(x =) 70

x

26d.

Markscheme85 (A1)(ft) (C1)

Note: Follow through from their Venn diagram and their answer to part (c), but only if the answer is a positive integer and less than120.

[1 mark]

[1 mark]

[2 marks]

[2 marks]

[1 mark]

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International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

27a.

Markscheme

(A1)(A1) (C2)

Note: Award (A1) for each correct pair of probabilities.

[2 marks]

27b.

Markscheme (A1)(ft)(M1)(M1)

Note: Award (A1)(ft) for two correct products from part (a), (M1) for adding their products, (M1) for equating the sum of any two

probabilities to .

(A1)(ft) (C4)

Note: Award the final (A1)(ft) only if . Follow through from part (a).

[4 marks]

p + × =45

15

14

35

35

(p =) (0.688, 0.6875)1116

0 ⩽ p ⩽ 1

[2 marks]

[4 marks]