IB Math Studies Year 2 Name:__________________________Topic 3: Probability – Formative Quiz 13

1. It is known that 5% of all AA batteries made by Power Manufacturers are defective. AA batteries are sold in packs of 4.

Find the probability that a pack of 4 has

(a) exactly two defective batteries;

(3)(b) at least one defective battery.

(2)(Total 5 marks)

2. Note: For this question, it is important that you show your working and explain your method clearly.A box contains 10 coloured light bulbs, 5 green, 3 red and 2 yellow. One light bulb is selected at random and put into the light fitting of room A.

(a) What is the probability that the light bulb selected is

(i) green?

(1)(ii) not green?

(1)A second light bulb is selected at random and put into the light fitting in room B.

(b) What is the probability that

(i) the second light bulb is green given the first light bulb was green?

(l)

1

(ii) both light bulbs are not green?

(2)(iii) one room has a green light bulb and the other room does not have a green light bulb?

(3)A third light bulb is selected at random and put in the light fitting of room C.

(c) What is the probability that

(i) all three rooms have green light bulbs?

(2)(ii) only one room has a green light bulb?

(3)(iii) at least one room has a green light bulb?

(2)(Total 15 marks)

3. In a club with 60 members, everyone attends either on Tuesday for Drama (D) or on Thursday for Sports (S) or on both days for Drama and Sports.One week it is found that 48 members attend for Drama and 44 members attend for Sports and x members attend for both Drama and Sports.

(a) (i) Draw and label fully a Venn diagram to illustrate this information.

(3)

2

(ii) Find the number of members who attend for both Drama and Sports.

(2) (iii) Describe, in words, the set represented by (D S)'.

(2)(iv) What is the probability that a member selected at random attends for Drama only or Sports only?

(3)The club has 28 female members, 8 of whom attend for both Drama and Sports.

(b) What is the probability that a member of the club selected at random

(i) is female and attends for Drama only or Sports only?

(2)(ii) is male and attends for both Drama and Sports?

(2)(Total 14 marks)

3

4. 100 students were asked which television channel (MTV, CNN or BBC) they had watched the previous evening. The results are shown in the Venn diagram below.

U

M T V C N N

B B C

3 5

6 5 2

3

2 3 1 9

7

From the information in the Venn diagram, write down the number of students who watched

(a) both MTV and BBC;

(b) MTV or BBC;

(c) CNN and BBC but not MTV;

(d) MTV or CNN but not BBC.

(Total 4 marks)

5. Let F be the set of all families that have exactly 2 children.

(a) Assuming P(boy) = P(girl), copy and complete the following tree diagram, for families with 2 children.

B o y

G irl

B o y

G irlB o y

G irl

12

(2)

4

(b) What is the probability that a family chosen at random from F has exactly

(i) 2 boys?

(ii) 2 boys, if it is known that the first child is a boy?

(iii) 2 boys, if it is known that there is a boy in the family?

(3)(Total 5 marks)

6. Children in a class of 30 students are asked whether they can swim (S) or ride a bicycle (B).There are 12 girls in the class. 8 girls can swim, 6 girls can ride a bicycle and 4 girls can do both.16 boys can swim, 13 boys can ride a bicycle and 12 boys can do both. This information is represented in a Venn diagram.

U B o y s G irls

S

B

41 2

b

a

2

(a) Find the values of a and b.

(2)(b) Calculate the number of students who can do neither.

(2)

5

(c) Write down the probability that a student chosen at random can swim.

(2)(d) Given that the student can ride a bicycle, write down the probability that the student is a girl.

(2)(Total 8 marks)

7. Neil has three dogs. Two are brown and one is grey. When he feeds the dogs, Neil uses three bowls and gives them out randomly. There are two red bowls and one yellow bowl. This information is shown on the tree diagram below.

23

13

23

23

13

13

B ro w n

G rey

R e d

Ye llo w

R e d

Ye llo w

(a) One of the dogs is chosen at random.

(i) Find P (the dog is grey and has the yellow bowl).

(ii) Find P (the dog does not get the yellow bowl).

(3)

6

(b) Neil often takes the dogs to the park after they have eaten. He has noticed that the grey dog plays with a stick for a quarter of the time and both brown dogs play with sticks for half of the time. This information is shown on the tree diagram below.

23

13

B ro w n

G rey

S tick

N ostick

S tick

N ostick

(i) Copy the tree diagram and add the four missing probability values on the branches that refer to playing

with a stick.

During a trip to the park, one of the dogs is chosen at random.

