Nuevo examen - 02 de Febrero de 2017 [280 marks]

1a. [5 marks]

Jar A contains three red marbles and five green marbles. Two marbles are drawn from the jar, one after the other, withoutreplacement.

Find the probability that

(i) none of the marbles are green;

(ii) exactly one marble is green.

Markscheme(i) attempt to find

(M1)

eg , ,

A1 N2

(ii) attempt to find (M1)

eg , ,

recognizing two ways to get one red, one green (M1)

eg ,

,

A1 N2

[5 marks]

P(red) × P(red)

×38

27

×38

38

×38

28

P(none green) = 656

(= )328

P(red) × P(green)

×58

37

×38

58

1556

2P(R) × P(G)× + ×5

837

38

57

× × 238

58

P(exactly one green) = 3056

(= )1528

[3 marks]1b. Find the expected number of green marbles drawn from the jar.

Markscheme (seen anywhere) (A1)

correct substitution into formula for A1

eg ,

expected number of green marbles is

A1 N2

[3 marks]

P(both green) = 2056

E(X)

0 × + 1 × + 2 ×656

3056

2056

+3064

5064

7056(= )5

4

1c. [2 marks]

Jar B contains six red marbles and two green marbles. A fair six-sided die is tossed. If the score is or, a marble is drawn from jar A. Otherwise, a marble is drawn from jar B.

(i) Write down the probability that the marble is drawn from jar B.

(ii) Given that the marble was drawn from jar B, write down the probability that it is red.

Markscheme(i)

A1 N1

(ii)

A1 N1

[2 marks]

12

P(jar B) = 46

(= )23

P(red| jar B) = 68

(= )34

[6 marks]1d. Given that the marble is red, find the probability that it was drawn from jar A.

Markschemerecognizing conditional probability (M1)

eg ,

, tree diagram

attempt to multiply along either branch (may be seen on diagram) (M1)

eg

attempt to multiply along other branch (M1)

eg

adding the probabilities of two mutually exclusive paths (A1)

eg

correct substitution

eg

,

A1

A1 N3

[6 marks]

P(A|R)P(jar A and red)

P(red)

P(jar A and red) = ×13

38

(= )18

P(jar B and red) = ×23

68

(= )12

P(red) = × + ×13

38

23

68

P(jar A|red) =×1

338

× + ×13

38

23

68

18

58

P(jar A|red) = 15

2a. [4 marks]

A running club organizes a race to select girls to represent the club in a competition.

The times taken by the group of girls to complete the race are shown in the table below.

Find the value of and of .

p

q

Markschemeattempt to find (M1)

eg ,

A1 N2

attempt to find (M1)

eg ,

A1 N2

[4 marks]

p

120 − 7050 + 20 + x = 120

p = 50

q

180 − 20200 − 20 − 20

q = 160

2b. [3 marks]A girl is chosen at random.

(i) Find the probability that the time she takes is less than minutes.

(ii) Find the probability that the time she takes is at least minutes.

Markscheme(i)

A1 N1

(ii) valid approach (M1)

eg ,

A1 N2

[3 marks]

14

26

70200(= )7

20

20 + 20200 − 160

40200(= )1

5

2c. [4 marks]A girl is selected for the competition if she takes less than minutes to complete the race.

Given that of the girls are not selected,

(i) find the number of girls who are not selected;

(ii) find .

x

40%

x

Markscheme(i) attempt to find number of girls (M1)

eg ,

are not selected A1 N2

(ii) are selected (A1)

A1 N2

[4 marks]

0.4× 20040

100

80

120

x = 20

2d. [4 marks]Girls who are not selected, but took less than minutes to complete the race, are allowed another chance to be selected. The new times taken by these girls are shown in the

cumulative frequency diagram below.

(i) Write down the number of girls who were allowed another chance.

(ii) Find the percentage of the whole group who were selected.

25

Markscheme(i)

given second chance A1 N1

(ii) took less than minutes (A1)

attempt to find their selected total (may be seen in calculation) (M1)

eg

, their answer from (i)

() A1 N3

[4 marks]

30

2020

%

120 + 20(= 140)120+

70%

3a. [3 marks]

Bill and Andrea play two games of tennis. The probability that Bill wins the first game is.

If Bill wins the first game, the probability that he wins the second game is.

If Bill loses the first game, the probability that he wins the second game is.

Copy and complete the following tree diagram. (Do not write on this page.)

45

56

23

Markscheme

A1A1A1 N3

Note: Award A1 for each correct bold probability. [3 marks]

3b. [2 marks]Find the probability that Bill wins the first game and Andrea wins the second game.

Markschememultiplying along the branches (may be seen on diagram) (M1)eg

A1 N2

[2 marks]

×45

16

( )430

215

3c. [4 marks]Find the probability that Bill wins at least one game.

3c.

