Probability: Venn Diagrams & Rules (Grade 11) — Worksheet & Memo

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Grade 11 · Probability

Probability: Venn Diagrams & Rules (Grade 11)

MARKING GUIDELINE

Marks

Duration

1 hour 10 minutes

Questions

Name:

Class:

Date:

Mark

/ 46

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
P(A or B) = …
(1)
A)P(A) + P(B) − P(A and B)✓
B)P(A) + P(B)
C)P(A) × P(B)
D)P(A) + P(B) + P(A and B)
Answer: A — Without the subtraction the overlap would be counted twice.
- B — that holds only when the events are MUTUALLY EXCLUSIVE
- C — that gives P(A and B) for INDEPENDENT events, not P(A or B)
- D — the overlap must be SUBTRACTED, not added again
1.2
Events A and B are INDEPENDENT when …
(1)
A)P(A and B) = P(A) × P(B)✓
B)P(A and B) = 0
C)P(A) + P(B) = 1
D)P(A) = P(B)
Answer: A — Independence means one event happening does not change the probability of the other.
- B — that is MUTUALLY EXCLUSIVE, which is a different idea
- C — that describes COMPLEMENTARY events
- D — equal probabilities say nothing about independence
1.3
P(A) = 0,6 and P(B) = 0,5, and A and B are independent. Then P(A and B) = …
(1)
A)0,3✓
B)1,1
C)0,1
D)0,8
Answer: A — For independent events the probabilities multiply: 0,6 × 0,5 = 0,3.
- B — ADDED the probabilities instead of multiplying them
- C — subtracted them
- D — used the addition rule, which gives P(A or B)
1.4
Mutually exclusive events with non-zero probabilities are …
(1)
A)never independent✓
B)always independent
C)independent only if P(A) = P(B)
D)always complementary
Answer: A — If A happens then B cannot — which is exactly a change in B's probability.
- B — P(A and B) = 0 while P(A) × P(B) is not 0, so they cannot be independent
- C — equal probabilities do not help — the overlap is still zero
- D — complementary events must fill the whole sample space, which these need not do
1.5
Of 100 people, 60 like tea, 45 like coffee and 20 like both. The number liking NEITHER is …
(1)
A)15✓
B)5
C)25
D)35
Answer: A — 60 + 45 − 20 = 85 like at least one, so 100 − 85 = 15 like neither.
- B — subtracted the overlap twice over
- C — never subtracted the 20 counted in both groups
- D — used 60 + 45 = 105 and then subtracted the wrong figure
1.6
For that same group, P(likes tea ONLY) = …
(1)
A)0,4✓
B)0,6
C)0,2
D)0,85
Answer: A — Tea only is 60 − 20 = 40 out of 100.
- B — used all 60 tea-drinkers instead of only those who like tea alone
- C — used the 20 who like both
- D — that is P(likes at least one of the two)
1.7
If P(A) = 0,4 and P(A and B) = 0,1, then P(A but NOT B) = …
(1)
A)0,3✓
B)0,5
C)0,1
D)0,4
Answer: A — P(A only) = P(A) − P(A and B) = 0,4 − 0,1 = 0,3.
- B — ADDED the overlap instead of subtracting it
- C — that is the overlap itself
- D — that is P(A), before the overlap is removed
1.8
A bag holds 4 red and 6 blue marbles. Two are drawn WITHOUT replacement. P(both red) = …
(1)
A)215✓
B)425
C)25
D)12100
Answer: A — 4/10 × 3/9 = 12/90 = 2/15 — the second draw has one fewer red and one fewer marble.
- B — treated the draws as WITH replacement: 4/10 × 4/10
- C — gave the probability of ONE red only
- D — reduced the reds but not the total: 4/10 × 3/10
1.9
From that same bag, but WITH replacement, P(both red) = …
(1)
A)425✓
B)215
C)25
D)810
Answer: A — With replacement the bag is unchanged, so it is 4/10 × 4/10 = 4/25.
- B — that is the WITHOUT-replacement answer
- C — gave the probability of ONE red only
- D — ADDED the two probabilities instead of multiplying them
1.10
If P(not A) = 0,35, then P(A) = …
(1)
A)0,65✓
B)0,35
C)1,35
D)0,3
Answer: A — P(A) = 1 − P(not A) = 0,65.
- B — that is P(not A) itself
- C — ADDED to 1 instead of subtracting from it
- D — subtracted 0,35 from 0,65 rather than from 1