(ii) Find P (the dog is grey or is playing with a stick, but not both).

(iii) Find P (the dog is grey given that the dog is playing with a stick).

(iv) Find P (the dog is grey and was fed from the yellow bowl and is notplaying with a stick).

(9)(Total 12 marks)

7

IB Math Studies Year 2 Name:__________________________Topic 3: Probability – Formative Quiz 14

1. A group of 30 children are surveyed to find out which of the three sports cricket (C), basketball (B) or volleyball (V) they play. The results are as follows:

3 children do not play any of these sports2 children play all three sports6 play volleyball and basketball3 play cricket and basketball6 play cricket and volleyball16 play basketball12 play volleyball.

(a) Draw a Venn diagram to illustrate the relationship between the three sports played.

(1) (b) On your Venn diagram indicate the number of children that belong to each region.

(3)(c) How many children play only cricket?

(2)(Total 6 marks)

8

2. A group of 25 females were asked how many children they each had. The results are shown in the histogram below.

1 09876543210

0 1 2 3 4N u m b er o f C h ild ren

F req u en cy

N u m b er o f C h ild ren p e r F em ale

(a) Show that the mean number of children per female is 1.4.

(2)(b) Show clearly that the standard deviation for this data is approximately 1.06.

(3) (c) Another group of 25 females was surveyed and it was found that the mean number of children per

female was 2.4 and the standard deviation was 2. Use the results from parts (a) and (b) to describe the differences between the number of children the two groups of females have.

(2)(d) A female is selected at random from the first group. What is the probability that she has more than two

children?

(2)

9

(e) Two females are selected at random from the first group. What is the probability that

(i) both females have more than two children?

(2)(ii) only one of the females has more than two children?

(3)(iii) the second female selected has two children given that the first female selected had no children?

(1)(Total 15 marks)

3. In a survey of 52 students it was found that 30 study Spanish and 15 have computers. Seven of the students who study Spanish also have computers.

(a) Copy and complete this table.

StudySpanish

Do not studySpanish Total

Have computersDo not have computersTotal 52

(3) (b) Draw and label fully a Venn diagram to illustrate this information. Use U to represent the set of all

students surveyed, S the set of students who study Spanish and C the set of students who have computers.

(2)

10

(b) Describe, in words, the set represented by C S .

(2) (d) Find n(C S ).

(1)A student is selected at random to attend a computer workshop given in Spanish.

(e) What is the probability that the student

(i) has a computer and studies Spanish?

(2)(ii) as a computer but does not study Spanish?

(2)(iii) as a computer if he/she studies Spanish?

(2)(Total 14 marks)

4. The Venn diagram below shows the number of students studying Science (S), Mathematics (M) and History (H) out of a group of 20 college students. Some of the students do not study any of these subjects, 8 study Science, 10 study Mathematics and 9 study History.

U

S M

H

41

21 3

3

A

11

(a) (i) How many students belong to the region labelled A?

(ii) Describe in words the region labelled A.

(iii) How many students do not study any of the three subjects?

(5)(b) Draw a sketch of the Venn diagram above and shade the region which represents S H.

(1)(c) Calculate n(S H).

(2)This group of students is to compete in an annual quiz evening which tests knowledge of Mathematics, Science and History. The names of the twenty students are written on pieces of paper and then put into a bag.

(c) One name is randomly selected from the bag. Calculate the probability that the student selected studies

(i) all three subjects;

(ii) History or Science.

(2)

12

(e) A team of two students is to be randomly selected to compete in the quiz evening. The first student selected will be the captain of the team. Calculate the probability that

(i) the captain studies all three subjects and the other team member does not study any of the three subjects;

(ii) one student studies Science only and the other student studies History only;

(iii) the second student selected studies History, given that the captain studies History and Mathematics.

(5)(Total 15 marks)

5. There are two biscuit tins on a shelf. The red tin contains three chocolate biscuits and seven plain biscuits. The blue tin contains one chocolate biscuit and nine plain biscuits.

(a) A child reaches into the red tin and randomly selects a biscuit. The child returns that biscuit to the tin, shakes the tin, and then selects another biscuit.Find the probability that

(i) both biscuits chosen are chocolate.

(2)(ii) one of the biscuits is plain and the other biscuit is chocolate.

(3)

13

(b) A second child chooses a biscuit from the blue tin. The child eats the biscuit and chooses another one from the blue tin. The tree diagram below represents the possible outcomes for this event.

11 0

91 0

ab

C

C

C

P

P

P

(i) Write down the values of a and b.

(2)(ii) Find the probability that both biscuits are chocolate.