MarkschemeMETHOD 1multiplying along the branches (may be seen on diagram) (M1)eg

adding their probabilities of three mutually exclusive paths (M1)eg

correct simplification (A1)eg

A1 N3

METHOD 2recognizing “Bill wins at least one” is complement of “Andrea wins 2” (R1)eg finding P (Andrea wins 2)

P (Andrea wins both)

(A1)

evidence of complement (M1)eg

A1 N3

[4 marks]

× , × , ×45

56

45

16

15

23

× + × + × , + ×45

56

45

16

15

23

45

15

23

+ + , + +2030

430

215

23

215

215

(= )2830

1415

= ×15

13

1 − p, 1 − 115

1415

3d. [5 marks]Given that Bill wins at least one game, find the probability that he wins both games.

MarkschemeP (B wins both)

A1

evidence of recognizing conditional probability (R1)eg

correct substitution (A2)eg

A1 N3

[5 marks]

× (= )45

56

23

P(A |B), P (Bill wins both |Bill wins at least one), tree diagram

×45

56

1415

(= )2028

57

4a. [2 marks]

Let

and

be independent events, where

and

.

Find

.

A

B

P(A) = 0.3

P(B) = 0.6

P(A ∩ B)

Markschemecorrect substitution (A1)eg

A1 N2[2 marks]

0.3 × 0.6

P(A ∩ B) = 0.18

4b. [2 marks]Find

.

Markschemecorrect substitution (A1)eg

A1 N2[2 marks]

P(A ∪ B)

P(A ∪ B) = 0.3 + 0.6 − 0.18

P(A ∪ B) = 0.72

4c. [1 mark]On the following Venn diagram, shade the region that represents

.

Markscheme

A1 N1

A ∩ B′

4d. [2 marks]Find

.

Markschemeappropriate approach (M1)eg

(may be seen in Venn diagram) A1 N2[2 marks]

P(A ∩ )B′

0.3 − 0.18, P(A) × P( )B′

P(A ∩ ) = 0.12B′

5a. [2 marks]

Two events

and

are such that

and

.

Given that

and

are mutually exclusive, find

.

Markschemecorrect approach (A1)eg

A1 N2[2 marks]

A

B

P(A) = 0.2

P(A ∪ B) = 0.5

A

B

P(B)

0.5 = 0.2 + P(B), P(A ∩ B) = 0

P(B) = 0.3

5b. [4 marks]Given that

and

are independent, find

.

MarkschemeCorrect expression for

(seen anywhere) A1eg

attempt to substitute into correct formula for

(M1)eg

correct working (A1)eg

A1 N3

[4 marks]

A

B

P(B)

P(A ∩ B)

P(A ∩ B) = 0.2P(B), 0.2x

P(A ∪ B)

P(A ∪ B) = 0.2 + P(B) − P(A ∩ B), P(A ∪ B) = 0.2 + x − 0.2x

0.5 = 0.2 + P(B) − 0.2P(B), 0.8x = 0.3

P(B) = (= 0.375, exact)38

[4 marks]6a.

Samantha goes to school five days a week. When it rains, the probability that she goes to school by bus is 0.5. When it does not rain,

the probability that she goes to school by bus is 0.3. The probability that it rains on any given day is 0.2.

On a randomly selected school day, find the probability that Samantha goes to school by bus.

Markschemeappropriate approach (M1)eg

, tree diagram,

one correct multiplication (A1)eg

correct working (A1)eg

A1 N3[4 marks]

P(R ∩ B) + P( ∩ B)R′

0.2 × 0.5, 0.24

0.2 × 0.5 + 0.8 × 0.3, 0.1 + 0.24

P(bus) = 0.34(exact)

[3 marks]6b. Given that Samantha went to school by bus on Monday, find the probability that it was raining.

Markschemerecognizing conditional probability (R1)eg

correct working A1eg

A1 N2

[3 marks]

P(A|B) = P(A∩B)

P(B)

0.2×0.50.34

P(R|B) = , 0.294517

[2 marks]6c. In a randomly chosen school week, find the probability that Samantha goes to school by bus on exactly three days.

Markschemerecognizing binomial probability (R1)eg

,

A1 N2[2 marks]

X ∼ B(n, p)

( )53

(0.34 , (0.34 (1 − 0.34)3 )3 )2

P(X = 3) = 0.171

6d. [5 marks]After

school days, the probability that Samantha goes to school by bus at least once is greater than

. Find the smallest value of

.

n

0.95

n

MarkschemeMETHOD 1evidence of using complement (seen anywhere) (M1)eg

valid approach (M1)eg

correct inequality (accept equation) A1eg

(A1) A1 N3

METHOD 2valid approach using guess and check/trial and error (M1)eg finding

for various values of n

seeing the “cross over” values for the probabilities A1A1

recognising

(R1) A1 N3

[5 marks]

1 − P (none), 1 − 0.95

1 − P (none) > 0.95, P (none) < 0.05, 1 − P (none) = 0.95

1 − (0.66 > 0.95, (0.66 = 0.05)n )n

n > 7.209 (accept n = 7.209)

n = 8

P(X ⩾ 1)

n = 7, P(X ⩾ 1) = 0.9454, n = 8, P(X ⩾ 1) = 0.939

0.9639 > 0.95

n = 8

[2 marks]7a.

