Practice Maths

Venn Diagrams and Two-Way Tables

Key Ideas

Key Terms

Venn diagram
Shows two (or more) sets as overlapping circles within a rectangle (the universal set).
Union
A ∪ B = all elements in A or B or both. P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
Intersection
A ∩ B = elements in both A and B.
Complement
A′ = all elements not in A. P(A′) = 1 − P(A).
Mutually exclusive events
Cannot both occur: A ∩ B = ∅, so P(A ∪ B) = P(A) + P(B).
two-way table
Organises data by two categorical variables. Read row and column totals carefully when finding probabilities.
Hot Tip Fill in the intersection first in a Venn diagram, then work outward. Always check that all regions sum to the total.

Worked Example

Question: 50 students were surveyed. 30 like Maths (M), 25 like Science (S), and 15 like both. Draw a Venn diagram and find:

(a) P(Maths only)   (b) P(Maths or Science)   (c) P(neither)

Step 1 — Fill in the Venn diagram:

• Both (M ∩ S): 15

• Maths only: 30 − 15 = 15

• Science only: 25 − 15 = 10

• Neither: 50 − 15 − 15 − 10 = 10

(a) P(Maths only) = 15/50 = 3/10

(b) P(M ∪ S) = P(M) + P(S) − P(M ∩ S) = 30/50 + 25/50 − 15/50 = 40/50 = 4/5

(c) P(neither) = 10/50 = 1/5

Venn Diagrams

A Venn diagram uses overlapping circles inside a rectangle (the universal set, ξ) to show how events relate. Each circle represents an event. Elements that belong to both events sit in the intersection (the overlapping region). Elements in only one circle belong to that event but not the other.

Key regions: A ∩ B (A and B) = the intersection (overlap). A ∪ B (A or B) = everything inside either circle. A' (not A) = everything outside circle A. Elements outside both circles belong to neither event.

Example: In a class of 30, 18 play sport (S) and 12 play music (M). 7 play both. Venn diagram: S only = 18 − 7 = 11, M only = 12 − 7 = 5, both = 7, neither = 30 − 11 − 7 − 5 = 7.

Two-Way Tables

A two-way table (also called a contingency table) organises the same information as a Venn diagram into rows and columns. Rows represent one category, columns represent another, and cells show the count for each combination.

Using the example above:

SportNo SportTotal
Music7512
No Music11718
Total181230

Two-way tables make it easy to read off counts for any combination and to calculate probabilities by dividing by the total.

Converting Between the Two Representations

Both Venn diagrams and two-way tables show the same information — knowing how to convert between them is essential. From a two-way table, the intersection cell gives you the overlap for the Venn diagram. From a Venn diagram, the overlap region fills the intersection cell in the table.

Always check that all counts add up to the total (the grand total in the bottom-right corner of the table, or the total of all regions in the Venn diagram). This is a built-in check for errors.

Conditional Probability

Conditional probability is the probability of an event given that another event has already occurred. The notation P(A|B) means "the probability of A given B."

Formula: P(A|B) = P(A ∩ B) ÷ P(B)

Or equivalently using counts: P(A|B) = (number in both A and B) ÷ (number in B).

Example: What is the probability a randomly chosen student plays sport, given they play music? P(S|M) = 7/12, because there are 12 music students and 7 of them also play sport. We restrict our sample space to the music players only.

Independent Events

Two events are independent if knowing one occurred doesn't change the probability of the other. Formally, A and B are independent if P(A|B) = P(A). In the table above, P(S|M) = 7/12 but P(S) = 18/30 = 3/5. Since these are different, sport and music are not independent in this class.

Key tip: For conditional probability questions, re-read to identify the "given" condition — this tells you which row or column to restrict your focus to. The denominator in P(A|B) is always the total count for the given condition (B), not the overall total.

Mastery Practice

  1. In a group of 40 students, 22 play soccer (S), 18 play basketball (B), and 7 play both sports. Fluency

    1. Draw a Venn diagram and fill in the number of students in each region.
    2. How many students play neither sport?
    3. Find P(plays soccer only).
    4. Find P(plays at least one sport).
    5. Find P(plays basketball but not soccer).

