Sets and Venn Diagrams worksheet for GCSE Higher level, featuring exercises on set operations and probability with traveler activity choices.

Let's solve each part of this Sets and Venn Diagrams (B) worksheet step by step.

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✔ Section A

#### 🔹 Left Diagram: Two sets A and B

The universal set includes all numbers shown:
{0, 1, 2, 3, 5, 7, 9, 11, 13, 14, 17, 20}

- Set A: {1, 3, 7, 11, 2, 14}
- Set B: {5, 20, 9, 13, 2, 14}

##### 1. $ A \cap B $ = Elements in both A and B
- Common elements: 2, 14
- So, $ A \cap B = \{2, 14\} $

##### 2. $ A' \cup B $ = Complement of A union B
- $ A' $ = All elements not in A
- Universal set: {0, 1, 2, 3, 5, 7, 9, 11, 13, 14, 17, 20}
- A = {1, 3, 7, 11, 2, 14}
- So $ A' = \{0, 5, 9, 13, 17, 20\} $
- Now $ A' \cup B = \{0, 5, 9, 13, 17, 20\} \cup \{5, 20, 9, 13, 2, 14\} $
- Combine: {0, 2, 5, 9, 13, 14, 17, 20}
- So $ A' \cup B = \{0, 2, 5, 9, 13, 14, 17, 20\} $

##### 3. $ (A \cap B)' $ = Complement of $ A \cap B $
- $ A \cap B = \{2, 14\} $
- So complement is everything except 2 and 14
- $ (A \cap B)' = \{0, 1, 3, 5, 7, 9, 11, 13, 17, 20\} $

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#### 🔹 Right Diagram: Three sets A, B, C

Universal set: {a, b, c, d, e, f, g, h, i, j, k, m, p, r, s, z} — all letters shown.

- A = Letters in circle A: {f, z, e, i, b}
- B = Letters in circle B: {z, p, r, m, e, h}
- C = Letters in circle C: {e, h, g, i, b}

##### 1. $ A = \{f, z, e, i, b\} $

##### 2. $ B' $ = Complement of B → all elements not in B
- B = {z, p, r, m, e, h}
- Universal set: {s, d, c, f, z, p, r, k, j, m, e, h, g, i, b}
- So $ B' = \{s, d, c, f, k, j, g, i, b\} $ — wait! Wait: check if b is in B?

Wait: Look at diagram:
- B has: z, p, r, m, e, h → so b is not in B, but b is in A and C
- So yes, b is in B'? Yes.
- But let’s list full universal set from diagram:

All elements outside or inside circles:
- Outside: s, d, c, k, j
- Inside: f, z, p, r, m, e, h, g, i, b

So universal set: {s, d, c, k, j, f, z, p, r, m, e, h, g, i, b}

Now B = {z, p, r, m, e, h}

So $ B' = $ all except those: {s, d, c, k, j, f, g, i, b}

So $ B' = \{s, d, c, k, j, f, g, i, b\} $

##### 3. $ A \cap B \cap C $ = Elements in all three sets
- A ∩ B = {z, e} (common to A and B)
- Then intersect with C: {e} is in C, z is not in C
- So only e is in all three
- $ A \cap B \cap C = \{e\} $

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✔ Section B

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#### 🔹 HOLIDAY 1: Camping vs Fishing

Venn diagram:
- Only Camping: 11
- Both: 5
- Only Fishing: 8
- Outside (neither): 24

Total travelers = 11 + 5 + 8 + 24 = 48

##### 1) How many chose camping?
- Camping = only camping + both = 11 + 5 = 16

✔ Answer: 16

##### 2) How many did not choose fishing?
- Not fishing = only camping + neither = 11 + 24 = 35

✔ Answer: 35

##### 3) How many did not choose camping or fishing?
- That means neither = 24

✔ Answer: 24

##### 4) How many only chose fishing?
- That’s the part of fishing not overlapping = 8

✔ Answer: 8

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#### 🔹 HOLIDAY 2: Camping, Fishing, Water Sports

Venn diagram with three sets:

- Only Camping: 10
- Only Fishing: 3
- Only Water Sports: 18
- Camping & Fishing only: 1
- Camping & Water Sports only: 6
- Fishing & Water Sports only: 1
- All three: 2
- Outside: 9

