IEB practice · Paper I
IEB Probability and counting practice
4 newly generated questions across all 9 probability and counting topics the IEB examines in Paper I, 62 marks in all. Work each one, then open its answer underneath to check yourself. Nothing here is taken from a past paper.
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The fundamental counting principle
114 marks
- (a)A lock code is made of 5 digits chosen from 1 to 9. How many different lock codes can be made if digits may be repeated?(2)
- (b)How many different lock codes can be made if no digit may be used more than once?(2)
- (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)
- (d)How many lock codes, digits allowed to repeat, start with a digit greater than 5 and end with an odd digit?(3)
- (e)How many 5-letter arrangements can be made from the letters of the word CRYSTAL, if no letter is used more than once?(2)
- (f)How many of these arrangements start with a vowel?(2)
Show the answers
(a) Answer: 59049
- 9 × 9 × 9 × 9 × 9
- = 59049
(b) Answer: 15120
- 9 × 8 × 7 × 6 × 5
- = 15120
(c) Answer: 0.7439
- P(nothing repeated) = 15120/59049
- P(at least one repeated) = 1 − 15120/59049 = 0.7439
(d) Answer: 14580
- first: 4 choices; last: 5 choices; the middle 3: 9 each
- 4 × 9 × 9 × 9 × 5 = 14580
(e) Answer: 2520
- 7 × 6 × 5 × 4 × 3
- = 2520
(f) Answer: 360
- 1 vowels for the first letter, then 6 × 5 × 4 × 3
- 1 × 360 = 360
216 marks
- (a)In how many different ways can the letters of the word BALLOON be arranged?(3)
- (b)The letters of BALLOON are arranged at random. What is the probability that the arrangement starts with A and ends with O?(3)
- (c)5 different people are arranged in a queue. In how many ways can this be done?(1)
- (d)In how many ways can this be done if 3 particular people must be next to each other, in any order?(3)
- (e)The digits 1; 2; 3; 5; 8 are used to form 5-digit numbers, without repetition (a number cannot start with 0). How many such numbers can be formed?(2)
- (f)How many of these numbers are even and greater than 80000?(4)
Show the answers
(a) Answer: 1260
- 7 letters, with L 2 times, O 2 times
- 7!/(2! × 2!) = 1260
(b) Answer: 1/21
- fix A first and O last: 60 arrangements of the other 5 letters
- P = 60/1260 = 1/21
(c) Answer: 120
- 5! = 120
(d) Answer: 36
- treat the 3 as one block: 3 units, 3! ways
- the block itself in 3! ways
- 3! × 3! = 36
(e) Answer: 120
- 5 × 4 × 3 × 2 × 1
- = 120
(f) Answer: 6
- the first digit is at least 8, the last digit even
- count each choice of first and last digit, then fill the middle
- = 6
The probability rules
318 marks
- (a)P(A) = 0.45 and P(B) = 0.4. Determine P(A or B) if A and B are mutually exclusive.(2)
- (b)Determine P(A or B) if A and B are independent.(3)
- (c)A spinner lands on red (R), blue (B) or green (G) with probabilities 0.2, 0.45 and 0.35. It is spun twice, and the spins are independent. Draw a tree diagram to show all the outcomes and their probabilities.(3)
- (d)Calculate the probability of one red and one blue, in any order.(3)
- (e)Calculate the probability of at least one red.(2)
- (f)In a group of 48 learners, 21 take chess, 9 take drama and 2 take both. A learner is chosen at random. Calculate the probability that the learner takes chess or drama.(2)
- (g)Are the events 'takes chess′ and 'takes drama′ independent? Show your calculation.(3)
Show the answers
(a) Answer: 0.85
- mutually exclusive: P(A and B) = 0
- P(A or B) = 0.45 + 0.4 = 0.85
(b) Answer: 0.67
- independent: P(A and B) = 0.45 × 0.4 = 0.18
- P(A or B) = 0.45 + 0.4 − 0.18 = 0.67
(c) Answer: See the tree below.
- three branches for the first spin, three from each for the second
- each outcome's probability is the product along its branches
(d) Answer: 0.18
- P(RB) + P(BR) = 0.2 × 0.45 + 0.45 × 0.2
- = 0.18
(e) Answer: 0.36
- 1 − P(no red) = 1 − (0.8)^2
- = 0.36
(f) Answer: 7/12
- n(chess or drama) = 21 + 9 − 2 = 28
- P = 28/48 = 7/12
(g) Answer: No, they are not independent
- P(chess) × P(drama) = 7/16 × 3/16 = 21/256
- P(both) = 1/24
- not equal: not independent
414 marks
- (a)For two events A and B, P(A) = 0.55, P(B) = 0.15 and P(A or B) = 0.6175. Determine P(A and B).(2)
- (b)Are A and B mutually exclusive? Are they independent? Give reasons.(3)
- (c)A bag holds 2 red and 5 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)
- (d)Calculate the probability that the two balls are the same colour.(2)
- (e)In a group of 36 learners, 22 take History, 14 take Geography and 3 take neither. How many take both?(2)
- (f)A learner is chosen at random. Calculate the probability that the learner takes exactly one of the two subjects.(2)
Show the answers
(a) Answer: 0.0825
- P(A or B) = P(A) + P(B) − P(A and B)
- P(A and B) = 0.55 + 0.15 − 0.6175 = 0.0825
(b) Answer: Not mutually exclusive, and independent
- P(A and B) = 0.0825 ≠ 0
- P(A) × P(B) = 0.55 × 0.15 = 0.0825 = P(A and B)
- not mutually exclusive, and independent
(c) Answer: See the tree below.
- first draw: red 2/7, blue 5/7
- second draw: 6 balls are left, one fewer of the colour taken
(d) Answer: 11/21
- P(RR) + P(BB) = 1/21 + 10/21
- = 11/21
(e) Answer: 3
- 22 + 14 − x + 3 = 36
- x = 3
(f) Answer: 5/6
- only History: 19; only Geography: 11
- P = 30/36 = 5/6
Download this booklet
The same questions as a booklet to print, with the memorandum as a separate file so you can work without the answers in front of you. In English and Afrikaans.
- Questions (English) 4 pages, 0.1 MB
- Memorandum (English) 7 pages, 0.1 MB
- Vrae (Afrikaans) 4 pages, 0.1 MB
- Memorandum (Afrikaans) 7 pages, 0.1 MB
Every question on this page was generated for this site by a computer engine, and its answer was confirmed by an independent method at the moment the question was made.