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IEB practice · Paper I

IEB Probability and counting practice

12 newly generated questions across all 9 probability and counting topics the IEB examines in Paper I, 165 marks in all. Work each one, then open its answer underneath to check yourself. Nothing here is taken from a past paper.

Download the booklet as PDF · Sit this paper under time · Afrikaans

The fundamental counting principle

3 questions · 44 marks

Paper I · Section B · 14 marks

QUESTION 1
  1. (a)
    A lock code is made of 4 digits chosen from 1 to 9. How many different lock codes can be made if digits may be repeated?
    (2)
  2. (b)
    How many different lock codes can be made if no digit may be used more than once?
    (2)
  3. (c)
    A lock code is chosen at random, digits allowed to repeat. What is the probability that at least one digit appears more than once? Give your answer as a decimal, correct to four decimal places where needed.
    (3)
  4. (d)
    How many lock codes, digits allowed to repeat, start with a digit greater than 7 and end with an odd digit?
    (3)
  5. (e)
    How many 4-letter arrangements can be made from the letters of the word PLUMBING, if no letter is used more than once?
    (2)
  6. (f)
    How many of these arrangements start with a vowel?
    (2)

[14]

Show the answers

(a) Answer: 6561

  1. 9 × 9 × 9 × 9
  2. = 6561

(b) Answer: 3024

  1. 9 × 8 × 7 × 6
  2. = 3024

(c) Answer: 0.5391

  1. P(nothing repeated) = 3024/6561
  2. P(at least one repeated) = 1 − 3024/6561 = 0.5391

(d) Answer: 810

  1. first: 2 choices; last: 5 choices; the middle 2: 9 each
  2. 2 × 9 × 9 × 5 = 810

(e) Answer: 1680

  1. 8 × 7 × 6 × 5
  2. = 1680

(f) Answer: 420

  1. 2 vowels for the first letter, then 7 × 6 × 5
  2. 2 × 210 = 420

Paper I · Section B · 17 marks

QUESTION 2
  1. (a)
    In how many different ways can the letters of the word BALLOON be arranged?
    (3)
  2. (b)
    The letters of BALLOON are arranged at random. What is the probability that the arrangement starts and ends with the same letter?
    (4)
  3. (c)
    6 different people are arranged in a queue. In how many ways can this be done?
    (1)
  4. (d)
    In how many ways can this be done if 2 particular people must be next to each other, in any order?
    (3)
  5. (e)
    The digits 0; 1; 3; 4; 7 are used to form 5-digit numbers, without repetition (a number cannot start with 0). How many such numbers can be formed?
    (2)
  6. (f)
    How many of these numbers are odd and less than 30 000?
    (4)

[17]

Show the answers

(a) Answer: 1260

  1. 7 letters, with L 2 times, O 2 times
  2. 7!/(2! × 2!) = 1260

(b) Answer: 2/21

  1. only a letter that appears more than once can be at both ends: L, O
  2. L at both ends: 60 arrangements of the other 5 letters
  3. O at both ends: 60 arrangements of the other 5 letters
  4. P = (60 + 60)/1260 = 2/21

(c) Answer: 720

  1. 6! = 720

(d) Answer: 240

  1. treat the 2 as one block: 5 units, 5! ways
  2. the block itself in 2! ways
  3. 5! × 2! = 240

(e) Answer: 96

  1. 4 × 4 × 3 × 2 × 1
  2. = 96

(f) Answer: 12

  1. the first digit is less than 3 and not 0, the last digit odd
  2. count each choice of first and last digit, then fill the middle
  3. = 12

Paper I · Section B · 13 marks

QUESTION 3

5 boys and 3 girls are to sit next to one another in a row of 8 seats at the cinema. Kabelo is one of the boys and Ayanda one of the girls.