Total travelers on Holiday 2 = sum of all regions:
= 10 (C only) + 3 (F only) + 18 (W only) + 1 (C∩F) + 6 (C∩W) + 1 (F∩W) + 2 (all three) + 9 (outside)

= 10+3+18+1+6+1+2+9 = 48

##### 1) How many travellers are on Holiday 2?
- Total = 48

✔ Answer: 48

##### 2) What is the probability that a traveller chooses camping, fishing, and water sports?
- Number who chose all three = 2
- Total = 48
- So $ P(\text{all three}) = \frac{2}{48} = \frac{1}{24} $

✔ Answer: $ \frac{1}{24} $

##### 3) Work out $ P((C \cup W)') $

This is the probability that a traveler is not in C or W, i.e., not in Camping OR Water Sports

So:
- $ C \cup W $ = everyone in Camping or Water Sports or both
- $ (C \cup W)' $ = everyone not in C or W, i.e., only in Fishing or outside

From diagram:
- Only Fishing: 3
- Only Fishing & Water Sports: 1
- Only Fishing & Camping: 1
- But wait: we need not in C or W

So only people not in C and not in W

- Who is not in C? → not in Camping
- Who is not in W? → not in Water Sports
- So $ (C \cup W)' $ = only in Fishing and not in C or W

But note: Fishing only = 3 (this person is only in F, not in C or W)

Also: Is there anyone in Fishing but not in C or W? Yes: only Fishing = 3

And also: The region outside all circles = 9

But outside is not in any activity → definitely not in C or W

So $ (C \cup W)' $ = only Fishing + outside = 3 + 9 = 12

So $ P((C \cup W)') = \frac{12}{48} = \frac{1}{4} $

✔ Answer: $ \frac{1}{4} $

##### 4) Work out $ P(C \cup W') $

This is P(C ∪ W'), where:
- $ W' $ = complement of Water Sports = not in Water Sports
- So $ C \cup W' $ = people in Camping OR not in Water Sports

We can use formula:
$ P(C \cup W') = P(C) + P(W') - P(C \cap W') $

But easier to count directly.

Total = 48

We want: all people who are either in Camping OR not in Water Sports

Let’s find how many are in Camping:
- Camping only: 10
- Camping & Fishing only: 1
- Camping & Water Sports only: 6
- All three: 2
- Total in Camping = 10 + 1 + 6 + 2 = 19

Now, not in Water Sports ($ W' $):
- People not in Water Sports = total – those in Water Sports
- Water Sports = only W: 18, C&W: 6, F&W: 1, all three: 2 → total = 18+6+1+2 = 27
- So $ W' = 48 - 27 = 21 $

But now $ C \cup W' $ = people in C OR not in W

Use inclusion:
$ |C \cup W'| = |C| + |W'| - |C \cap W'| $

Now $ C \cap W' $ = people in Camping AND NOT in Water Sports
- From diagram: Camping only (10), Camping & Fishing only (1) → these are in C but not in W
- So $ C \cap W' = 10 + 1 = 11 $

So:
$ |C \cup W'| = 19 + 21 - 11 = 29 $

So $ P(C \cup W') = \frac{29}{48} $

✔ Answer: $ \frac{29}{48} $

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✔ Final Answers Summary

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#### Section A

Left Diagram:
- $ A \cap B = \{2, 14\} $
- $ A' \cup B = \{0, 2, 5, 9, 13, 14, 17, 20\} $
- $ (A \cap B)' = \{0, 1, 3, 5, 7, 9, 11, 13, 17, 20\} $

Right Diagram:
- $ A = \{f, z, e, i, b\} $
- $ B' = \{s, d, c, k, j, f, g, i, b\} $
- $ A \cap B \cap C = \{e\} $

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#### Section B

Holiday 1:
1. Chose camping: 16
2. Did not choose fishing: 35
3. Did not choose camping or fishing: 24
4. Only chose fishing: 8

Holiday 2:
1. Travellers on Holiday 2: 48
2. Probability chose all three: $ \frac{1}{24} $
3. $ P((C \cup W)') = \frac{1}{4} $
4. $ P(C \cup W') = \frac{29}{48} $

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