Probability Practice 1 1.users.desertacademy.org/balei/Math/HL/HL.ProbPractice1.pdf · IB Math – High Level Year 1 – Probability Practice 1 Alei - Desert Academy Macintosh HD:Users:bobalei:Dropbox:Desert:HL1:6StatProb:Practice:HLProbPractice1.docx

IBMath–HighLevelYear1–ProbabilityPractice1 Alei-DesertAcademy

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ProbabilityPractice11. A bag contains 2 red balls, 3 blue balls and 4 green balls. A ball is chosen at random from the bag

and is not replaced. A second ball is chosen. Find the probability of choosing one green ball and one blue ball in any order.

Working:

Answer: …………………………………………..

(Total 4 marks) 2. In a bilingual school there is a class of 21 pupils. In this class, 15 of the pupils speak Spanish as their

first language and 12 of these 15 pupils are Argentine. The other 6 pupils in the class speak English as their first language and 3 of these 6 pupils are Argentine.

A pupil is selected at random from the class and is found to be Argentine. Find the probability that the pupil speaks Spanish as his/her first language.

Working:

Answer: …………………………………………..

(Total 4 marks) 3. A new blood test has been shown to be effective in the early detection of a disease. The probability

that the blood test correctly identifies someone with this disease is 0.99, and the probability that the blood test correctly identifies someone without that disease is 0.95. The incidence of this disease in the general population is 0.0001.

A doctor administered the blood test to a patient and the test result indicated that this patient had the disease. What is the probability that the patient has the disease?

(Total 6 marks) 4. Given that events A and B are independent with P(A ∩ B) = 0.3 and P(A ∩ B′) = 0.3,

find P(A ∪ B). Working:

Answer: ..................................................................

(Total 3 marks)

5. A girl walks to school every day. If it is not raining, the probability that she is late is . If it is

raining, the probability that she is late is . The probability that it rains on a particular day is .

On one particular day the girl is late. Find the probability that it was raining on that day. Working:

Answer: …………………………………………..

(Total 3 marks)

51

32

41



6. Given that P(X) = P(Y | X) = and P(Y | X ′) = find

(a) P(Y ′); (b) P(X ′ ∪ Y ′).

Working:

Answers: (a) .................................................................. (b) ..................................................................

(Total 3 marks)

7. In a game, the probability of a player scoring with a shot is . Let X be the number of shots the

player takes to score, including the scoring shot. (You can assume that each shot is independent of the others.) (a) Find P(X = 3).

(2) (b) Find the probability that the player will have at least three misses before scoring twice.

(6) (c) Prove that the expected value of X is 4. (You may use the result (1 – x)–2 = 1 + 2x + 3x2 + 4x3......)

(5) (Total 13 marks)

8. The probability that a man leaves his umbrella in any shop he visits is . After visiting two shops in

succession, he finds he has left his umbrella in one of them. What is the probability that he left his umbrella in the second shop?

Working:

Answer: ..........................................................................

(Total 3 marks) 9. Two women, Ann and Bridget, play a game in which they take it in turns to throw an unbiased six-

sided die. The first woman to throw a “6” wins the game. Ann is the first to throw. (a) Find the probability that

(i) Bridget wins on her first throw; (ii) Ann wins on her second throw; (iii) Ann wins on her nth throw.

(6)

(b) Let p be the probability that Ann wins the game. Show that

(4) (c) Find the probability that Bridget wins the game.

(2) (d) Suppose that the game is played six times. Find the probability that Ann wins more games than

Bridget. (5)

(Total 17 marks)

,32

52 ,

41

41

31

.3625

61 pp +=



10. The probability that it rains during a summer’s day in a certain town is 0.2. In this town, the probability that the daily maximum temperature exceeds 25°C is 0.3 when it rains and 0.6 when it does not rain. Given that the maximum daily temperature exceeded 25°C on a particular summer’s day, find the probability that it rained on that day.

Working:

Answer: ..........................................................................

(Total 6 marks) 11. Two children, Alan and Belle, each throw two fair cubical dice simultaneously. The score for each

child is the sum of the two numbers shown on their respective dice. (a) (i) Calculate the probability that Alan obtains a score of 9.

(ii) Calculate the probability that Alan and Belle both obtain a score of 9. (2)

(b) (i) Calculate the probability that Alan and Belle obtain the same score, (ii) Deduce the probability that Alan’s score exceeds Belle’s score.

(4) (c) Let X denote the largest number shown on the four dice.

(i) Show that for P(X ≤ x) = , for x = 1, 2,... 6

(ii) Copy and complete the following probability distribution table. x 1 2 3 4 5 6 P(X = x)

(iii) Calculate E(X). (7)

(Total 13 marks) 12. An integer is chosen at random from the first one thousand positive integers. Find the probability that

the integer chosen is (a) a multiple of 4; (b) a multiple of both 4 and 6.