  1. (a)
    In how many different ways can the 8 learners be arranged?
    (1)
  2. (b)
    In how many ways can they be arranged if there must be a boy at each end of the row?
    (3)
  3. (c)
    In how many ways can they be arranged if Kabelo and Ayanda may not be next to each other?
    (3)
  4. (d)
    In how many ways can they be arranged if all the girls must be next to one another?
    (3)
  5. (e)
    The learners take their places in a random order. What is the probability that all the boys are next to one another?
    (3)

[13]

Show the answers

(a) Answer: 40 320

  1. 8! = 40 320

(b) Answer: 14 400

  1. the two ends: 5 × 4 choices
  2. the other 6: 6!
  3. 5 × 4 × 6! = 14 400

(c) Answer: 30 240

  1. Kabelo and Ayanda together, as one unit: 7! × 2! = 10 080
  2. not together: 40 320 − 10 080 = 30 240

(d) Answer: 4320

  1. the 3 girls as one block: 6 units, 6! ways
  2. the girls inside the block: 3! ways
  3. 6! × 3! = 4320

(e) Answer: 1/14

  1. boys together: 4! × 5! = 2880
  2. P = 2880/40 320 = 1/14

The probability rules

4 questions · 56 marks

Paper I · Section B · 12 marks

QUESTION 4

A bag holds 7 red sweets and x green sweets, and nothing else. Two sweets are taken out at random, one after the other, without the first being put back.

  1. (a)
    Show that the probability that both sweets are red is 42/((x + 7)(x + 6)).
    (2)
  2. (b)
    The probability that the two sweets are the same colour is 27/55. Determine the value of x.
    (5)
  3. (c)
    Determine the probability that at least one of the two sweets is green.
    (2)
  4. (d)
    Suppose instead that the first sweet is put back before the second is taken. Determine the probability that the two sweets are different colours.
    (3)

[12]

Show the answers

(a) Answer: 7/(x + 7) × 6/(x + 6)

  1. first red: 7/(x + 7); then 6 of the red are left among x + 6
  2. 7/(x + 7) × 6/(x + 6) = 42/((x + 7)(x + 6))

(b) Answer: x = 4

  1. P(both red) + P(both green) = [42 + x(x − 1)]/((x + 7)(x + 6)) = 27/55
  2. 2x^2 − 29x + 84 = 0
  3. x = 4 or x = 21/2
  4. x = 4: the number of sweets is a whole number

(c) Answer: 34/55

  1. 1 − P(both red) = 1 − 42/110
  2. = 34/55

(d) Answer: 56/121

  1. red then green, or green then red
  2. 7/11 × 4/11 + 4/11 × 7/11 = 56/121

Paper I · Section B · 15 marks

QUESTION 5
  1. (a)
    A bag holds 4 red and 6 blue balls. Two balls are taken out one after the other, without replacement. Draw a tree diagram to show all the outcomes and their probabilities.
    (3)
  2. (b)
    Calculate the probability that the two balls are the same colour.
    (2)
  3. (c)
    For two events A and B, P(A) = 0.35, P(B′) = 0.85 and P(A′ and B′) = 0.6. Determine P(A or B) and P(A and B).
    (3)
  4. (d)
    Are A and B mutually exclusive? Are they independent? Give reasons.
    (3)
  5. (e)
    In a group of 36 learners, 22 take Accounting, 11 take Business Studies and 5 take neither. How many take both?
    (2)
  6. (f)
    A learner is chosen at random. Calculate the probability that the learner takes exactly one of the two subjects.
    (2)

[15]

Show the answers

(a) Answer: See the tree below.

2/5R1/3RRR: 2/152/3BRB: 4/153/5B4/9RBR: 4/155/9BBB: 1/3
  1. first draw: red 4/10, blue 6/10
  2. second draw: 9 balls are left, one fewer of the colour taken

(b) Answer: 7/15

  1. P(RR) + P(BB) = 2/15 + 1/3
  2. = 7/15

(c) Answer: P(A or B) = 0.4; P(A and B) = 0.1

  1. P(A or B) = 1 − P(A′ and B′) = 1 − 0.6 = 0.4
  2. P(B) = 1 − P(B′) = 0.15
  3. P(A and B) = P(A) + P(B) − P(A or B) = 0.35 + 0.15 − 0.4 = 0.1

(d) Answer: Not mutually exclusive, and not independent

  1. P(A and B) = 0.1 ≠ 0
  2. P(A) × P(B) = 0.35 × 0.15 = 0.0525 ≠ P(A and B)
  3. not mutually exclusive, and not independent