(1)(iii) What is the probability that at least one of the biscuits is chocolate?

(3) (c) Suppose that before the two children arrived, their brother randomly selected one of the biscuit tins and

took out one biscuit.Calculate the probability that this biscuit was chocolate.

(4)(Total 15 marks)

14

6. The data in the table below refers to a sample of 60 randomly chosen plants.

Growth rate Classification by environment dark light shady total

high 3 8 14 25low 8 9 18 35total 11 17 32 60

(a) (i) Find the probability of a plant being in a shady environment.

(ii) Find the probability of a plant having a low growth rate and being in a dark environment.

(ii) Find the probability of a plant not being in a dark environment.

(5)(b) A plant is chosen at random from the above group.

Find the probability that the chosen plant has

(i) a high growth rate or is in a dark environment, but not both

(ii) a light environment, given that it has a high growth rate.

(4)

15

(c) The 60 plants in the above group were then classified according to leaf type. It was found that 15 of the plants had type A leaves, 37 had type B leaves and 8 had type C leaves.Two plants were randomly selected from this group. Find the probability that

(i) both plants had type C leaves

(ii) neither of the plants had type B leaves.

(5)(Total 14 marks)

7. When Geraldine travels to work she can travel either by car (C), bus (B) or train (T). She travels by car on one day in five. She uses the bus 50 of the time. The probabilities of her being late (L) when travelling by car, bus or train are 0.05, 0.12 and 0.08 respectively.

(a) Copy the tree diagram below and fill in all the probabilities, where NL represents not late, to represent this information.

(5)L

L

LN L

N L

N L

C

B

T

0 .2

(b) Find the probability that Geraldine travels by bus and is late.

(1)(c) Find the probability that Geraldine is late.

(3)(d) Find the probability that Geraldine travelled by train, given that she is late.

(3)(Total 12 marks)

16

IB Math Studies Year 2 Topic 3: Probability – Formative Quiz 13 – MS

1. (a) p(two defective) = 6×0.052×0.952 (M2)Note: Award (M1) for 0.052×0.952, (M1) for (×6)

= 0.0135375= 0.0135 (3 s.f.) (A1)

(b) p(at least one defective) = 1 – 0.954 (M1)

= 0.18549375...= 0.0185 (3 s.f.) (A1)

[5]

2. (a) (i) p(green) = 105

(A1)

(ii) p(not green) = 105

(A1)

Note: Accept 21

, 0.5 or 50% for either answer

(b) (i) p(GG) = 94

or 0.444 (3 s.f.) (A1)(ii) p(not green then not green)

= 94

105

(M1)

= 9020

or 92

or 0.222 (3 s.f.) (A1) (iii) p(one green and one not green)

= 95

105

95

105

(M2)

Note: Award (M1) for 95

105

, (M1) for (×2)

= 9050

or 95

or 0.556 (3 s.f.) (A1)

(c) (i) p(3 green) = 83

94

105

(M1)

= 72060

or 121

or 0.0833 (3 s.f.) (A1)

(ii) p(only one green) = 84

95

1053

(M2)

Note: Award (M1) for 84

95

105

, (M1) for (×3)

= 720300

or 0.417 (3 s.f.) or 2410

or 125

(A1)

17

(iii) p(at least one green) = 1 – p(no green)

= 1 – 83

94

105

(M1)

= 1 – 72060

= 720660

or 1211

or 0.917 (3 s.f.) (A1)[15]

3. (a) (i)

D S U

4 8 – x 4 4 – xx

(A3) 3Note: Award (A1) for a correct diagram (labelled), (A1) for x in the correct position, (A1) for either (48 – x) or (44 – x) correctly positioned.

(ii) 48 – x + x + 44 – x = 60 (or equivalent), allow ft from (i) (M1) x = 32 (A1) 2

(iii) The set of members who did not attend for both (A2) 2

Drama and Sports (or equivalent)

(iv) P(D or S) =

603244

603248

(M1)(M1)

Note: Award (M1) for either 603248

or 603244

, (M1) for adding.

= 6028

or 157

or 0.467 (3 s.f.) or 46.7% (3s.f.) (A1) 3

(b) (i) P(Female and (S or D)) = 6020

(M1)

= 31

or 0.333 (3 s.f.) or 33.3% (3s.f.) (A1) 2

(ii) P(Male and both D and S) =

60832

(M1)

= 52

or 0.4 or 40% (A1) 2[14]

18

4. (a) n(MTV BBC) = 11 (A1)(b) n(MTV BBC) = 74 (A1)(c) n(CNN BBC MTV) = 2 (A1)(d) n(MTV CNN BBC) = 77 (A1)

[4]

5. (a)

B o y

G irl

B o y

G irl

B o y

G irl

12

12

12

12

12

12

(A2) 2Note: Award (A2) for 5 or 4 correct probabilities, (A1) for 3 or 2, (A0) for 1 or 0.