Working:

Answers: (a) .................................................................. (b) ..................................................................

(Total 6 marks) 13. (a) At a building site the probability, P(A), that all materials arrive on time is 0.85. The

4

6⎟⎠⎞⎜

⎝⎛ x

12961

129615

1296671



probability, P(B), that the building will be completed on time is 0.60. The probability that the materials arrive on time and that the building is completed on time is 0.55. (i) Show that events A and B are not independent. (ii) All the materials arrive on time. Find the probability that the building will not be

completed on time. (5)

(b) There was a team of ten people working on the building, including three electricians and two plumbers. The architect called a meeting with five of the team, and randomly selected people to attend. Calculate the probability that exactly two electricians and one plumber were called to the meeting.

(2) (c) The number of hours a week the people in the team work is normally distributed with a mean

of 42 hours. 10% of the team work 48 hours or more a week. Find the probability that both plumbers work more than 40 hours in a given week.

(8) (Total 15 marks)

14. The random variable X has a Poisson distribution with mean λ.

(a) Given that P(X = 4) = P(X = 2) + P(X = 3), find the value of λ. (3)

(b) Given that λ = 3.2, find the value of (i) P(X ≥ 2); (ii) P(X ≤ 3 | X ≥ 2).

(5) (Total 8 marks)

15. Robert travels to work by train every weekday from Monday to Friday. The probability that he

catches the 08.00 train on Monday is 0.66. The probability that he catches the 08.00 train on any other weekday is 0.75. A weekday is chosen at random. (a) Find the probability that he catches the train on that day. (b) Given that he catches the 08.00 train on that day, find the probability that the chosen day is

Monday. Working:

Answers: (a) .................................................................. (b) ..................................................................

(Total 6 marks)

IBMath–HighLevelYear1–ProbabilityPractice1-MarkScheme Alei-DesertAcademy


ProbabilityPractice1-MarkScheme1. Using a tree diagram,

p(BG or GB) = (M1)(M1)

= = (A1)(A1)

OR p(BG or GB) = 2 × (M1)(M1)

= (A2) (C4)

[4] 2. Using a tree diagram,

(M2) Let p(S) be the probability that the pupil speaks Spanish.

Let p(A) be the probability that the pupil is Argentine. Then, from diagram,

p(S|A) = (A1)

= (A1)

OR p(S|A) = (M1)

= (M1)(A1)

= (A1)

RB

GR

B

G R

B

G

R

B

G

39

49

48

38

⎟⎠⎞⎜

⎝⎛ ×+⎟

⎠⎞⎜

⎝⎛ ×

83

94

84

93

61

61 +

31

83

94 ×

31

621

36

1521

1215

English(6)

Spanish(15)

Argentine(3)

Argentine(12)

1512

54

)()(

ApASp ∩

2115

2112

54



OR

(M2)

p(S | A) = (A1)

= (A1) (C4)

[4] 3. Let D be the event that the patient has the disease and S be the event that

the new blood test shows that the patient has the disease. Let D′ be the complement of D, ie the patient does not have the disease. Now the given probabilities can be written as p(S | D) = 0.99, p(D) = 0.0001, p(S | D′) = 0.05. (A1)(A1)(A1)

Since the blood test shows that the patient has the disease,

we are required to find p(D | S). By Bayes’ theorem,

p(D | S) = (M1)

= (M1)

= 0.001976...= 0.00198 (3 sf) (A1) OR

(A3) Note: Award (A1) for 0.99, (A1) for 0.0001, (A1) for 0.05

Therefore p(S) = 0.0001 × 0.99 + 0.9999 × 0.05 = 0.0500939 (A1)

p(D | S) = (M1)

= 0.00198 (3 sf) (A1) [6]

E(6) A S(15)

3 3 0 12 3

1512

54

)()()()()()(

DpDSpDpDSpDpDSp

′′+

)0001.01)(05.0()0001.0)(99.0()0001.0)(99.0(−+

0.0001

0.9999

D

D′

S′

S′

S

S0.99

0.01

0.05

0.95

0500939.099.00001.0 ×



4. Method 1: (Venn diagram) (M1)

P(A ∩ B) = P(A)P(B) (M1)

0.3 = 0.6 × P(B) P(B) = 0.5 Therefore, P(A ∪ B) = 0.8 (A1) (C3)

Method 2: P(A ∩ B′) = P(A) – P(A ∩ B)

0.3 = P(A) – 0.3 P(A) = 0.6 (A1) P(A ∩ B) = P(A)P(B) since A, B are independent 0.3 = 0.6 × P(B) P(B) = 0.5 (A1) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.6 + 0.5 – 0.3 = 0.8 (A1) (C3)

[3] 5. Let P(R | L) be the probability that it is raining given that the girl is late.

