Counting Principles & Probability (Grade 12) — Worksheet & Memo

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Grade 12 · Probability

Counting Principles & Probability (Grade 12)

MARKING GUIDELINE

Marks

Duration

50 minutes

Questions

Name:

Class:

Date:

Mark

/ 33

Instructions and Information

Answer ALL the questions in this question paper.
Answer QUESTION 1 by circling the letter (A–D) in the answer grid at the end of that section.
Show ALL calculations clearly.
Show all units where applicable.
Number the answers correctly according to the numbering system used in this question paper.
A non-programmable calculator may be used, unless stated otherwise.
Write neatly and legibly.

Question 1

[10 MARKS]

Four options are given as possible answers to the following questions. Choose the correct answer and circle the letter (A–D) in the grid at the end of this section. If you want to change your choice, put a cross through the wrong letter and circle your new choice.

1.1
The number of ways to arrange 5 different books in a row is …
(1)
A)120✓
B)25
C)5
D)10
Answer: A — 5! = 5 × 4 × 3 × 2 × 1 = 120.
- B — squared the 5 instead of taking its factorial
- C — gave the number of books, not the number of arrangements
- D — 5 × 2 is not a factorial
1.2
The FUNDAMENTAL COUNTING PRINCIPLE says the total number of ways is found by …
(1)
A)multiplying the number of choices at each stage✓
B)adding the number of choices at each stage
C)taking the largest number of choices
D)taking the factorial of the number of stages
Answer: A — Every choice at one stage can pair with every choice at the next.
- B — adding counts alternatives, not sequences
- C — every stage contributes, not only the largest
- D — the factorial applies only when ALL the items are being arranged
1.3
A PIN has 4 digits, each from 0 to 9, and repeats ARE allowed. The number of possible PINs is …
(1)
A)10 000✓
B)5 040
C)40
D)24
Answer: A — There are 10 choices for each of the 4 positions, giving 10⁴.
- B — that is 10 × 9 × 8 × 7, which assumes NO repeats
- C — multiplied 10 by 4 instead of raising to the power
- D — that is 4!, the arrangements of four fixed digits
1.4
If repeats are NOT allowed, the number of 4-digit PINs is …
(1)
A)5 040✓
B)10 000
C)24
D)40
Answer: A — Each digit used removes one choice from the next position: 10 × 9 × 8 × 7.
- B — that allows repeats
- C — that is 4!, ignoring the ten available digits
- D — multiplied 10 by 4
1.5
The number of ways to arrange the letters of the word LEVEL is …
(1)
A)30✓
B)120
C)60
D)20
Answer: A — Five letters with L repeated twice and E repeated twice: 5! ÷ (2! × 2!) = 30.
- B — treated all five letters as different
- C — divided by only one of the two repeats
- D — divided by 3! instead of by 2! × 2!
1.6
Five people sit in a row, but two particular people must sit TOGETHER. The number of arrangements is …
(1)
A)48✓
B)120
C)24
D)240
Answer: A — Treat the pair as one block: 4! ways to arrange the blocks, times 2! within the pair.
- B — that is the number with no restriction at all
- C — forgot to let the pair swap places
- D — multiplied by 10 instead of by 2
1.7
Four letters are arranged at random. The probability of one particular arrangement is …
(1)
A)124✓
B)14
C)116
D)112
Answer: A — There are 4! = 24 equally likely arrangements.
- B — that is 1 ÷ the number of letters, not of arrangements
- C — that is 1 ÷ 4², which would allow repeats
- D — that halves the correct count
1.8
Three coins are tossed and a die is rolled. The sample space has …
(1)
A)48 outcomes✓
B)18 outcomes
C)12 outcomes
D)36 outcomes
Answer: A — 2 × 2 × 2 × 6 = 48.
- B — ADDED the three coins' outcomes to get 3, then multiplied by 6
- C — used 2 × 6 and ignored two of the coins
- D — used 6 × 6 instead of 2³ × 6
1.9
In how many ways can a chairperson and a secretary be chosen from 8 people?
(1)
A)56✓
B)64
C)28
D)40 320
Answer: A — There are 8 choices for the chair and then 7 for the secretary.
- B — allowed the same person to hold both posts
- C — that is 8 × 7 ÷ 2, which would apply if the two posts were identical
- D — that is 8!, which arranges all eight people
1.10
Two events are INDEPENDENT when …
(1)
A)P(A and B) = P(A) × P(B)✓
B)P(A and B) = 0
C)P(A or B) = P(A) + P(B)
D)P(A) = P(B)
Answer: A — One event happening leaves the other's probability unchanged.
- B — that is MUTUALLY EXCLUSIVE
- C — that also describes mutually exclusive events
- D — equal probabilities say nothing about independence

Answer grid — marking guideline

1.1	A	B	C	D
1.2	A	B	C	D
1.3	A	B	C	D
1.4	A	B	C	D
1.5	A	B	C	D
1.6	A	B	C	D
1.7	A	B	C	D
1.8	A	B	C	D
1.9	A	B	C	D
1.10	A	B	C	D

Question 2

[11 MARKS]

The letters of the word NUMBERS are arranged in a row. The word has 7 different letters.

2.1
Determine the number of different arrangements possible, and how many of them begin with the letter N.
(3)
7! (1)
= 5 040 (1)
Fixing N in the first position leaves 6 letters to arrange: 6! = 720 (1)
2.2
Determine the number of arrangements in which the letters N and U are next to each other.
(4)
Treat NU as one item, leaving 6 items: 6! (2)
N and U can swap: × 2 (1)
= 1 440 (1)
2.3
Hence determine the probability that in a random arrangement N and U are NOT next to each other.
(4)
Not together = 5 040 − 1 440 = 3 600 (2)
P = 3 6005 040 (1) = 57 (1)

Question 3

[12 MARKS]

A code consists of 4 digits chosen from 0 to 9.

3.1
Determine how many codes are possible if digits may be repeated, and how many of those begin with an EVEN digit.
(3)
10⁴ (1)
= 10 000 (1)
There are 5 even digits (0; 2; 4; 6; 8) for the first position, so 5 × 10³ = 5 000 (1)
3.2
Determine how many codes are possible if no digit may be repeated.
(3)
10 × 9 × 8 × 7 (2) = 5 040 (1)
3.3
Determine the probability that a randomly chosen 4-digit code has no repeated digit.
(3)
5 04010 000 (2) = 0,504 (1)
3.4
Explain why the answer to the previous part would be smaller for a 5-digit code, and calculate that probability.
(3)
Each extra position has FEWER digits still available, so the fraction of codes with no repeat keeps shrinking (1)
P = 10 × 9 × 8 × 7 × 610⁵ = 30 240100 000 (1)
= 0,3024, which is indeed less than 0,504 (1)

TOTAL: 33 marks

This question paper consists of 3 questions.