Math SL Day 66 Probability Practice [196 marks]

[2 marks]1a.

Events and are independent with and .

Find .

Markschemevalid interpretation (may be seen on a Venn diagram) (M1)

eg

A1 N2

[2 marks]

A B P(A ∩ B) = 0.2 P(A′ ∩ B) = 0.6

P(B)

P(A ∩ B) + P(A′ ∩ B), 0.2 + 0.6

P(B) = 0.8

[4 marks]1b. Find .

Markschemevalid attempt to find (M1)

eg

correct working for (A1)

eg

correct working for (A1)

eg

A1 N3

[4 marks]

P(A ∪ B)

P(A)

P(A ∩ B) = P(A) × P(B), 0.8 × A = 0.2

P(A)

0.25, 0.20.8

P(A ∪ B)

0.25 + 0.8 − 0.2, 0.6 + 0.2 + 0.05

P(A ∪ B) = 0.85

2a. [3 marks]

The following Venn diagram shows the events and , where and

. The values and are probabilities.

(i) Write down the value of .

(ii) Find the value of .

A B

P(A) = 0.4, P(A ∪ B) = 0.8P(A ∩ B) = 0.1 p q

q

p

Markscheme(i)

A1 N1

(ii) appropriate approach (M1)

eg

A1 N2

[3 marks]

q = 0.1

P(A) − q, 0.4 − 0.1

p = 0.3

[3 marks]2b. Find .

Markschemevalid approach (M1)

eg

correct values (A1)

eg

A1 N2

[3 marks]

P(B)

P(A ∪ B) = P(A) + P(B) − P(A ∩ B), P(A ∩ B) + P(B ∩ A′)

0.8 = 0.4 + P(B) − 0.1, 0.1 + 0.4

P(B) = 0.5

3a. [5 marks]

In a class of 21 students, 12 own a laptop, 10 own a tablet, and 3 own neither.

The following Venn diagram shows the events “own a laptop” and “own a tablet”.

The values , , and represent numbers of students.

(i) Write down the value of .

(ii) Find the value of .

(iii) Write down the value of and of .

Markscheme(i)

A1 N1

(ii) valid approach (M1)

eg

A1 N2

(iii) A1A1 N2

p q r s

p

q

r s

p = 3

(12 + 10 + 3) − 21, 22 − 18

q = 4

r = 8, s = 6

3b. [4 marks]

A student is selected at random from the class.

(i) Write down the probability that this student owns a laptop.

(ii) Find the probability that this student owns a laptop or a tablet but not both.

Markscheme(i)

A2 N2

(ii) valid approach (M1)

eg

A1 N2

(= )1221

47

8 + 6, r + s

(= )1421

23

3c. [4 marks]

Two students are randomly selected from the class. Let be the event a “student owns a laptop”.

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

(ii) Write down the probability that the second student owns a laptop given that the first owns a laptop.

Markscheme

(i) A1A1A1 N3

(ii) A1 N1

[4 marks]

L

1120

4a. [2 marks]

Let and be independent events, with and , where .

Write down an expression for in terms of.

C

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

P(C ∩ D)k

Markscheme (A1)

A1 N2

[2 marks]

P(C ∩ D) = 2k × 3k2

P(C ∩ D) = 6k3

[3 marks]4b. 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(C′|D)

P(C′ ∩ D)

0.4 × 0.27,0.27 − 0.162,0.108

P(C′|D) = , P(C ′∩D)

0.27

(1−2k)(3k2)

3k2

P(C′|D) = 0.4

P(C′|D) = P(C′)

P(C′) = 1 − P(C)

1 − 2k, 1 − 0.6

P(C′|D) = 0.4

[2 marks]5a.

Ann and Bob play a game where they each have an eight-sided die. Ann’s die has three green faces and five red faces; Bob’s die hasfour green faces and four red faces. They take turns rolling their own die and note what colour faces up. The first player to roll greenwins. Ann rolls first. Part of a tree diagram of the game is shown below.

Find the probability that Ann wins on her first roll.

Markschemerecognizing Ann rolls green (M1)

eg

A1 N2

[2 marks]

P(G)

38

Find the probability that Ann wins the game.