  2. A Venn diagram for events A and B from a sample of 60 shows: A only = 14, B only = 19, A ∩ B = 11, neither = 16. Fluency

    ξ = 60 A B 14 11 19 16
    1. Verify the four regions sum to 60.
    2. Find P(A).
    3. Find P(B).
    4. Find P(A ∪ B).
    5. Find P(A′ ∩ B′) (i.e. neither A nor B).
    6. Verify using the addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

  3. The two-way table below shows the results of a survey of 100 Year 9 students about their preferred after-school activity and whether they own a pet. Fluency

    Has PetNo PetTotal
    Sport241640
    Gaming182240
    Reading81220
    Total5050100
    1. Find P(a randomly selected student prefers Sport).
    2. Find P(the student has a pet).
    3. Find P(the student prefers Gaming AND has no pet).
    4. Find P(the student prefers Reading OR has a pet).
    5. A student who prefers Sport is chosen at random. Find P(this student has a pet).

  4. In a school of 200 students: 120 study Music, 80 do not study Music; 90 study Art, 110 do not study Art. 50 students study both Music and Art. Fluency

    1. Construct a fully completed two-way table for Music and Art.
    2. Find P(studies Music only).
    3. Find P(studies neither Music nor Art).
    4. Find P(studies Art given that they study Music) — this is a conditional probability P(Art | Music).

  5. Events A and B satisfy: P(A) = 0.45, P(B) = 0.38, P(A ∪ B) = 0.62. Understanding

    1. Find P(A ∩ B).
    2. Find P(A only — i.e. in A but not B).
    3. Find P(B only).
    4. Find P(neither A nor B).
    5. Are A and B mutually exclusive? Explain.

  6. Events C and D are mutually exclusive. P(C) = 0.3 and P(D) = 0.5. Understanding

    1. Find P(C ∪ D).
    2. Find P(C ∩ D) and explain why.
    3. Find P(neither C nor D).
    4. Can C and D together be exhaustive (cover all outcomes)? Explain.
    5. If a third mutually exclusive event E has P(E) = 0.2, confirm that C, D and E together are exhaustive.

  7. A survey of 80 Year 9 students asked whether they watch Movies (M), Sport (S), or News (N) on TV. The results were: M only = 18, S only = 14, N only = 9, M ∩ S only = 8, M ∩ N only = 6, S ∩ N only = 5, all three = 4, none = 16. Understanding

    ξ = 80 M S N 18 14 9 8 6 5 4 16
    1. Verify the counts sum to 80.
    2. Find P(watches Movies).
    3. Find P(watches exactly one category).
    4. Find P(watches at least two categories).
    5. Find P(watches Movies and Sport but not News).

  8. The table below shows 150 gym members classified by gender and type of exercise preferred. Understanding

    CardioWeightsClassesTotal
    Male28351780
    Female32152370
    Total605040150
    1. Find P(member prefers Weights).
    2. Find P(member is Female AND prefers Classes).
    3. Find P(member prefers Cardio | member is Male).
    4. Find P(member is Male | member prefers Weights).
    5. Are Gender and Exercise preference independent for the Cardio category? Check using P(Male ∩ Cardio) = P(Male) × P(Cardio).

  9. In a year group of 120 students, P(plays a team sport) = 0.6 and P(plays a team sport AND plays an instrument) = 0.25. P(plays neither) = 0.1. Problem Solving

    1. Find the number of students who play a team sport.
    2. Find the number who play a team sport AND an instrument.
    3. Find P(plays an instrument).
    4. Draw a Venn diagram showing the four regions in terms of number of students.
    5. A student is chosen at random from those who play a team sport. Find P(this student also plays an instrument).

  10. A medical screening test for a disease gives the following results across 500 patients: Problem Solving

    Has DiseaseNo DiseaseTotal
    Test Positive453075
    Test Negative5420425
    Total50450500
    1. Find P(patient has disease).
    2. Find P(test is positive given patient has disease) — called the sensitivity of the test.
    3. Find P(test is negative given patient has no disease) — called the specificity.
    4. Find P(patient has disease given the test was positive) — called the positive predictive value.
    5. Discuss: if a patient tests positive, should they be very worried? Use your calculations to support your answer.
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