Counting principle & probability — Mathematics Grade 12

4 exam-style question sets on Counting principle & probability (Paper 1), each with a hint and a fully worked answer. The app holds 20 questions on this section in total, including variants of every set below, and lets you mark yourself part by part.

Independent events, tree diagrams & arrangementsVenn diagrams & independent eventsFundamental counting principleTwo-way tables, independence & arrangements

Practise this section in the app →

Question 1

Independent events, tree diagrams & arrangementsRoutine15 marksPaper 1

Parts (a) and (b) refer to the following situation:

On his way to work in Polokwane, Mr Maluleke drives through two intersections that are controlled by traffic lights (robots). The probability that the first robot shows green when he reaches it is $0, 4$ , and the probability that the second robot shows green when he reaches it is $0, 55$ . The two robots operate independently of each other.

Parts (c) and (d) refer to the following situation:

The probability that a randomly chosen Grade 12 learner at a school in Soweto attends the Saturday revision classes is $0, 65$ . A learner who attends these classes passes the preparatory examination with a probability of $0, 9$ , while a learner who does NOT attend passes with a probability of $0, 5$ .

Parts (e) and (f) refer to the following situation:

Seven friends, among them Palesa and Nandi, arrive together at the gate of a jazz festival and stand one behind the other in a single queue.

aCalculate the probability that BOTH robots show green when Mr Maluleke reaches them. (2)
bCalculate the probability that at least ONE of the two robots shows green when he reaches it. (2)
cDraw a tree diagram to represent this information. Show the probability on EACH branch and label ALL the outcomes. (3)
dCalculate the probability that a Grade 12 learner chosen at random from this school passes the preparatory examination. (3)
eIn how many different ways can the seven friends stand in the queue? (1)
fThe friends take their places in the queue at random. Calculate the probability that exactly TWO of the other friends stand between Palesa and Nandi. (4)

Hint

For independent events multiply: $P (A and B) = P (A) \times P (B)$ . For 'at least one' use $P (A or B) = P (A) + P (B) - P (A and B)$ , or subtract the probability that NEITHER happens from 1. On the tree, the two branches leaving any point must add up to 1, and the second-stage probabilities depend on which first-stage branch they follow. For the queue, number the places 1 to 7 and list the pairs of places that have exactly two places between them; then remember that Palesa and Nandi can swap places and that the other five friends can be arranged in the remaining places.

Worked answer

a) The events are independent, so P(both green) $=$ P(first green) $\times$ P(second green)
$= 0, 4 \times 0, 55$ ✓ $= 0, 22$ ✓

b) P(at least one green) $=$ P(first green) $+$ P(second green) $-$ P(both green)
$= 0, 4 + 0, 55 - 0, 22$ ✓ $= 0, 73$ ✓
(OR: $1 -$ P(neither green) $= 1 - (0, 6) (0, 45) = 1 - 0, 27 = 0, 73$ )

c) First stage (Saturday classes): attends $0, 65$ ; does NOT attend $1 - 0, 65 = 0, 35$ ✓
Second stage after 'attends': passes $0, 9$ ; fails $1 - 0, 9 = 0, 1$ ✓
Second stage after 'does NOT attend': passes $0, 5$ ; fails $1 - 0, 5 = 0, 5$ ✓

Saturday classes	Preparatory examination	Probability of the path
Attends ( $0, 65$ )	Passes ( $0, 9$ )	$0, 65 \times 0, 9 = 0, 585$
Attends ( $0, 65$ )	Fails ( $0, 1$ )	$0, 65 \times 0, 1 = 0, 065$
Does NOT attend ( $0, 35$ )	Passes ( $0, 5$ )	$0, 35 \times 0, 5 = 0, 175$
Does NOT attend ( $0, 35$ )	Fails ( $0, 5$ )	$0, 35 \times 0, 5 = 0, 175$

(The four path probabilities add up to 1.)