5b. [7 marks]

Markschemerecognize the probability is an infinite sum (M1)

eg Ann wins on her roll or roll or roll…,

recognizing GP (M1)

(seen anywhere) A1

(seen anywhere) A1

correct substitution into infinite sum of GP A1

eg

correct working (A1)

eg

A1 N1

[7 marks]

Total [15 marks]

1st 2nd 3rd S∞

u1 = 38

r = 2064

, ( ) , 38

1− 516

38

1

1−( × )58

48

1

1− 516

, ×38

1116

38

1611

P (Ann wins) = (= )4888

611

[2 marks]6a.

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

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]6b. 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]6c. 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]6d. 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)

1123

12

P(C|L) = 13

6e. [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, ) , ( ) ( )2, ( 3

1)1

414

34

( 31

) ( )1( )2, × × + × × + × ×1

434

14

34

34

34

14

34

34

34

14

3 ( ) ( ) , + +14

916

964

964

964

2764

[2 marks]7a.

The following cumulative frequency graph shows the monthly income, dollars, of families.

Find the median monthly income.

I 2000

Markschemerecognizing that the median is at half the total frequency (M1)

eg

A1 N2

[2 marks]

20002

m = 2500 (dollars)

7b. [4 marks](i) Write down the number of families who have a monthly income of dollars or less.

(ii) Find the number of families who have a monthly income of more than dollars.

Markscheme(i) families have a monthly income less than A1 N1

(ii) correct cumulative frequency, (A1)

subtracting their cumulative frequency from (M1)

eg

families have a monthly income of more than dollars A1 N2

Note: If working shown, award M1A1A1 for , using the table.

[4 marks]

2000

4000

500 2000

1850

2000

2000 − 1850

150 4000

128 + 22 = 150

7c. [2 marks]The families live in two different types of housing. The following table gives information about the number of families living ineach type of housing and their monthly income .

Find the value of .

Markschemecorrect calculation (A1)

eg (A1)

A1 N2

[2 marks]

2000I

p

2000 − (436 + 64 + 765 + 28 + 122), 1850 − 500 − 765

p = 585

7d. [2 marks]A family is chosen at random.

(i) Find the probability that this family lives in an apartment.

(ii) Find the probability that this family lives in an apartment, given that its monthly income is greater than dollars.4000

Markscheme(i) correct working (A1)

eg

(exact) A1 N2

(ii) correct working/probability for number of families (A1)

eg

A1 N2

[4 marks]

436 + 765 + 28

0.6145

, 0.615 [0.614, 0.615]12292000

122 + 28, , 0.0751502000

0.186666

(= ) , 0.187 [0.186, 0.187]28150

1475

[2 marks]7e. Estimate the mean monthly income for families living in a villa.

Markschemeevidence of using correct mid-interval values ( ) (A1)

attempt to substitute into (M1)

eg

A1 N2

[3 marks]

Total [15 marks]

1500,3000,4500

∑ fx

∑ f

1500×64+3000×p+4500×12264+585+122

3112.84

3110 [3110, 3120] (dollars)

8. [6 marks]Celeste wishes to hire a taxicab from a company which has a large number of taxicabs.

The taxicabs are randomly assigned by the company.

The probability that a taxicab is yellow is 0.4.

The probability that a taxicab is a Fiat is 0.3.

The probability that a taxicab is yellow or a Fiat is 0.6.

Find the probability that the taxicab hired by Celeste is not a yellow Fiat.

Markschemerecognize need for intersection of Y and F (R1)

eg

valid approach to find

(M1)

eg

, Venn diagram

correct working (may be seen in Venn diagram) (A1)

eg

A1

recognize need for complement of

(M1)

eg

A1 N3

[6 marks]

P(Y ∩ F), 0.3 × 0.4

P(Y ∩ F)

P(Y ) + P(F) − P(Y ∪ F)

0.4 + 0.3 − 0.6

P(Y ∩ F) = 0.1

Y ∩ F

1 − P(Y ∩ F), 1 − 0.1

P ((Y ∩ F)′) = 0.9

9a. [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]

[2 marks]9b. 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

[4 marks]9c. Find the probability that Bill wins at least one game.

MarkschemeMETHOD 1

multiplying 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 2

recognizing “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

[5 marks]9d. 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

10a. [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

10b. [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

10c. [1 mark]On the following Venn diagram, shade the region that represents

.

Markscheme

A1 N1

A ∩ B′

10d. [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 ∩ B′) = 0.12

[4 marks]11a.

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(R′ ∩ B)

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]11b. Given that Samantha went to school by bus on Monday, find the probability that it was raining.