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

[13 MARKS]

150 Grade 11 learners were asked which of three streaming services they use: Netflix (N), Showmax (M) or YouTube (Y). The results are shown in the Venn diagram below, where x and y represent numbers of learners. It is further given that n(N) = 75.

2.1
Calculate the value of x.
(3)
n(N) = 31 + 18 + 14 + x (1)
75 = 63 + x (1)
x = 12 (1)
2.2
Hence calculate the value of y.
(3)
All eight regions must add up to 150 (1)
31 + 18 + 14 + 12 + 22 + 27 + 9 + y = 150 (1)
133 + y = 150, so y = 17 (1)
2.3
Calculate the probability that a learner chosen at random from this group uses EXACTLY ONE of the three services.
(3)
Only the three outer regions count: 31 + 17 + 27 (1)
= 75 learners (1)
P = 75/150 = 0,50 (1)
2.4
Calculate the probability that a learner chosen at random uses Showmax OR YouTube.
(4)
n(M) = 18 + 14 + 22 + 17 = 71 (1)
n(Y) = 27 + 22 + 14 + 12 = 75 (1)
n(M or Y) = 71 + 75 − 36 = 110 (the addition rule) (1)
P(M or Y) = 110/150 = 0,73 (1)

Question 3

[13 MARKS]

The Venn diagram below shows the PROBABILITIES associated with two events A and B in a sample space S. The probability that both A and B occur is m.

3.1
Calculate the value of m.
(3)
The four regions of a Venn diagram cover the whole sample space, so they add up to 1 (1)
0,28 + m + 0,32 + 0,19 = 1 (1)
m = 0,21 (1)
3.2
Determine, showing ALL calculations, whether A and B are independent events.
(4)
P(A) = 0,28 + 0,21 = 0,49 (1)
P(B) = 0,32 + 0,21 = 0,53 (1)
P(A) × P(B) = 0,2597 (1)
but P(A and B) = 0,21 ≠ 0,2597, so A and B are NOT independent (1)
3.3
Are A and B mutually exclusive? Give a reason for your answer.
(2)
No (1)
P(A and B) = 0,21 ≠ 0, so the two events CAN happen together — the circles overlap (1)
3.4
Calculate P(not A or not B).
(4)
"not A or not B" fails only when A and B BOTH occur (1)
∴ P(not A or not B) = 1 − P(A and B) (1)
= 1 − 0,21 (1)
= 0,79 (1)

Question 4

[10 MARKS]

Two events A and B are such that P(A) = 0,4 and P(A or B) = 0,7.

4.1
Determine P(B) if A and B are mutually exclusive.
(3)
Mutually exclusive means P(A and B) = 0 (1)
P(A or B) = P(A) + P(B): 0,7 = 0,4 + P(B) (1)
P(B) = 0,3 (1)
4.2
Determine P(B) if A and B are independent.
(4)
Independent means P(A and B) = P(A) × P(B) (1)
0,7 = 0,4 + P(B) − 0,4P(B) (1)
0,3 = 0,6P(B) (1)
P(B) = 0,5 (1)
4.3
Hence, for the independent case, calculate P(A and B) and the probability that B occurs but A does NOT.
(3)
P(A and B) = 0,4 × 0,5 = 0,2 (1)
P(only B) = P(B) − P(A and B) (1)
= 0,5 − 0,2 = 0,3 (1)

TOTAL: 46 marks

This question paper consists of 4 questions.