Sets and Venn Diagrams (B) Worksheet | Geometry PDF Worksheets - Free Printable

Let’s work through each part of the worksheet step by step.

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Section A — Left Diagram (Numbers)

We have two sets:
- Set A contains: 1, 3, 7, 11, 2, 14
- Set B contains: 5, 9, 13, 20, 2, 14
- Outside both sets: 0, 17

So total universe = {0, 1, 2, 3, 5, 7, 9, 11, 13, 14, 17, 20}

Now answer:

A ∩ B → Elements in BOTH A and B → only 2, 14

✔ So: A ∩ B = {2, 14}

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A’ ∪ B → “Not A” union “B”

First, A’ = everything NOT in A → that’s {0, 5, 9, 13, 17, 20}
Then union with B = {5, 9, 13, 20, 2, 14}

Combine them: {0, 2, 5, 9, 13, 14, 17, 20}

Wait — let’s double-check:
A’ is all elements except those in A → so remove 1, 3, 7, 11 from full set → left with 0, 2, 5, 9, 13, 14, 17, 20? Wait no — 2 and 14 are in A, so they should NOT be in A’.

Actually, A = {1, 3, 7, 11, 2, 14} → so A’ = {0, 5, 9, 13, 17, 20}

B = {2, 5, 9, 13, 14, 20}

So A’ ∪ B = {0, 5, 9, 13, 17, 20} ∪ {2, 5, 9, 13, 14, 20} = {0, 2, 5, 9, 13, 14, 17, 20}

✔ So: A’ ∪ B = {0, 2, 5, 9, 13, 14, 17, 20}

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(A ∩ B)’ → Complement of intersection → everything EXCEPT what’s in both A and B

We know A ∩ B = {2, 14}, so complement is all other elements in universe:
{0, 1, 3, 5, 7, 9, 11, 13, 17, 20}

✔ So: (A ∩ B)’ = {0, 1, 3, 5, 7, 9, 11, 13, 17, 20}

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Section A — Right Diagram (Letters)

Sets:
A = {f, z, e, b, i}
B = {z, p, r, m, e, h}
C = {b, i, e, h, g}
Outside: s, d, c, k, j

Total universe = {s, d, c, f, z, p, r, m, e, h, b, i, g, k, j}

Now answer:

A = → All letters inside circle A → {f, z, e, b, i}

✔ A = {f, z, e, b, i}

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B’ → Not in B → everything except elements in B

B = {z, p, r, m, e, h}
So B’ = {s, d, c, f, b, i, g, k, j}

Check: remove z,p,r,m,e,h from total → yes, leaves above.

✔ B’ = {s, d, c, f, b, i, g, k, j}

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A ∩ B ∩ C → Elements in ALL THREE sets

Look at center where all three overlap → only e

Check:
A has e? Yes
B has e? Yes
C has e? Yes
Any others? No — b,i are in A and C but not B; h is in B and C but not A; z is in A and B but not C.

✔ A ∩ B ∩ C = {e}

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Section B — Holiday 1

Venn diagram:
Camping only: 11
Fishing only: 8
Both: 5
Neither: 24

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

Questions:

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

✔ Answer: 16

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2) Did not choose fishing?
→ Those who did NOT pick fishing = Camping only + Neither = 11 + 24 = 35

✔ Answer: 35

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3) Did not choose camping or fishing?
→ That means neither → given as 24

✔ Answer: 24

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4) Only chose fishing?
→ Fishing only = 8

✔ Answer: 8

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Section B — Holiday 2

Three-circle Venn: Camping (C), Fishing (F), Water Sports (W)

Regions:
- Only C: 10
- Only F: 3
- Only W: 18
- C & F only: 1
- C & W only: 6
- F & W only: 1
- All three: 2
- None: 9

Total travelers = sum all regions:
10 + 3 + 18 + 1 + 6 + 1 + 2 + 9 = Let’s add:

10+3=13; +18=31; +1=32; +6=38; +1=39; +2=41; +9=50

✔ Total = 50

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Question 1: How many on Holiday 2?
→ Total = 50

✔ Answer: 50

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Question 2: Probability traveler chooses camping, fishing AND water sports?
→ That’s the center region: 2 people out of 50

Probability = 2/50 = 1/25 = 0.04

But since it says “probability”, we can leave as fraction unless specified.

✔ Answer: 2/50 or simplified 1/25

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Question 3: Calculate P((C ∪ W)')

This means probability of NOT being in Camping OR Water Sports → i.e., outside both C and W.

What’s in C ∪ W? Anyone in C or W or both.

So (C ∪ W)' = people in NEITHER C nor W → that’s only those in Fishing ONLY + None

From diagram:
Fishing only = 3
None = 9
Also, check if anyone else is outside C and W — yes, only Fishing-only and None.

Wait — actually, (C ∪ W)' means not in C and not in W → so must be in F only or none.

Yes: F only = 3, None = 9 → total = 12

Total people = 50

So P = 12/50 = 6/25

✔ Answer: 12/50 or 6/25

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Question 4: Calculate P(C ∪ W')

This means: Probability of being in Camping OR NOT in Water Sports.

Use formula: P(C ∪ W') = P(C) + P(W') - P(C ∩ W')

But easier to count directly.

W' = not in Water Sports → everyone except those in W

W includes:
Only W: 18
C&W only: 6
F&W only: 1
All three: 2
→ Total in W = 18+6+1+2 = 27
So W' = 50 - 27 = 23

But we want C ∪ W' → which is almost everyone except those who are NOT in C AND ARE in W → i.e., exclude only those in W but not in C.

Who is in W but not in C?
→ Only W: 18
→ F&W only: 1
Total = 19

So C ∪ W' = total - 19 = 50 - 19 = 31

Alternatively: list all who are in C OR not in W.

In C:
Only C: 10
C&F only: 1
C&W only: 6
All three: 2
→ Total in C = 19

Not in W:
Only C: already counted
Only F: 3
Only C: already counted
None: 9
C&F only: already counted
→ Add: Only F (3) + None (9) = 12 new ones

Total = 19 + 12 = 31

Yes.

So P = 31/50

✔ Answer: 31/50

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Final Answers Compiled:

Section A — Left:
- A ∩ B = {2, 14}
- A’ ∪ B = {0, 2, 5, 9, 13, 14, 17, 20}
- (A ∩ B)’ = {0, 1, 3, 5, 7, 9, 11, 13, 17, 20}

Section A — Right:
- A = {f, z, e, b, i}
- B’ = {s, d, c, f, b, i, g, k, j}
- A ∩ B ∩ C = {e}

Section B — Holiday 1:
1) 16
2) 35
3) 24
4) 8

Section B — Holiday 2:
1) 50
2) 2/50 (or 1/25)
3) 12/50 (or 6/25)
4) 31/50

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Final Answer:
A ∩ B = {2, 14}; A’ ∪ B = {0, 2, 5, 9, 13, 14, 17, 20}; (A ∩ B)’ = {0, 1, 3, 5, 7, 9, 11, 13, 17, 20}; A = {f, z, e, b, i}; B’ = {s, d, c, f, b, i, g, k, j}; A ∩ B ∩ C = {e}; Holiday 1 Q1: 16; Q2: 35; Q3: 24; Q4: 8; Holiday 2 Q1: 50; Q2: 2/50; Q3: 12/50; Q4: 31/50