(b) (i) P(2 boys) = 41

21

21

(A1)

(ii) P(2 boys first child is boy) = 21

2141

(A1)

(iii) P(2 boys boy in family) = 31

4341

(A1) 3Note: Answers can be obtained logically without using conditional probability.Award full marks for correct answers.

[5]

6. (a) a = 4, b = 1 (A1)(A1) 2 (b) 30 – (4 + 12 + 1 + 2 + 4 + 4) = 3 (M1)(A1)

(or (A2)) 2

(c)

54

3024

(A1)(A1)2

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

(d) 196

(A1)(A1)2

Note: Award (A1) for numerator, (A1) for denominator.19

[8]

7. (a) (i) P (a dog is grey and has the yellow bowl)

)111.0(91

31

31

(M1)(A1)(G2)

Note: The (M1) is for multiplying two values along any branch of the tree.

(ii) P (dog does not get yellow bowl) = )6.0orsf)3(667.0(

32

(A1) 3

(b) (i) The tree diagram should show the values 21,

21

for the brown branch (A1)

and 43,

41

in the correct positions for the grey branch. (A1)(ft)

Note: Follow through if the branches are interchanged.(ii) P (the dog is grey or is playing with a stick, but not both)

21

32

43

31

(M1)

)583.0(127

(A1)(ft)(G1)

Notes: The (M1) is for showing two correct products (whether added or not).Follow through from b(i).

Award (M1) for 41

31

(must be a sum).(iii) P (dog is grey given that it is playing with stick)

125

121or

41

31

21

32

41

31

)()(

SP

SGp

(M1)(A1)(ft)Note: (M1) for substituted conditional probability formula, (A1) for correct substitutions.

)2.0(51

(A1)(ft)(G2)

(iv) P (grey and fed from yellow bowl and not playing with stick)

= 121

43

31

31

sf).30833.0363(

(M1)(A1)(ft)(G1) 9

Note: (M1) is for product of 3 reasonable probability values.[12]

IB Math Studies Year 2 Topic 3: Probability – Formative Quiz 14 – MS

1. (a) (A1)Award (A1) for a rectangle containing 3 intersecting circles.

(b) (A3) 4

20

C B

V32

24 4

1 9

Notes:(b) Award (A3) for 6 or 7 correct numbers in the regions.

(A2) for 4 or 5 correct, (A1) for 2 or 3 correct. (c) 1 + 9 + 4 + 2 + 4 + 2 + 3 = 25 (M1)

n(C) = 30 – 25= 5 (A1)ORn(C) = 5 (C2) 2

[6]

2. (a) Mean = 2541332611005

= 1.4 (M2)(AG) 2Note: Award (M1) for the numerator and (M1) for the denominator.

(b) 2_)( xxf = 5(0 – 1.4)2 + 10(1 – 1.4)2 + 6(2 – 1.4)2

+ 3(3 – 1.4)2 + 1(4 – 1.4)2 = 28 (M2)

Note: Award (M1) for 2

_)( xx values, and (M1) for multiplying by the

appropriate frequencies.

S.D. = 2528

(M1)= 1.06 (AG) 3

(c) Award (R1) for each acceptable reason, e.g.

Group 2 has more children in total.Group 2 has a larger number of children per female.Group 2 generally have larger families. (R2) 2

(d) P(> 2 children) = 2513

(M1)

= 254

(A1) 2

(e) (i) P(both females have > 2 children) = 243

254

(M1)

= 60012

or 501

or 0.02 (A1)

(ii) P(only 1 female has > 2 children) = 2 × 2421

254

(M2)

21

Note: Award (M1) for 2421

254

, (M1) for multiplying by 2.

= 600168

or 7521

or 0.28 (A1)

(iii) P(second has 2 childrenfirst has 0) = 246

or 41

or 0.25 (A1) 6[15]

3. (a)Study Spanish Do not study

SpanishTotal

Have computers 7 8 15Do not have computers 23 14 37Total 30 22 52 (A3) 3

Notes: Award ½ mark for each correct bold entry.Round up to a maximum 3 marks.

