Frequency trees, two-way tables, Venn diagrams and tree diagrams

Welcome to Probability Tools: Diagrams, Tables, and Trees!

Probability can sometimes feel like trying to guess the future, but with the right visual tools, it becomes as simple as following a map. In this chapter, we will master four fantastic tools used to organise data and calculate probabilities:

• Frequency Trees — Great for breaking down real counts of people or items step-by-step.
• Two-Way Tables — Perfect for sorting information that has two different categories.
• Venn Diagrams — Ideal for seeing how groups overlap and who belongs where.
• Probability Tree Diagrams — The ultimate tool for finding probabilities of multiple events happening one after another.

Don't worry if this seems like a lot to take in right now! We will break down each tool step-by-step with clear examples.

1. Frequency Trees

What is a Frequency Tree?

A frequency tree uses branches to split a total group into smaller subgroups using actual numbers (frequencies), not probabilities or percentages. Think of it like a river splitting into smaller streams.

The Golden Rule of Frequency Trees

The sum of the branches must equal the number at the node they came from.
If \(100\) people split into two groups and one group has \(60\), the other branch must be \(100 - 60 = 40\).

Step-by-Step Example

Scenario: \(80\) students take a driving theory test.
• \(50\) of the students are girls, and the rest are boys.
• \(35\) of the girls pass the test.
• In total, \(20\) students fail the test.

Let's complete the frequency tree step-by-step:

Step 1: Find the number of boys.
Total students = \(80\). Girls = \(50\).
Number of boys = \(80 - 50 = 30\).

Step 2: Find the girls who failed.
Total girls = \(50\). Girls who pass = \(35\).
Girls who fail = \(50 - 35 = 15\).

Step 3: Find the boys who failed and passed.
Total students who fail = \(20\). Girls who fail = \(15\).
Boys who fail = \(20 - 15 = 5\).
Total boys = \(30\). Boys who fail = \(5\).
Boys who pass = \(30 - 5 = 25\).

Finding Probabilities from a Frequency Tree

To find a probability, use the standard probability formula:
\(P(\text{Event}) = \frac{\text{Number of successful outcomes}}{\text{Total number of possible outcomes}}\)

Example: A student is chosen at random. What is the probability that the student is a boy who passed?
• Number of boys who passed = \(25\)
• Total number of students = \(80\)
• \(P(\text{Boy who passed}) = \frac{25}{80} = \frac{5}{16}\)

Common Mistake to Avoid: Always check that the final branches add up to the total starting number. Here: \(35 + 15 + 25 + 5 = 80\). If they don't, re-check your subtraction!

Key Takeaway: Frequency trees record actual counts. Start with the information given and work forward by subtracting to fill in the missing blanks.

2. Two-Way Tables

What is a Two-Way Table?

A two-way table organises data involving two separate categorical variables (for example, Gender and Favourite Sport). One variable is shown across the rows, and the other is shown down the columns.

Key Feature: Row and Column Totals

Every row adds up to a row total on the far right. Every column adds up to a column total at the bottom. The bottom-right cell is the grand total (the total number of items or people overall).

Step-by-Step Example

Scenario: \(60\) students in Year 11 choose between French and Spanish.
• \(32\) students are boys.
• \(18\) girls study Spanish.
• \(24\) students in total study French.

Let's find the missing values:

1. Total girls: Grand total \(-\) Total boys = \(60 - 32 = 28\).
2. Girls studying French: Total girls \(-\) Girls studying Spanish = \(28 - 18 = 10\).
3. Boys studying French: Total French \(-\) Girls studying French = \(24 - 10 = 14\).
4. Boys studying Spanish: Total boys \(-\) Boys studying French = \(32 - 14 = 18\).
5. Total studying Spanish: Grand total \(-\) Total French = \(60 - 24 = 36\) (or \(18 + 18 = 36\)).

Calculating Probabilities from a Two-Way Table

• Simple Probability: A student is picked at random. What is the probability they study French?
\(P(\text{French}) = \frac{\text{Total French}}{\text{Grand Total}} = \frac{24}{60} = \frac{2}{5}\)

• Restricted / "Given that" Probability: A girl is picked at random. What is the probability she studies Spanish?
Notice the denominator changes! We are only choosing from girls.
\(P(\text{Spanish given Girl}) = \frac{\text{Girls studying Spanish}}{\text{Total Girls}} = \frac{18}{28} = \frac{9}{14}\)

Key Takeaway: Always read probability questions carefully. If the question says "Given that it is a girl..." or "Out of the boys...", the denominator is the category total, not the grand total.

3. Venn Diagrams

What is a Venn Diagram?

A Venn diagram uses overlapping circles inside a rectangle to show relationships between sets of items. The surrounding box represents the Universal Set (\(\xi\)), which contains everything being considered.

Important Set Notation

• \(\xi\) = Universal Set (everything in the box).
• \(A \cap B\) = Intersection (elements in both \(A\) AND \(B\) — the middle overlap).
• \(A \cup B\) = Union (elements in \(A\) OR \(B\) or both — everything inside circle \(A\) and circle \(B\) combined).
• \(A'\) = Complement of \(A\) (elements NOT in \(A\) — everything outside circle \(A\)).

Memory Trick:
• The symbol \(\cap\) looks like an n for aNd (Intersection).
• The symbol \(\cup\) looks like a u for Union (all United together).

How to Fill in a Venn Diagram

Top Tip: Always start from the inside out (fill in the middle intersection \(A \cap B\) first!).

Example: In a class of \(30\) pupils:
• \(17\) pupils have a Dog (\(D\)).
• \(12\) pupils have a Cat (\(C\)).
• \(5\) pupils have both a Dog and a Cat (\(D \cap C\)).