(e) Answer: 2

  1. 22 + 11 − x + 5 = 36
  2. x = 2

(f) Answer: 29/36

  1. only Accounting: 20; only Business Studies: 9
  2. P = 29/36 = 29/36

Paper I · Section B · 18 marks

QUESTION 6
  1. (a)
    P(A) = 0.25 and P(B) = 0.35. Determine P(A or B) if A and B are mutually exclusive.
    (2)
  2. (b)
    Determine P(A or B) if A and B are independent.
    (3)
  3. (c)
    A spinner lands on red (R), blue (B) or green (G) with probabilities 0.15, 0.25 and 0.6. It is spun twice, and the spins are independent. Draw a tree diagram to show all the outcomes and their probabilities.
    (3)
  4. (d)
    Calculate the probability of one red and one green, in any order.
    (3)
  5. (e)
    Calculate the probability of at least one green.
    (2)
  6. (f)
    In a group of 30 learners, 7 take chess, 10 take drama and 2 take both. A learner is chosen at random. Calculate the probability that the learner takes chess or drama.
    (2)
  7. (g)
    Are the events 'takes chess' and 'takes drama' independent? Show your calculation.
    (3)

[18]

Show the answers

(a) Answer: 0.6

  1. mutually exclusive: P(A and B) = 0
  2. P(A or B) = 0.25 + 0.35 = 0.6

(b) Answer: 0.5125

  1. independent: P(A and B) = 0.25 × 0.35 = 0.0875
  2. P(A or B) = 0.25 + 0.35 − 0.0875 = 0.5125

(c) Answer: See the tree below.

0.15R0.15RRR: 0.02250.25BRB: 0.03750.6GRG: 0.090.25B0.15RBR: 0.03750.25BBB: 0.06250.6GBG: 0.150.6G0.15RGR: 0.090.25BGB: 0.150.6GGG: 0.36
  1. three branches for the first spin, three from each for the second
  2. each outcome's probability is the product along its branches

(d) Answer: 0.18

  1. P(RG) + P(GR) = 0.15 × 0.6 + 0.6 × 0.15
  2. = 0.18

(e) Answer: 0.84

  1. 1 − P(no green) = 1 − (0.4)^2
  2. = 0.84

(f) Answer: 1/2

  1. n(chess or drama) = 7 + 10 − 2 = 15
  2. P = 15/30 = 1/2

(g) Answer: No, they are not independent

  1. P(chess) × P(drama) = 7/30 × 1/3 = 7/90
  2. P(both) = 1/15
  3. not equal: not independent

Paper I · Section B · 11 marks

QUESTION 7

A fair die numbered 1 to 8 and a fair die numbered 1 to 6 are thrown. The number on the first die is a and the number on the second is b, and they give the point P(a; b) on the Cartesian plane.

  1. (a)
    How many different points P are possible?
    (1)
  2. (b)
    Determine the probability that P lies on the line y = x + 2.
    (2)
  3. (c)
    Determine the probability that P lies inside (not on) the circle x^2 + y^2 = 58.
    (3)
  4. (d)
    Determine the probability that P lies above the line y = 3x − 2.
    (2)
  5. (e)
    Are the events 'P lies inside the circle' and 'a is even' independent? Show your calculation.
    (3)

[11]

Show the answers

(a) Answer: 48

  1. 8 × 6 = 48

(b) Answer: 1/12

  1. b = a + 2: (1; 3); (2; 4); (3; 5); (4; 6), 4 points
  2. P = 4/48 = 1/12

(c) Answer: 35/48

  1. inside: a^2 + b^2 < 58
  2. the values of b that work for a = 1, 2, …, 8: 6 + 6 + 6 + 6 + 5 + 4 + 2 + 0 = 35
  3. P = 35/48

(d) Answer: 7/48

  1. above: b > 3a − 2: 7 points
  2. P = 7/48

(e) Answer: No, they are not independent

  1. P(inside) × P(a even) = 35/48 × 1/2 = 35/96
  2. P(inside and a even) = 16/48 = 1/3
  3. not equal: not independent

Independent events

3 questions · 37 marks

Paper I · Section B · 13 marks

QUESTION 8

Lerato and Johan take turns to kick a ball at a goal from the penalty spot, Lerato first. On each turn Lerato succeeds with probability 0.5 and Johan with probability 0.2, independently of every other turn. The first one to succeed wins the game.