(b)

US C

2 3 7 8

1 4 (A2) 2Notes: Award 1 mark for correctly labelled diagram indicating all of S, C and U.Award ½ mark for any two correct entries from 23, 7, 8 and 14, up to a maximum of 1 mark.

(c) “Students who have computers or do not study Spanish”

(or equivalent statements) (C2) 2Notes: Award (R0) for “Students who have computers but do not study Spanish”. It is incorrect.

(d) n(C S) = 29 (A1) 1

(e) (i) 527

(M1)(A1)2

(ii) 528

(M1)(A1) 2

(iii) 307

(M1)(A1) 2Notes: If no method is shown, award (M1)(A1) if and only if answer is correct, otherwise award zero marks. However, award (M1) if correct method is shown; even if final answer is wrong.Accept probabilities expressed as decimals or percentages.

[14]

4. (a) (i) 10 – 6 = 4 (M1)(A1)OR

22

4 (A2) (ii) Students who study Mathematics only. (A1)(iii) 20 – (4 + l + 2 + l + 3 + 3 + 4) = 2 (M1)(A1)

OR2 (A2) 5

(b)

S M

H

(A1) 1 (c) n(S H) = 8 + 9 – 3 = 14 (M1)(A1)

ORn(S H) = 14 (A2) 2

(d) (i) P(studies all 3 subjects) =

1.0

101

202

(A1)

(ii) P(History or Science) =

7.0

107

2014

(A1) 2

(e) (i) P(captain studies all, other student studies none) = 192

202

(M1)

=

0105.0

951

3804

(A1) (ii) P(one studies Science only and the other studies History only)

= 193

204

× 2 (M1)

Note: Award (M1) for 193

204

=

0632.0

956

38024

(A1)

(iii) P(HistoryHistory and Maths) = 198

(A1) 5[15]

5. (a) (i) P(chocolate, chocolate) = 1009

103

103

= (0.09) (M1)(A1) 2

(ii) P(one is plain) = P(chocolate, plain) + P (plain, chocolate) (M1)

23

= 10042

103

107

107

103

= (0.42) (M1)(A1) 3 (b) (i) a = 8, b = 9 (A1)(A1)

2 (ii) P(chocolate, chocolate) = 0 (A1) 1(iii) P(at least one chocolate) = P(one is chocolate) (M1)

= 9018

91091

91019

= 0.2 (M1)(A1) 3 (c) P(tin, chocolate)

101

21

103

21

(M1)(M1)(M1)

= 204

= (0.2) (A1) 4[15]

6. Accept all answers given as percentages. If a correct fraction is seen as answer, ignore subsequent cancellations or decimal reductions (including AP’s) if incorrect.

Note In each of (a)(i), (a)(iii), (b)(i) and (b)(ii) award (A1) for numerator and (A1) for denominator. Ft if the denominator is incorrect.

(a) (i) P(shady) =

0.533or

158

6032

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

(ii) P(dark and low growth rate) =

0.133or

152

608

(A1)

(iii) P(not dark) = 6049

601160

(= 0.817) (A1)(A1) 5 (b) (i) P (high growth rate or dark environment)

=

0.5or

21

6030

608148or

606–2511

(A1)(A1)

(ii) P (in light, given high growth rate) = 258

(= 0.32) (A1)(A1) 4

(c) (i) P(CC) =

0158.0

88514

354056

597

608

(M1)(A1)or (G2)

Note: Award (M1) for multiplication of two fractions, (A1) for the answer.

(ii) Three alternative valid approaches might be seen:

5922

6023

24

OR 597

608

5915

608

598

6015

5914

6015

OR 1 –

5937

608

598

6037

5937

6015

5915

6037

5936

6037

(M1)(A1)

=

0.143or

1770253or

3540506

(A1)or (G3) 5

Note: If = 5922

6023

is doubled, award (M1)(A0)(A0).

If non-replacement is treated incorrectly in (i), ft to (ii) if consistent[14]

7. (a)

(A5) 5

Note: Award (A1) for 0.5 at B, (A1) for 0.3 at T, then (A1) for each correct pair. Accept fractions or percentages.

(b) 0.06 (A1)(ft) 1

Note: Accept 0.5 0.12 or 6% (c) for a relevant two-factor product, either C L or T L

for summing three two-factor products(0.2 0.05 + 0.06 + 0.3 0.08) (M1)(M1)0.094 (A1)(ft)(G2) 3

(d) 094.008.03.0

(M1)(A1)(ft)Note: Award (M1) for substituted conditional probability formula seen, (A1)(ft) for correct substitution

= 0.255 (A1)(ft)(G2) 3[12]

25