Adam travels to school by car ( ) or by bicycle ( ). On any particular day he is equally likely to travel by car or bybicycle.

The probability of being late ( ) for school is if he travels by car.

The probability of being late for school is if he travels by bicycle.

This information is represented by the following tree diagram.

Find the value of .

Markschemecorrect working (A1)

eg

A1 N2

[2 marks]

C B

L 16

13

p

1 − 16

p = 56

[2 marks]7b. Find the probability that Adam will travel by car and be late for school.

Markschememultiplying along correct branches (A1)

eg

A1 N2

[2 marks]

×12

16

P(C ∩ L) = 112

[4 marks]7c. Find the probability that Adam will be late for school.

Markschememultiplying along the other branch (M1)

eg

adding probabilities of their mutually exclusive paths (M1)

eg

correct working (A1)

eg

A1 N3

[4 marks]

×12

13

2

× + ×12

16

12

13

+112

16

P(L) = (= )312

14

[3 marks]7d. Given that Adam is late for school, find the probability that he travelled by car.

Markschemerecognizing conditional probability (seen anywhere) (M1)

eg

correct substitution of their values into formula (A1)

eg

A1 N2

[3 marks]

P(C|L)

112312

P(C|L) = 13

7e. [4 marks]Adam will go to school three times next week.

Find the probability that Adam will be late exactly once.

Markschemevalid approach (M1)

eg , three ways it could happen

correct substitution (A1)

eg

correct working (A1)

eg

A1 N2

[4 marks]

Total [15 marks]

X ∼ B (3, ) , ( ) , ( )14

14

( )34

2 31

( ) , × × + × × + × ×31

( )14

1( )3

42 1

434

34

34

14

34

34

34

14

3 ( ) ( ) , + +14

916

964

964

964

2764

8a. [2 marks]

Let and be independent events, with and , where .

Write down an expression for in terms of.

Markscheme (A1)

A1 N2

[2 marks]

C

D P(C) = 2k P(D) = 3k2 0 < k < 0.5

P(C ∩ D)k

P(C ∩ D) = 2k × 3k2

P(C ∩ D) = 6k3

[3 marks]8b. Find .

MarkschemeMETHOD 1

finding their (seen anywhere) (A1)

eg

correct substitution into conditional probability formula (A1)

eg

A1 N2

METHOD 2

recognizing A1

finding their (only if first line seen) (A1)

eg

A1 N2

[3 marks]

Total [7 marks]

P( |D)C ′

P( ∩ D)C ′

0.4 × 0.27, 0.27 − 0.162, 0.108

P( |D) = , C ′ P( ∩D)C ′

0.27(1−2k)(3 )k2

3k2

P( |D) = 0.4C ′

P( |D) = P( )C ′ C ′

P( ) = 1 − P(C)C ′

1 − 2k, 1 − 0.6

P( |D) = 0.4C ′

9a. [2 marks]At a large school, students are required to learn at least one language, Spanish or French. It is known that of the students learn Spanish, and learn French.

Find the percentage of students who learn both Spanish and French.

Markschemevalid approach (M1)

e.g. Venn diagram with intersection, union formula,

(accept) A1 N2

[2 marks]

75%40%

P(S ∩ F) = 0.75 + 0.40 − 1

1515%

9b. [2 marks]At a large school, students are required to learn at least one language, Spanish or French. It is known that of the students learn Spanish, and learn French.

Find the percentage of students who learn Spanish, but not French.

Markschemevalid approach involving subtraction (M1)

e.g. Venn diagram,

60 (accept) A1 N2

[2 marks]

75%40%

75 − 15

60%

9c. [5 marks]At a large school, students are required to learn at least one language, Spanish or French. It is known that of the students learn Spanish, and learn French.

At this school, of the students are girls, and of the girls learn Spanish.

A student is chosen at random. Let G be the event that the student is a girl, and let S be the event that the student learns Spanish.

(i) Find .

(ii) Show that G and S are not independent.

75%40%

52%85%

P(G ∩ S)

Markscheme(i) valid approach (M1)

e.g. tree diagram, multiplying probabilities,

correct calculation (A1)

e.g.

(exact) A1 N3

(ii) valid reasoning, with words, symbols or numbers (seen anywhere) R1

e.g. ,

, not equal,

one correct value A1

e.g. ,

,

G and S are not independent AG N0

[5 marks]

P(S|G) × P(G)

0.52 × 0.85

P(G ∩ S) = 0.442

P(G) × P(S) ≠ P(G ∩ S)P(S|G) ≠ P(S)

P(G) × P(S) = 0.39P(S|G) = 0.850.39 ≠ 0.442

9d. [6 marks]At a large school, students are required to learn at least one language, Spanish or French. It is known that of the students learn Spanish, and learn French.

At this school, of the students are girls, and of the girls learn Spanish.

A boy is chosen at random. Find the probability that he learns Spanish.