d) P(passes) $=$ P(attends and passes) $+$ P(does NOT attend and passes) ✓
$= (0, 65) (0, 9) + (0, 35) (0, 5)$ ✓
$= 0, 585 + 0, 175 = 0, 76$ ✓

e) $7! = 5040$ different ways ✓

f) Number the places in the queue 1 to 7. With exactly TWO friends between them, Palesa and Nandi must occupy places $(1; 4)$ , $(2; 5)$ , $(3; 6)$ or $(4; 7)$ : 4 pairs of places ✓
In each pair of places they can stand in 2 orders: $4 \times 2 = 8$ ✓
The other 5 friends fill the remaining places in $5!$ ways, so there are $8 \times 5! = 960$ favourable arrangements ✓
P(exactly two between them) $= \frac{960}{5 040} = \frac{4}{21} \approx 0, 19$ ✓

Modelled on Nov 2023 Paper 1, Question 10 — same skills, new scenario.

Question 2

Venn diagrams & independent eventsRoutine9 marksPaper 1

A survey of the 150 Grade 12 learners at a school looked at three subjects: Mathematics ( $M$ ), Physical Sciences ( $P$ ) and Life Sciences ( $L$ ). EVERY learner takes at least ONE of these subjects.

90 learners take Mathematics
70 learners take Physical Sciences
50 learners take Life Sciences
25 learners take Mathematics and Physical Sciences but NOT Life Sciences
15 learners take Mathematics and Life Sciences but NOT Physical Sciences
10 learners take Physical Sciences and Life Sciences but NOT Mathematics

aDraw a Venn diagram to represent the information above and use it to calculate how many learners take ALL THREE subjects. (3)
bCalculate the probability that a randomly chosen learner takes at least TWO of these subjects. (2)
cDetermine, with the necessary calculations, whether the events "takes Mathematics" and "takes Physical Sciences" are independent. (4)

Hint

Let the number taking all three be x, fill the diagram from the middle outwards, and remember all seven regions must add up to 150.

Worked answer

a) Let $x$ take all three. Then $150 = 90 + 70 + 50 - 25 - 15 - 10 - 2 x$ ✓, so $2 x = 10$ and $x = 5$ ✓. Venn regions: all three $= 5$ ; $M$ and $P$ only $= 25$ ; $M$ and $L$ only $= 15$ ; $P$ and $L$ only $= 10$ ; only $M = 90 - 25 - 15 - 5 = 45$ ; only $P = 70 - 25 - 10 - 5 = 30$ ; only $L = 50 - 15 - 10 - 5 = 20$ ✓ (regions total $45 + 30 + 20 + 25 + 15 + 10 + 5 = 150$ )

b) P(at least two) $= \frac{25 + 15 + 10 + 5}{150}$ ✓ $= \frac{55}{150} = \frac{11}{30} \approx 0, 37$ ✓

c) $P (M) = \frac{90}{150} = \frac{3}{5}$ and $P (P) = \frac{70}{150} = \frac{7}{15}$ , so $P (M) \times P (P) = \frac{7}{25} = 0, 28$ ✓✓. $P (M and P) = \frac{25 + 5}{150} = \frac{1}{5} = 0, 2$ ✓. Since $0, 28 \neq = 0, 2$ , $P (M and P) \neq = P (M) \times P (P)$ , so the events are NOT independent ✓

Modelled on Nov 2024 Paper 1, Question 11 — same skills, new scenario.

Question 3

Fundamental counting principleComplex8 marksPaper 1

A gym gives every member a 4-character locker code in the pattern letter–digit–letter–digit, for example K3M8. The letters are chosen from the 26 letters of the alphabet and the digits from 0 to 9.

aHow many different locker codes are possible if letters and digits MAY be repeated? (2)
bFor security the gym now insists that: no vowel may be used; the FIRST letter may also not be Q, X or Z; no letter or digit may be repeated; and the code must end on an EVEN digit. How many codes are possible now? (4)
cThe gym later allows vowels again, while ALL the other rules in (b) stay in force. Calculate the percentage increase in the number of possible codes. Round off to TWO decimal places. (2)

Hint

Multiply the number of choices position by position, and in the restricted count fill the final digit and the first letter before the unrestricted positions.