Let's place the numbers:
1. Middle overlap (\(D \cap C\)): Write \(5\) in the intersection.
2. Dog ONLY: Total dog owners \(-\) both = \(17 - 5 = 12\). Write \(12\) in the Dog-only section.
3. Cat ONLY: Total cat owners \(-\) both = \(12 - 5 = 7\). Write \(7\) in the Cat-only section.
4. Neither: Total pupils \(-\) (\(12 + 5 + 7\)) = \(30 - 24 = 6\). Write \(6\) outside the circles but inside the box.

Calculating Probabilities from the Venn Diagram

• \(P(\text{Dog and Cat}) = P(D \cap C) = \frac{5}{30} = \frac{1}{6}\)
• \(P(\text{Dog or Cat}) = P(D \cup C) = \frac{12 + 5 + 7}{30} = \frac{24}{30} = \frac{4}{5}\)
• \(P(\text{Does not own a Dog}) = P(D') = \frac{7 + 6}{30} = \frac{13}{30}\)
• \(P(\text{Owns a Cat given that they own a Dog}) = \frac{\text{Both}}{\text{Total Dog}} = \frac{5}{17}\)

Common Mistake to Avoid: Forgetting to subtract the intersection from the group totals! If you write \(17\) in the Dog circle and \(5\) in the middle, you have accidentally counted \(22\) dog owners.

Key Takeaway: Start with the overlap in the centre, subtract to find the "only" sections, and don't forget the items that belong outside both circles.

4. Probability Tree Diagrams

What is a Tree Diagram?

A probability tree diagram shows all possible outcomes of two or more successive events along with their probabilities on each branch.

The Two Golden Rules of Tree Diagrams

1. Branches from any single point must add up to \(1\).
For example, if \(P(\text{Rain}) = 0.3\), then \(P(\text{No Rain}) = 1 - 0.3 = 0.7\).

2. Multiply along the branches, Add the resulting outcomes.
• Multiply to find the probability of one event AND another event happening: \(P(A \text{ and } B) = P(A) \times P(B)\).
• Add to find the probability of one outcome OR another outcome happening: \(P(\text{Outcome 1 or Outcome 2}) = P(\text{Outcome 1}) + P(\text{Outcome 2})\).

Independent vs Dependent Events

• Independent Events (With Replacement): The outcome of the first event does not affect the second event. Probabilities stay the same.
• Dependent Events (Without Replacement): The outcome of the first event does affect the second event. The total number of items and the number of specific items decrease for the second pick!

Worked Example 1: Independent Events (With Replacement)

A bag contains \(3\) Red counters and \(7\) Blue counters (Total = \(10\)). A counter is picked, its colour recorded, and then replaced. A second counter is then picked.

Probabilities on branches:
• First pick: \(P(R) = \frac{3}{10}\), \(P(B) = \frac{7}{10}\)
• Second pick: \(P(R) = \frac{3}{10}\), \(P(B) = \frac{7}{10}\) (since the counter was replaced)

Calculating the combined outcomes:
• \(P(R, R) = \frac{3}{10} \times \frac{3}{10} = \frac{9}{100}\)
• \(P(R, B) = \frac{3}{10} \times \frac{7}{10} = \frac{21}{100}\)
• \(P(B, R) = \frac{7}{10} \times \frac{3}{10} = \frac{21}{100}\)
• \(P(B, B) = \frac{7}{10} \times \frac{7}{10} = \frac{49}{100}\)

Question: What is the probability of picking at least one Red counter?
"At least one Red" means \((R, R)\), \((R, B)\), or \((B, R)\):
\(P(\text{At least one Red}) = \frac{9}{100} + \frac{21}{100} + \frac{21}{100} = \frac{51}{100}\)

Clever Alternative Method: Use the complement rule!
\(P(\text{At least one Red}) = 1 - P(\text{No Reds}) = 1 - P(B, B) = 1 - \frac{49}{100} = \frac{51}{100}\)

Worked Example 2: Dependent Events (Without Replacement)

A box has \(4\) Green sweets and \(6\) Orange sweets (Total = \(10\)). A sweet is taken and eaten (not replaced). A second sweet is taken.

Probabilities for the second pick change:
• If Green was picked first: Remaining total is \(9\). Sweets left: \(3\) Green, \(6\) Orange.
Branch probabilities: \(P(G) = \frac{3}{9}\), \(P(O) = \frac{6}{9}\).
• If Orange was picked first: Remaining total is \(9\). Sweets left: \(4\) Green, \(5\) Orange.
Branch probabilities: \(P(G) = \frac{4}{9}\), \(P(O) = \frac{5}{9}\).

Question: Find the probability of picking two sweets of the same colour.
"Same colour" means \((G, G)\) or \((O, O)\):
• \(P(G, G) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90}\)
• \(P(O, O) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90}\)
• \(P(\text{Same Colour}) = \frac{12}{90} + \frac{30}{90} = \frac{42}{90} = \frac{7}{15}\)

Key Takeaway:
• Along branches = MULTIPLY (\(\times\)).
• Down columns of outcomes = ADD (\(+\)).
• "Without replacement" = Reduce both the numerator and denominator for the second set of branches!

Quick Comparison: Which Tool Should I Use?

• Frequency Tree: Use when you have real counts (frequencies) splitting into two or more successive subgroups.
• Two-Way Table: Use when you have counts for two distinct categories (e.g., Year Group and Choice of Subject).
• Venn Diagram: Use when items can belong to one group, both groups, or neither group.
• Tree Diagram: Use when calculating probabilities of combined, consecutive events (especially with/without replacement).