Markschemerecognizing conditional probability (R1)

eg

correct working A1

eg

A1 N2

[3 marks]

P(A|B) =P(A∩B)

P(B)

0.2×0.50.34

P(R|B) = , 0.294517

11c. [2 marks]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)3, (0.34)3(1 − 0.34)2

P(X = 3) = 0.171

11d. [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 1

evidence of using complement (seen anywhere) (M1)

eg

valid approach (M1)

eg

correct inequality (accept equation) A1

eg

(A1)

A1 N3

METHOD 2

valid 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)n > 0.95, (0.66)n = 0.05

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

12a. [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

12b. [4 marks]Given that

and

are independent, find

.

A

B

P(B)

MarkschemeCorrect expression for

(seen anywhere) A1

eg

attempt to substitute into correct formula for

(M1)

eg

correct working (A1)

eg

A1 N3

[4 marks]

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

13a. [5 marks]

Jar A contains three red marbles and five green marbles. Two marbles are drawn from the jar, one after the other,without replacement.

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]13b. 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

(= )54

13c. [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]13d. 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

1858

P(jar A|red) = 15

14a. [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

.pq

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

14b. [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

(= )720

20 + 20200 − 16040200

(= )15

14c. [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

14d. [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%

[2 marks]15a.

A Ferris wheel with diameter metres rotates clockwise at a constant speed. The wheel completes rotations every hour. The bottom of the wheel is

metres above the ground.

A seat starts at the bottom of the wheel.

Find the maximum height above the ground of the seat.

Markschemevalid approach (M1)

eg ,

maximum height (m) A1 N2

[2 marks]

1222.413

13 + diameter13 + 122

= 135

15b. [2 marks]

After t minutes, the height metres above the ground of the seat is given by

(i) Show that the period of is minutes.

(ii) Write down the exact value of .

Markscheme(i) period

A1

period minutes AG N0

(ii)

A1 N1

[2 marks]

h

h = 74 + a cos bt.

h25

b

= 602.4

= 25

b = 2π25

(= 0.08π)

15c. [3 marks]Find the value of .

MarkschemeMETHOD 1

valid approach (M1)

eg ,

,

(accept ) (A1)

A1 N2

METHOD 2

attempt to substitute valid point into equation for h (M1)

eg

correct equation (A1)

eg ,

A1 N2

[3 marks]

a

max − 74|a| = 135−13

274 − 13

|a| = 61a = 61

a = −61

135 = 74 + a cos( )2π×12.525

135 = 74 + a cos(π)13 = 74 + a

a = −61

15d. [4 marks]Sketch the graph of , for

.h0 ≤ t ≤ 50

Markscheme

A1A1A1A1 N4

Note: Award A1 for approximately correct domain, A1 for approximately correct range,

A1 for approximately correct sinusoidal shape with cycles.

Only if this last A1 awarded, award A1 for max/min in approximately correct positions.

[4 marks]

2

15e. [5 marks]In one rotation of the wheel, find the probability that a randomly selected seat is at least metres above the ground.

Markschemesetting up inequality (accept equation) (M1)

eg ,

, sketch of graph with line

any two correct values for t (seen anywhere) A1A1

eg ,

, ,

valid approach M1

eg ,

,

,

A1 N2

[5 marks]

105

h > 105105 = 74 + a cos bty = 105

t = 8.371 …t = 16.628 …t = 33.371 …t = 41.628 …

16.628−8.37125

t1−t2

252×8.257

502(12.5−8.371)

25

p = 0.330

16a. [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.

75%40%

Markschemevalid approach (M1)

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

(accept) A1 N2

[2 marks]

P(S ∩ F) = 0.75 + 0.40 − 1

1515%

16b. [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%

16c. [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 learnsSpanish.

(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

16d. [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]

[1 mark]17a.

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

17b. [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

17c. [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

18a. [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

18b. [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,the probability 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 arechosen from bag B.

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

18c. [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,the probability 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 arechosen from bag B.

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

27

Printed for North Hills Preparatory

© International Baccalaureate Organization 2018

International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markschemerecognizing conditional probability (M1)

e.g.

,

correct numerator (A1)

e.g.

correct denominator (A1)

e.g.

probability

A1 N3

[4 marks]

P(A∩B)P(B)

P(A|WW) =P(WW∩A)

P(WW)

P(A ∩ WW) = × ,642

26

121

,6252

521

(= )84420

15