  1. (a)
    Determine the probability that neither of them succeeds in the first round (one turn each).
    (2)
  2. (b)
    Determine the probability that Lerato wins the game on Lerato's second turn.
    (2)
  3. (c)
    Write the probability that Lerato wins the game as an infinite geometric series, and hence calculate it correct to four decimal places.
    (5)
  4. (d)
    Is the game fair to both players? Explain.
    (2)
  5. (e)
    Instead, Lerato and Johan each kick at the goal once, at the same time. Determine the probability that at least one of them succeeds.
    (2)

[13]

Show the answers

(a) Answer: 0.4

  1. (1 − 0.5)(1 − 0.2) = 0.5 × 0.8 = 0.4

(b) Answer: 0.2

  1. a first round with no winner, then Lerato succeeds
  2. 0.4 × 0.5 = 0.2

(c) Answer: 0.8333

  1. Lerato wins on turn 1, 2, 3, …: 0.5 + 0.4(0.5) + 0.4^2(0.5) + …
  2. a = 0.5 and r = 0.4, with −1 < r < 1
  3. S∞ = a/(1 − r) = 0.5/0.6 = 0.8333

(d) Answer: No: Lerato wins with probability 0.8333 and Johan with 0.1667, so Lerato is more likely to win

  1. P(Johan wins) = 1 − 0.8333 = 0.1667
  2. a fair game would give each 0.5

(e) Answer: 0.6

  1. P(at least one) = 1 − P(neither)
  2. = 1 − 0.4 = 0.6

Paper I · Section B · 12 marks

QUESTION 9

A and B are independent events. In the Venn diagram P(A and B) = 0.1 and P(A and B′) = 0.1; x and y are the probabilities of the other two regions.

SAB0.10.1xy
  1. (a)
    Determine P(A).
    (1)
  2. (b)
    Use the independence of A and B to determine P(B).
    (3)
  3. (c)
    Determine the values of x and y.
    (4)
  4. (d)
    Determine P(A or B).
    (2)
  5. (e)
    Determine the probability that exactly one of A and B occurs.
    (2)

[12]

Show the answers

(a) Answer: 0.2

  1. P(A) = 0.1 + 0.1 = 0.2

(b) Answer: 0.5

  1. independent: P(A and B) = P(A) × P(B)
  2. 0.1 = 0.2 × P(B)
  3. P(B) = 0.1/0.2 = 0.5

(c) Answer: x = 0.4 and y = 0.4

  1. x = P(B) − P(A and B) = 0.5 − 0.1 = 0.4
  2. y = 1 − (0.1 + 0.1 + 0.4) = 0.4

(d) Answer: 0.6

  1. P(A or B) = P(A) + P(B) − P(A and B)
  2. = 0.2 + 0.5 − 0.1 = 0.6

(e) Answer: 0.5

  1. A only or B only: 0.1 + 0.4 = 0.5

Paper I · Section B · 12 marks

QUESTION 10

120 members of a gym were asked when they train and whether they run or swim. A is the event that a member trains in the mornings and B the event that a member runs. The results are in the table below, whose last row and last column are the totals; a, b and c are missing numbers.

BB′AA′ab304545c6060120
  1. (a)
    Determine the values of a, b and c.
    (3)
  2. (b)
    A member is chosen at random. Determine the probability that the member trains in the evenings and swims.
    (2)
  3. (c)
    Determine the probability that a member chosen at random trains in the mornings or swims.
    (2)
  4. (d)
    A member is chosen at random from those who run. Determine the probability that this member trains in the mornings.
    (2)
  5. (e)
    Are the events 'training in the mornings' and 'running' independent? Show your calculation.
    (3)

[12]