Worked answer

a) $26 \times 10 \times 26 \times 10 = 67600$ codes ✓✓

b) Fill the most restricted positions first. Last digit (even): $5$ choices ✓; other digit: $9$ remaining ✓; first letter (21 consonants minus Q, X, Z): $18$ ✓; second letter: $20$ remaining consonants. Total $= 18 \times 9 \times 20 \times 5 = 16200$ codes ✓

c) With vowels back: first letter $26 - 3 = 23$ , second letter $25$ , digits as before: $23 \times 9 \times 25 \times 5 = 25875$ codes ✓. Percentage increase $= \frac{25 875 - 16 200}{16 200} \times 100 = 59, 72%$ ✓

Modelled on Nov 2024 Paper 1, Question 12 — same skills, new scenario.

Question 4

Two-way tables, independence & arrangementsProblem-solving16 marksPaper 1

A coffee shop kept a record of the age group of each of the 200 customers it served on a particular Saturday, and whether the customer ordered a COLD drink or a HOT drink. Some of the results are summarised in the table below.

	Cold drink	Hot drink	Total
Younger than 30	20	$m$	$k$
30 years or older	$r$	$s$	$t$
Total	50	$u$	200

The events 'the customer is younger than 30' and 'the customer orders a cold drink' are independent.

aUse the given information to show that $k = 80$ . (3)
bComplete the table and hence determine the probability that a randomly chosen customer aged 30 years or older ordered a HOT drink. (3)
cA community swimming pool had 1 500 visitors in January. Each visitor came on either a weekday or a weekend day. The probability that a visitor who came on a weekday rented a locker was $0, 2$ , and the probability that a visitor who came on a weekend day rented a locker was $0, 35$ . The probability that a randomly selected visitor from the whole of January rented a locker was $0, 26$ . Calculate how many visitors came on weekend days. (4)
dSeven cyclists, including Anele and Musa, take part in a road race and all seven finish at different times. In how many different ways can the seven cyclists fill the finishing positions if Musa finishes immediately after Anele? (2)
eIf the seven cyclists finish in a random order, calculate the probability that Anele finishes ahead of Musa with AT LEAST FOUR of the other cyclists finishing between them. (4)

Hint

Use P(A and B) = P(A) × P(B) to find the missing total, let x be the number of weekend visitors and balance the weighted equation against 0,26 × 1 500, glue the two cyclists into one block, and count favourable position pairs before multiplying by the arrangements of the rest.

Worked answer

a) For independent events: P(younger than 30 AND cold drink) $=$ P(younger than 30) $\times$ P(cold drink) ✓, so $\frac{20}{200} = \frac{k}{200} \times \frac{50}{200}$ ✓. This gives $\frac{1}{10} = \frac{k}{800}$ , so $k = 80$ ✓

b) Completing the table: $m = 80 - 20 = 60$ ; $r = 50 - 20 = 30$ ; $t = 200 - 80 = 120$ ; $s = 120 - 30 = 90$ ; $u = 200 - 50 = 150$ ✓. P(hot drink from the 30-or-older group) $= \frac{90}{120}$ ✓ $= \frac{3}{4} = 0, 75$ ✓

c) Let $x$ be the number of weekend visitors; then $1500 - x$ came on weekdays ✓. $0, 35 x + 0, 2 (1500 - x) = 0, 26 \times 1500$ ✓ $\Rightarrow 0, 15 x + 300 = 390$ ✓ $\Rightarrow x = 600$ . So 600 visitors came on weekend days ✓

d) Treat Anele followed directly by Musa as ONE unit, so 6 objects must be arranged ✓: $6! = 720$ ways ✓

e) Consider the finishing positions (Anele; Musa). With Anele ahead and AT LEAST FOUR others between them, the only position pairs are $(1; 6)$ , $(1; 7)$ and $(2; 7)$ ✓✓. The other five cyclists fill the remaining places in $5!$ ways ✓, so P $= \frac{3 \times 5 !}{7 !} = \frac{360}{5 040} = \frac{1}{14} \approx 0, 07$ ✓

Modelled on Nov 2025 Paper 1, Question 11 — same skills, new scenario.

Open Counting principle & probability in the app →

Grade 12 Mathematics: Counting principle & probability

Question 1

Question 2

Question 3

Question 4

About this material