75%40%

52%85%

MarkschemeMETHOD 1

are boys (seen anywhere) A1

e.g.

appropriate approach (M1)

e.g.

correct approach to find P(boy and Spanish) (A1)

e.g. ,

, 0.308

correct substitution (A1)

e.g. ,

correct manipulation (A1)

e.g.

,

A1 N3

[6 marks]

METHOD 2

are boys (seen anywhere) A1

e.g. 0.48 used in tree diagram

appropriate approach (M1)

e.g. tree diagram

correctly labelled branches on tree diagram (A1)

e.g. first branches are boy/girl, second branches are Spanish/not Spanish

correct substitution (A1)

e.g.

correct manipulation (A1)

e.g. ,

,

[6 marks]

48%

P(B) = 0.48

P(girl and Spanish) + P(boy and Spanish) = P(Spanish)

P(B ∩ S)= P(S) − P(G ∩ S)P(B ∩ S)= P(S|B) × P(B)

0.442 + 0.48x = 0.750.48x = 0.308

P(S|B) = 0.3080.48

P(Spanish|boy) = 0.641666…0.6416̄

P(Spanish|boy) = 0.642[0.641, 0.642]

48%

0.442 + 0.48x = 0.75

0.48x = 0.308P(S|B) = 0.308

0.48

P(Spanish|boy) = 0.641666…0.6416̄

P(Spanish|boy) = 0.642[0.641, 0.642]

10a. [5 marks]Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.

(i) Copy and complete the following tree diagram.

(ii) Find the probability that two white balls are chosen.

Markscheme(i)

A1A1A1 N3

(ii) multiplying along the correct branches (may be seen on diagram) (A1)

e.g.

A1 N2

[5 marks]

, and ( , and )46

36

36

23

12

12

×37

26

(= )642

17

10b. [5 marks]Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.

Bag B contains four white balls and three red balls. When two balls are chosen at random without replacement from bag B, theprobability that they are both white is

.

A standard die is rolled. If 1 or 2 is obtained, two balls are chosen without replacement from bag A, otherwise they are chosen from bagB.

Find the probability that the two balls are white.

27

Markscheme , (seen anywhere) (A1)(A1)

appropriate approach (M1)

e.g.

correct calculation A1

e.g. ,

A1 N3

[5 marks]

P(bagA) = (= )26

13

P(bagB) = (= )46

23

P(WW ∩ A) + P(WW ∩ B)

× + ×13

17

23

27

+242

842

P(2W) = (= )60252

521

10c. [4 marks]Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.

Bag B contains four white balls and three red balls. When two balls are chosen at random without replacement from bag B, theprobability that they are both white is

.

A standard die is rolled. If 1 or 2 is obtained, two balls are chosen without replacement from bag A, otherwise they are chosen from bagB.

Given that both balls are white, find the probability that they were chosen from bag A.

Markschemerecognizing conditional probability (M1)

e.g. ,

correct numerator (A1)

e.g.

correct denominator (A1)

e.g.

probability A1 N3

[4 marks]

27

P(A∩B)

P(B)

P(A|WW) = P(WW∩A)

P(WW)

P(A ∩ WW) = × ,642

26

121

,6252

521

(= )84420

15

[1 mark]11a.

Events A and B are such that , and

.

The values q , r , s and t represent probabilities.

Write down the value of t .

Markscheme A1 N1

[1 mark]

P(A) = 0.3P(B) = 0.6P(A ∪ B) = 0.7

t = 0.3

11b. [3 marks](i) Show that .

(ii) Write down the value of q and of s .

Markscheme(i) correct values A1

e.g. ,

AG N0

(ii) , A1A1 N2

[3 marks]

r = 0.2

0.3 + 0.6 − 0.70.9 − 0.7

r = 0.2

q = 0.1s = 0.4

11c. [3 marks](i) Write down .

(ii) Find .

Markscheme(i)

A1 N1

(ii) A2 N2

[3 marks]

P( )B′

P(A| )B′

0.4

P(A| ) =B′ 14

[1 mark]12a.

A box contains six red marbles and two blue marbles. Anna selects a marble from the box. She replaces the marble and then selects asecond marble.

Write down the probability that the first marble Anna selects is red.

MarkschemeNote: In this question, method marks may be awarded for selecting without replacement, as noted in the examples.

A1 N1

[1 mark]

P(R) = (= )68

34

[2 marks]12b. Find the probability that Anna selects two red marbles.

Markschemeattempt to find

(M1)

e.g. ,

,

A1 N2

[2 marks]

P(Red) × P(Red)

P(R) × P(R)×3

434

×68

57

P(2R) = (= )3664

916

[3 marks]12c. Find the probability that one marble is red and one marble is blue.