P(R | L) =

P(R | L) = (M1)(A1)

(using a tree diagram or by calculation)

= (A1)

[3] 6.

(a) P(Y′) = (M1)

0.3 0.3

UA B

R

R’

L

L

L’

L’

14

34

23

13

15

45

––

–

–

–

–15–

34–

3–20× =

16–

23–

14–× =

)P()P(

LLR∩

20/36/16/1

+

1910

X

X’

Y

Y

Y’

Y’

23

13

25

34

35 1

4

–

––

–

–

–

43

31

53

32 ×+×



= (A1) (C2)

(b) P(X′ ∪ Y′) = 1 – P(X ∩ Y) = 1 –

= (A1) (C1)

[3]

7. (a) P(X = 3) = (= 0.141 to 3 sf) (M1)(A1) 2

(b) Let the probability of at least three misses before scoring twice = P(3 m) Let S mean “Score” and M mean “Miss”.

P(3 m) = 1 – [P(0 misses) + P(1 miss) + P(2 misses)] (M1) = 1 – [P(SS) + P(SMS or MSS) + P(MMSS or MSMS or SMMS)] (M2)

= 1 – (A2)

= (= 0.738 to 3 sf) (A1) 6

(c) E(x) = (M1)(A1)

= (A1)

= (using the given result) (M1)

= (4)2 = 4 (A1)(AG) 5

[13] 8.

(M1)(A1)

Required probability = . (A1)(C3)

[3] 9. (a) (i) P(Bridget wins on her first throw)

= P(Ann does not throw a “6”) × P(Bridget throws a “6”) (M1)

2013

154

1511

649

41

43 2

=×⎟⎠⎞⎜

⎝⎛

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛

2222

43

413

43

412

41

256189

∑ +⎟⎟⎠

⎞⎜⎜⎝

⎛××+××+×=

xallfor

2

...4

3413

43

412

41 1)P(xx

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛×+×+ ...433

4321

41 2

2

431

41 −

⎟⎠⎞⎜

⎝⎛ −

41

41

41 2

=⎟⎠⎞⎜

⎝⎛

−

Left umbrella

Left umbrella

Did notleave umbrella

Did notleave umbrella

First shop Second shop Probability

13

29

49

13

23

13

23

–

–

––

––

–

52

31

9292

=+



=

= (C1)

(ii) P(Ann wins on her second throw) = P(Ann does not throw a “6”) × P(Bridget does not throw a “6”) × P(Ann throws a “6”) (M1)

=

= (C1)

(iii) P(Ann wins on her nth throw) = P(neither Ann nor Bridget win on their first (n – 1) throws) × P(Ann throws a “6” on her nth throw) (M1)

= . (C1) 6

(b) p = P(Ann wins) = P(Ann wins on her first throw) ++ P(both Ann and Bridget do not win on their first throws) × P(Ann wins from then on) (M1)(R2)

= × p (C1)

= (AG)

OR p = P(Ann wins on first throw) + P(Ann wins on second throw)

+ P(Ann wins on third throw) + … . (M1)

= + ... (C2)

= (C1)

= , as required. (AG) 4

(c) From part (b), . (C1)

Therefore, P(Bridget wins) = 1 – p = . (C1) 2

(d) P(Ann wins more games than Bridget) = P(Ann wins 4 games) + P(Ann wins 5 games) ++ P(Ann wins 6 games) (M2)

= (M2)

= (15 × 25 + 36 × 5 + 36)

= 0.432. (A1) 5 [17]

61

65 ×

365

61

65

65 ××

21625

61

65 )1(2

×⎟⎠⎞⎜

⎝⎛

−n

2

65

61

⎟⎠⎞⎜

⎝⎛+

p3625

61 +

⎟⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛+

61

65

61

65

61 42

)

3625–1

61

or (...61

65

61

65

61

3625

61 42

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+⎟⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛++

p3625

61 +

116

61

3611 =⇒= pp

115

6524

116

115

116

56

115

116

46

⎟⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

6

4

116



10.

(M2) P(> 25°) = 0.2 × 0.3 + 0.8 × 0.6 = 0.54 (M1)(A1)

P(R | >25°) = = (or 0.111) (M1)(A1)(C6)

[6]

11. (a) (i) P(Alan scores 9) = (= 0.111) (A1)

(ii) P(Alan scores 9 and Belle scores 9) =

(= 0.0123) (A1) 2

(b) (i) P(Same score) = + + … + + … +

+ (M1)

= (= 0.113) (A1)

(ii) P(A>B) = (M1)

= (= 0.444) (A1) 4

(c) (i) P(One number ≤ x) = (with some explanation) (R1)

P(X ≤ x) = P(All four numbers ≤ x) = (M1)(AG)

(ii) P(X = x) = P(X ≤ x) – P(X ≤ x – l) = –

x 1 2 3 4 5 6 P(X = x)

(A1)(A1)(A1) Note: Award (A3) if table is not completed but calculation of E(X) in part (iii) is correct.