Show the answers

(a) Answer: a = 15, b = 15 and c = 90

  1. c = 45 + 45 = 90
  2. a = 60 − 45 = 15
  3. b = 30 − 15 = 15

(b) Answer: 3/8

  1. P = 45/120 = 3/8

(c) Answer: 5/8

  1. n = 30 + 60 − 15 = 75
  2. P = 75/120 = 5/8

(d) Answer: 1/4

  1. only the 60 in that column count
  2. P = 15/60 = 1/4

(e) Answer: Yes, they are independent

  1. P(A) × P(B) = 30/120 × 60/120 = 1/8
  2. P(A and B) = 15/120 = 1/8
  3. equal: independent

Tree diagrams and events in stages

1 questions · 13 marks

Paper I · Section B · 13 marks

QUESTION 11

The probability that Sam's bus is on time is 0.65. If the bus is on time, the probability that Sam catches his train is 0.9; if it is late, the probability is 0.6. Let B be the event that the bus is on time and C the event that Sam catches his train.

  1. (a)
    Draw a tree diagram to show this information, with the probability on every branch.
    (3)
  2. (b)
    Determine the probability that Sam catches his train.
    (3)
  3. (c)
    Determine the probability that the bus is on time and Sam misses his train.
    (2)
  4. (d)
    Are the events B and C independent? Give a reason.
    (2)
  5. (e)
    What happens on one day is independent of any other day. Determine the probability that Sam catches his train on exactly one of two days. Give your answer correct to four decimal places.
    (3)

[13]

Show the answers

(a) Answer: See the tree below.

0.65B0.9C0.5850.1C′0.0650.35B′0.6C0.210.4C′0.14
  1. first stage: P(B) = 0.65, P(B′) = 0.35
  2. second stage: after B, 0.9 and 0.1; after B′, 0.6 and 0.4

(b) Answer: 0.795

  1. P(C) = P(B and C) + P(B′ and C)
  2. = 0.65 × 0.9 + 0.35 × 0.6
  3. = 0.795

(c) Answer: 0.065

  1. P(B and C′) = 0.65 × 0.1
  2. = 0.065

(d) Answer: No, they are not independent

  1. P(B) × P(C) = 0.65 × 0.795 = 0.51675
  2. P(B and C) = 0.585
  3. not equal: not independent

(e) Answer: 0.3260

  1. P(C) = 0.795 on each, independently
  2. 2 × 0.795 × 0.205 = 0.3260

Venn diagrams

1 questions · 15 marks

Paper I · Section B · 15 marks

QUESTION 12

99 learners were asked which of three sports they play: tennis (T), golf (G) and hockey (H).

  • 54 play tennis, 33 play golf and 41 play hockey.
  • 12 play tennis and golf but not hockey; 12 play tennis and hockey but not golf; 4 play golf and hockey but not tennis.
  • x play all three sports, and 11 play none of them.
  1. (a)
    Draw a Venn diagram to represent this information. Write the number of learners in each region, in terms of x where necessary.
    (4)
  2. (b)
    Determine the value of x.
    (3)
  3. (c)
    A learner is chosen at random. Determine the probability that the learner plays exactly one of the three sports.
    (3)
  4. (d)
    Determine the probability that the learner plays tennis or hockey.
    (2)
  5. (e)
    Are the events 'plays tennis' and 'plays golf' independent? Show your calculation.
    (3)

[15]

Show the answers

(a) Answer: See the Venn diagram below.

STGH30 − x17 − x25 − x12124x11
  1. all three: x; the three pairs as given: 12, 12 and 4
  2. tennis only: 54 − 12 − 12 − x = 30 − x; likewise golf only 17 − x and hockey only 25 − x
  3. none of them: 11

(b) Answer: x = 6

  1. the regions add up to 99: 111 − 2x = 99
  2. 2x = 12

(c) Answer: 6/11

  1. only one sport: 24 + 11 + 19 = 54
  2. P = 54/99 = 6/11

(d) Answer: 7/9

  1. n(tennis or hockey) = 54 + 41 − 18 = 77
  2. P = 77/99 = 7/9

(e) Answer: Yes, they are independent

  1. P(tennis) × P(golf) = 54/99 × 33/99 = 2/11
  2. P(tennis and golf) = 18/99 = 2/11
  3. equal: independent

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