MarkschemeMETHOD 1

attempt to find (M1)

e.g. ,

,

recognizing two ways to get one red, one blue (M1)

e.g. ,

,

A1 N2

[3 marks]

METHOD 2

recognizing that is

(M1)

attempt to find and (M1)

e.g. ,

; ,

A1 N2

[3 marks]

P(Red) × P(Blue)

P(R) × P(B)×6

828

×68

27

P(RB) + P(BR)2 ( )12

64× + ×6

827

28

67

P(1R,1B) = (= )2464

38

P(1R,1B)1 − P(2B) − P(2R)

P(2R)P(2B)

P(2R) = ×34

34

×68

57

P(2B) = ×14

14

×28

17

P(1R,1B) = (= )2464

38

13a. [4 marks]

Let , where

.

Find the values of k such that has two equal roots.

f(x) = + kx + 812x2

k ∈ Z

f(x) = 0

MarkschemeMETHOD 1

evidence of discriminant (M1)

e.g. , discriminant = 0

correct substitution into discriminant A1

e.g. ,

A1A1 N3

METHOD 2

recognizing that equal roots means perfect square (R1)

e.g. attempt to complete the square,

correct working

e.g. ,

A1

A1A1 N3

[4 marks]

− 4acb2

− 4 × × 8k2 12

− 16 = 0k2

k = ±4

( + 2kx + 16)12

x2

(x + k12

)2

= 812k2

k = ±4

13b. [4 marks]Each value of k is equally likely for . Find the probability that

has no roots.

Markschemeevidence of appropriate approach (M1)

e.g.

correct working for k A1

e.g. ,

, list all correct values of k

A2 N3

[4 marks]

−5 ≤ k ≤ 5f(x) = 0

− 4ac < 0b2

−4 < k < 4< 16k2

p = 711

[1 mark]14a.

The diagram below shows the probabilities for events A and B , with .

Write down the value of p .

Markscheme A1 N1

[1 mark]

P( ) = pA′

p = 45

14b. [3 marks]Find .

Markschememultiplying along the branches (M1)

e.g. ,

adding products of probabilities of two mutually exclusive paths (M1)

e.g. ,

A1 N2

[3 marks]

P(B)

×15

14

1240

× + ×15

14

45

38

+120

1240

P(B) = 1440

(= )720

14c. [3 marks]Find .P( |B)A′

Markschemeappropriate approach which must include

(may be seen on diagram) (M1)

e.g.

(do not accept

)

(A1)

A1 N2

[3 marks]

A′

P( ∩B)A′

P(B)P(A∩B)

P(B)

P( |B) =A′×4

538

720

P( |B) =A′ 1214

(= )67

15a. [3 marks]

Consider the events A and B, where , and

.

The Venn diagram below shows the events A and B, and the probabilities p, q and r.

Write down the value of

(i) p ;

(ii) q ;

(iii) r.

Markscheme(i)

A1 N1

(ii) A1 N1

(iii) A1 N1

[3 marks]

P(A) = 0.5P(B) = 0.7P(A ∩ B) = 0.3

p = 0.2

q = 0.4

r = 0.1

15b. [2 marks]Find the value of .P(A| )B′

Markscheme A2 N2

Note: Award A1 for an unfinished answer such as .

[2 marks]

P(A| ) =B′ 23

0.20.3

[1 mark]15c. Hence, or otherwise, show that the events A and B are not independent.

Markschemevalid reason R1

e.g. ,

thus, A and B are not independent AG N0

[1 mark]

≠ 0.5230.35 ≠ 0.3

16a. [4 marks]

José travels to school on a bus. On any day, the probability that José will miss the bus is .

If he misses his bus, the probability that he will be late for school is .

If he does not miss his bus, the probability that he will be late is .

Let E be the event “he misses his bus” and F the event “he is late for school”.

The information above is shown on the following tree diagram.

Find

(i) ;

(ii) .

13

78

38

P(E ∩ F)

P(F)

Markscheme(i)

A1 N1

(ii) evidence of multiplying along the branches (M1)

e.g. ,

adding probabilities of two mutually exclusive paths (M1)

e.g. ,

A1 N2

[4 marks]

724

×23

58

×13

78

( × ) + ( × )13

78

23

38

( × ) + ( × )13

18

23

58

P(F) = 1324

16b. [5 marks]Find the probability that

(i) José misses his bus and is not late for school;

(ii) José missed his bus, given that he is late for school.

Markscheme(i)

(A1)

A1

(ii) recognizing this is (M1)

e.g.

A2 N3

[5 marks]

×13

18

124

P(E|F)

÷724

1324

168312(= )7

13

16c. [3 marks]The cost for each day that José catches the bus is 3 euros. José goes to school on Monday and Tuesday.

Copy and complete the probability distribution table.

Markscheme

A2A1 N3

[3 marks]

16d. [2 marks]The cost for each day that José catches the bus is 3 euros. José goes to school on Monday and Tuesday.

Find the expected cost for José for both days.

Markschemecorrect substitution into

formula (M1)

e.g. ,

(euros) A1 N2

[2 marks]

E(X)

0 × + 3 × + 6 ×19

49

49

+129

249

E(X) = 4

17a. [3 marks]

In a class of 100 boys, 55 boys play football and 75 boys play rugby. Each boy must play at least one sport from football and rugby.

(i) Find the number of boys who play both sports.