(iii) E(X) = 1 × + 2 × + … + 6 × (M1)

= (= 5.24) (A1) 7

R(0.2)

R’(0.8)

>25°(0.3)

>25°(0.6)

54.006.0

91

91

⎟⎠⎞⎜

⎝⎛=⎟

⎠⎞⎜

⎝⎛

811

91 2

2

361⎟⎠⎞⎜

⎝⎛

2

362⎟⎠⎞⎜

⎝⎛

2

366⎟⎠⎞⎜

⎝⎛

2

362⎟⎠⎞⎜

⎝⎛

2

361⎟⎠⎞⎜

⎝⎛

64873

21

⎟⎠⎞⎜

⎝⎛

64873–1

1296575

6x

4

6⎟⎠⎞⎜

⎝⎛ x

4

6⎟⎠⎞⎜

⎝⎛ x 4

61–⎟⎠⎞⎜

⎝⎛ x

12961

129615

129665

1296175

1296369

1296671

12961

129615

1296671

12966797



[13] 12. (a) The number of multiples of 4 is 250. (M1)

Required probability = 0.25. (A1)(C2) (b) The number of multiples of 4 and 6 is (M1)

the number of multiples of 12 (A1) = 83. (A1) Required probability = 0.083 (A1) (C4)

[6] 13. (a) (i) To be independent P(A ∩ B) = P(A) × P(B) (R1)

P(A) × P(B) = (0.85)(0.60) = 0.51 but P(A ∩ B) = 0.55 (A1) P(A ∩ B) ≠ P(A) × P(B) Hence A and B are not independent. (AG)

(ii)

P(B′ | A) = (M1)

= (M1)

= (= 0.353) (A1) 5

(b) Probability of 2 electricians and 1 plumber = (M1)

= (A1)

OR Probability of 2 electricians and 1 plumber = (M1)

= (= 0.238) (A1) 2

(c) X = number of hours worked. X ~ N (42, σ2) P(X ≥ 48) = 0.10 (AG) P(X < 48) = 0.90 (M1) Φ(z) = 0.90 z = 1.28 (z = 1.28155) (A1) (Answers given to more than 3 significant figures will be accepted.)

A B

0.30 0.55 0.05

0.10

)P()'( P

AAB∩

85.030.0

176

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

510

25

12

23

⎟⎠⎞⎜

⎝⎛ == 238.0215

25260

⎟⎠⎞⎜

⎝⎛⎟⎠⎞⎜

⎝⎛⎟⎠⎞⎜

⎝⎛⎟⎠⎞⎜

⎝⎛⎟⎠⎞⎜

⎝⎛

64

75

82

92

103

!2!2!5

215



z = => 1.28 = (M1)

=> σ = 4.69 (Accept σ = 4.68) (A1)

P(X > 40) = P (M1)

= 0.665 (A1) OR P(X > 40) = 0.665 (G2) Therefore, the probability that one plumber works more than 40 hours per week is 0.665. The probability that both plumbers work more than 40 hours per week = (0.665)2 (M1) = 0.443 (Accept 0.442 or 0.444) (A1) 8

[15]

14. (a) (M1)

λ2 – 4λ – 12 = 0 ⇒ λ = 6 (A1)(A1) 3 (b) (i) P(X ≥ 2) = 1 – e–3.2 – e–3.2 × 3.2 = 0.829 (M1)(A1)

OR P(X ≥ 2) = 0.829 (G2)

(ii) P(X ≤ 3⎪X ≤ 2) = (M1)

= (A1)

= 0.520 (A1) OR P(X ≤ 3⎪X ≥ 2) = 0.520 (G3) 5

[8] 15. (a) Probability = 0.2 × 0.66 + 0.8 × 0.75 (M1)(A1)

= 0.732 (A1)(C3)

(b) Probability = (M1)

= (A1)

= 0.180 (A1) (C3)

[6]

σµ–X

σ42–48

⎟⎠⎞⎜

⎝⎛ >

69.442–40Z

!3e

!2e

!4e 324 λλλ λλλ ×+×=× −−−

)2P()3P(2

≥≤≤

XX

2.3

32.322.3

e2.4162.3e

22.3e

−

−−

−

×+×

train)P(catches train)catchesP(Mon ∩

732.066.02.0 ×

⎟⎠⎞⎜

⎝⎛=6111

Documents

Probability Practice 1 1.users.desertacademy.org/balei/Math/HL/HL.ProbPractice1.pdf · IB Math – High Level Year 1 – Probability Practice 1 Alei - Desert Academy Macintosh HD:Users:bobalei:Dropbox:Desert:HL1:6StatProb:Practice:HLProbPractice1.docx