(ii) Write down the number of boys who play only rugby.

Markscheme(i) evidence of substituting into

(M1)

e.g. , Venn diagram

30 A1 N2

(ii) 45 A1 N1

[3 marks]

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

75 + 55 − 100

17b. [4 marks]One boy is selected at random.

(i) Find the probability that he plays only one sport.

(ii) Given that the boy selected plays only one sport, find the probability that he plays rugby.

Markscheme(i) METHOD 1

evidence of using complement, Venn diagram (M1)

e.g. ,

A1 N2

METHOD 2

attempt to find P(only one sport) , Venn diagram (M1)

e.g.

A1 N2

(ii)

A2 N2

[4 marks]

1 − p

100 − 30

70100(= )7

10

+25100

45100

70100(= )7

10

4570(= )9

14

17c. [2 marks]Let A be the event that a boy plays football and B be the event that a boy plays rugby.

Explain why A and B are not mutually exclusive.

Markschemevalid reason in words or symbols (R1)

e.g. if mutually exclusive, if not mutually exclusive

correct statement in words or symbols A1 N2

e.g. ,

, , some students play both sports, sets intersect

[2 marks]

P(A ∩ B) = 0P(A ∩ B) ≠ 0

P(A ∩ B) = 0.3P(A ∪ B) ≠ P(A) + P(B)P(A) + P(B) > 1

[3 marks]17d. Show that A and B are not independent.

Markschemevalid reason for independence (R1)

e.g. ,

correct substitution A1A1 N3

e.g. ,

[3 marks]

P(A ∩ B) = P(A) × P(B)P(B|A) = P(B)

≠ ×30100

75100

55100

≠3055

75100

[3 marks]18a.

Jan plays a game where she tosses two fair six-sided dice. She wins a prize if the sum of her scores is 5.

Jan tosses the two dice once. Find the probability that she wins a prize.

Markscheme36 outcomes (seen anywhere, even in denominator) (A1)

valid approach of listing ways to get sum of 5, showing at least two pairs (M1)

e.g. (1, 4)(2, 3), (1, 4)(4, 1), (1, 4)(4, 1), (2, 3)(3, 2) , lattice diagram

A1 N3

[3 marks]

P(prize) = 436

(= )19

[2 marks]18b. Jan tosses the two dice 8 times. Find the probability that she wins 3 prizes.

Markschemerecognizing binomial probability (M1)

e.g. , binomial pdf,

A1 N2

[2 marks]

B(8, )19

( )83

( )19

3( )8

9

5

P(3 prizes) = 0.0426

19a. [5 marks]

In a group of 16 students, 12 take art and 8 take music. One student takes neither art nor music. The Venn diagram below shows theevents art and music. The values p , q , r and s represent numbers of students.

(i) Write down the value of s .

(ii) Find the value of q .

(iii) Write down the value of p and of r .

Markscheme(i)

A1 N1

(ii) evidence of appropriate approach (M1)

e.g. ,

A1 N2

(iii) , A1A1 N2

[5 marks]

s = 1

21 − 1612 + 8 − q = 15

q = 5

p = 7r = 3

19b. [4 marks](i) A student is selected at random. Given that the student takes music, write down the probability the student takes art.

(ii) Hence, show that taking music and taking art are not independent events.

Markscheme(i)

A2 N2

(ii) METHOD 1

A1

evidence of correct reasoning R1

e.g.

the events are not independent AG N0

METHOD 2

A1

evidence of correct reasoning R1

e.g.

the events are not independent AG N0

[4 marks]

P(art|music) = 58

P(art) = 1216

(= )34

≠34

58

P(art) × P(music) = 96256

(= )38

× ≠1216

816

516

19c. [4 marks]Two students are selected at random, one after the other. Find the probability that the first student takes only music and thesecond student takes only art.

Markscheme (seen anywhere) A1

(seen anywhere) A1

evidence of valid approach (M1)

e.g.

A1 N2

[4 marks]

P(first takes only music) = 316

P(second takes only art) = 715

×316

715

P(music and art) = 21240

(= )780

[3 marks]20a.

A company uses two machines, A and B, to make boxes. Machine A makes of the boxes.

of the boxes made by machine A pass inspection.

of the boxes made by machine B pass inspection.

A box is selected at random.

Find the probability that it passes inspection.

Markschemeevidence of valid approach involving A and B (M1)

e.g. , tree diagram

correct expression (A1)

e.g.

A1 N2

[3 marks]

60%

80%

90%

P(A ∩ pass) + P(B ∩ pass)

P(pass) = 0.6 × 0.8 + 0.4 × 0.9

P(pass) = 0.84

20b. [4 marks]The company would like the probability that a box passes inspection to be 0.87.

Find the percentage of boxes that should be made by machine B to achieve this.

Markschemeevidence of recognizing complement (seen anywhere) (M1)

e.g. ,

, ,

,

evidence of valid approach (M1)

e.g. ,

correct expression A1

e.g. ,

,

from B A1 N2

[4 marks]

P(B) = x

P(A) = 1 − x

1 − P(B)100 − x

x + y = 1

0.8(1 − x) + 0.9x

0.8x + 0.9y

0.87 = 0.8(1 − x) + 0.9x

0.8 × 0.3 + 0.9 × 0.7 = 0.870.8x + 0.9y = 0.87

70%

21a. [4 marks]

The Venn diagram below shows events A and B where ,

and . The values m , n , p and q are probabilities.

(i) Write down the value of n .

(ii) Find the value of m , of p , and of q .

Markscheme(i)

A1 N1

(ii) ,

, A1A1A1 N3

[4 marks]

P(A) = 0.3P(A ∪ B) = 0.6P(A ∩ B) = 0.1

n = 0.1

m = 0.2p = 0.3q = 0.4

21b. [2 marks]Find .P( )B′

Markschemeappropriate approach

e.g. ,

, (M1)

A1 N2

[2 marks]

P( ) = 1 − P(B)B′

m + q

1 − (n + p)

P( ) = 0.6B′

[3 marks]22a.

Two fair 4-sided dice, one red and one green, are thrown. For each die, the faces are labelled 1, 2, 3, 4. The score for each die is thenumber which lands face down.

List the pairs of scores that give a sum of 6.

Markschemethree correct pairs A1A1A1 N3

e.g. (2, 4), (3, 3), (4, 2) , R2G4, R3G3, R4G2

[3 marks]

22b. [3 marks]The probability distribution for the sum of the scores on the two dice is shown below.

Find the value of p , of q , and of r .

Markscheme , , A1A1A1 N3

[3 marks]

p = 116

q = 216

r = 216

22c. [6 marks]Fred plays a game. He throws two fair 4-sided dice four times. He wins a prize if the sum is 5 on three or more throws.

Find the probability that Fred wins a prize.

Markschemelet X be the number of times the sum of the dice is 5

evidence of valid approach (M1)

e.g. , tree diagram, 5 sets of outcomes produce a win

one correct parameter (A1)

e.g. ,

,

Fred wins prize is (A1)

appropriate approach to find probability M1

e.g. complement, summing probabilities, using a CDF function

correct substitution (A1)

e.g. ,

, ,

A1 N3

[6 marks]

X ∼ B(n, p)

n = 4p = 0.25q = 0.75

P(X ≥ 3)

1 − 0.949…1 − 243

2560.046875 + 0.00390625

+12256

1256

probability of winning = 0.0508( )13

256

23a. [1 mark]

Let A and B be independent events, where and

.

Write down an expression for .

Markscheme A1 N1

[1 mark]

P(A) = 0.6P(B) = x

P(A ∩ B)

P(A ∩ B) = P(A) × P(B)(= 0.6x)

23b. [4 marks]Given that ,

(i) find x ;

(ii) find .

P(A ∪ B) = 0.8

P(A ∩ B)

Markscheme(i) evidence of using

(M1)

correct substitution A1

e.g. ,

A1 N2

(ii) A1 N1

[4 marks]

P(A ∪ B) = P(A) + P(B) − P(A)P(B)

0.8 = 0.6 + x − 0.6x

0.2 = 0.4x

x = 0.5

P(A ∩ B) = 0.3

[1 mark]23c. Hence, explain why A and B are not mutually exclusive.

Markschemevalid reason, with reference to

R1 N1

e.g.

[1 mark]

P(A ∩ B)

P(A ∩ B) ≠ 0

24a. [4 marks]

There are 20 students in a classroom. Each student plays only one sport. The table below gives their sport and gender.

One student is selected at random.

(i) Calculate the probability that the student is a male or is a tennis player.

(ii) Given that the student selected is female, calculate the probability that the student does not play football.

Markscheme(i) correct calculation (A1)

e.g. ,

A1 N2

(ii) correct calculation (A1)

e.g. ,

A1 N2

[4 marks]

+ −920

520

220

4+2+3+320

P(male or tennis) = 1220

÷620

1120

3+311

P(not football|female) = 611

[3 marks]24b. Two students are selected at random. Calculate the probability that neither student plays football.

24b.

Markscheme ,

A1

A1

A1 N1

[3 marks]

P(first not football) = 1120

P(second not football) = 1019

P(neither football) = ×1120

1019

P(neither football) = 110380

25. [7 marks]Consider the independent events A and B . Given that , and

, find .

P(B) = 2P(A)P(A ∪ B) = 0.52P(B)

MarkschemeMETHOD 1

for independence (R1)

expression for , indicating

(A1)

e.g. ,

substituting into (M1)

correct substitution A1

e.g. ,

correct solutions to the equation (A2)

e.g., (accept the single answer)

A1 N6

[7 marks]

METHOD 2

for independence (R1)

expression for , indicating

(A1)

e.g. ,

substituting into (M1)

correct substitution A1

e.g. ,

correct solutions to the equation (A2)

e.g. 0.4, 2.6 (accept the single answer 0.4)

(accept if x set up as

) A1 N6

[7 marks]

P(A ∩ B) = P(A) × P(B)

P(A ∩ B)P(B) = 2P(A)

P(A) × 2P(A)x × 2x

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

0.52 = x + 2x − 2x2

0.52 = P(A) + 2P(A) − 2P(A)P(A)

0.21.30.2

P(B) = 0.4

P(A ∩ B) = P(A) × P(B)

P(A ∩ B)P(A) = P(B)1

2

P(B) × P(B)12

x × x12

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

0.52 = 0.5x + x − 0.5x2

0.52 = 0.5P(B) + P(B) − 0.5P(B)P(B)

P(B) = 0.4x = 0.4P(B)

26a. [1 mark]

The letters of the word PROBABILITY are written on 11 cards as shown below.

Two cards are drawn at random without replacement.

Let A be the event the first card drawn is the letter A.

Let B be the event the second card drawn is the letter B.

Find .

Markscheme A1 N1

[1 mark]

P(A)

P(A) = 111

26b. [2 marks]Find .

Markscheme A2 N2

[2 marks]

P(B|A)

P(B|A) = 210

26c. [3 marks]Find .

Markschemerecognising that

(M1)

correct values (A1)

e.g.

A1 N3

[3 marks]

P(A ∩ B)

P(A ∩ B) = P(A) × P(B|A)

P(A ∩ B) = ×111

210

P(A ∩ B) = 2110

27a. [6 marks]

Two standard six-sided dice are tossed. A diagram representing the sample space is shown below.

Let be the sum of the scores on the two dice.

(i) Find .

(ii) Find .

(iii) Find .

Markscheme(i) number of ways of getting

is 5 A1

A1 N2

(ii) number of ways of getting is 21 A1

A1 N2

(iii) A2 N2

[6 marks]

X

P(X = 6)

P(X > 6)

P(X = 7|X > 6)

X = 6

P(X = 6) = 536

X > 6

P(X > 6) = (= )2136

712

P(X = 7|X > 6) = (= )621

27

27b. [8 marks]Elena plays a game where she tosses two dice.

If the sum is 6, she wins 3 points.

If the sum is greater than 6, she wins 1 point.

If the sum is less than 6, she loses k points.

Find the value of k for which the game is fair.

Markschemeattempt to find

M1

e.g.

A1

fair game if (may be seen anywhere) R1

attempt to substitute into formula M1

e.g.

correct substitution into A1

e.g.

work towards solving M1

e.g.

A1

A1 N4

[8 marks]

P(X < 6)

1 − −536

2136

P(X < 6) = 1036

E(W) = 0

E(X)

3 ( ) + 1 ( ) − k( )536

2136

1036

E(W) = 0

3 ( ) + 1 ( ) − k( ) = 0536

2136

1036

15 + 21 − 10k = 0

36 = 10k

k = (= 3.6)3610

28a. [2 marks]

A four-sided die has three blue faces and one red face. The die is rolled.

Let B be the event a blue face lands down, and R be the event a red face lands down.

Write down

(i) P(B);(ii) P(R).

Markscheme(i) P(B)

A1 N1

(ii) P(R) A1 N1

[2 marks]

= 34

= 14

28b. [2 marks]If the blue face lands down, the die is not rolled again. If the red face lands down, the die is rolled once again. This isrepresented by the following tree diagram, where p, s, t are probabilities.

Find the value of p, of s and of t.

Markscheme A1 N1

, A1 N1

[2 marks]

p = 34

s = 14

t = 34

28c. [3 marks]Guiseppi plays a game where he rolls the die. If a blue face lands down, he scores 2 and is finished. If the red face landsdown, he scores 1 and rolls one more time. Let X be the total score obtained.

(i) Show that .

(ii) Find .

Markscheme(i)

A1

AG N0

(ii) (A1)

A1 N2

[3 marks]

P(X = 3) = 316

P(X = 2)

P(X = 3)

= P (getting 1 and 2) = ×14

34

= 316

P(X = 2) = × + (or 1 − )14

14

34

316

= 1316

28d. [5 marks](i) Construct a probability distribution table for X.

(ii) Calculate the expected value of X.

Printed for Colegio Aleman de Barranquilla

© International Baccalaureate Organization 2017 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markscheme(i)

A2 N2

(ii) evidence of using (M1) (A1)

A1 N2

[5 marks]

E(X) = ∑ xP(X = x)E(X) = 2 ( ) + 3 ( )13

163

16= (= 2 )35

163

16

28e. [4 marks]If the total score is 3, Guiseppi wins. If the total score is 2, Guiseppi gets nothing.

Guiseppi plays the game twice. Find the probability that he wins exactly.

Markschemewin

scores 3 one time, 2 other time (M1)

(seen anywhere) A1

evidence of recognising there are different ways of winning (M1)

e.g. ,

,

A1 N3

[4 marks]

$10

$10

$10 ⇒

P(3) × P(2) = ×1316

316

$10

P(3) × P(2) + P(2) × P(3)2 ( × )13

163

16+ + +36

2563

25636

2563

256P(win $10) = (= )78

25639

128