GMAT Focus Edition — Data Insights: Table Analysis · Graphics Interpretation · Multi-Source Reasoning · Two-Part Analysis
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Data Insights Lesson 15 of 20

Lesson 15:
Tree Diagrams & Decision Logic

Use tree diagrams to map sequential decisions, compute path probabilities, and count conditional outcomes systematically.

⏱ 50 mins
🎯 DI 75 to 88
📚 Prereq: Lesson 13 (Probability)
Note: Lesson 15 of 20 — Tree Diagrams & Decision Logic.
1

Core Concepts: Tree Diagrams & Decision Logic

Use tree diagrams to map sequential decisions, compute path probabilities, and count conditional outcomes systematically.

Key Framework
Branch probability = product along the path
Multiply conditional probabilities from root to leaf.
All leaf probabilities sum to 1
Use this as a sanity check when building trees.
At least one = 1 − P(none)
The complement is always faster for "at least one" scenarios.
Tree diagrams make conditional logic visual
Especially useful for sequential events with changing probabilities.
2

Application Strategy

Step-by-Step Approach
①
Identify the question type and the specific metric being asked about.
②
Locate the relevant data in the chart, table, or tab.
③
Apply the key formula or logic rule from the framework above.
④
Verify that your answer satisfies the question as stated.
3

Visual Reference Diagram

Visual Framework for Lesson 15
📊
Branch probability = product along the path
📈
All leaf probabilities sum to 1
🔍
At least one = 1 − P(none)
📋
Tree diagrams make conditional logic visual
4

Quick Reference Rules

▸
Branch probability = product along the path: Multiply conditional probabilities from root to leaf.
▸
All leaf probabilities sum to 1: Use this as a sanity check when building trees.
▸
At least one = 1 − P(none): The complement is always faster for "at least one" scenarios.
▸
Tree diagrams make conditional logic visual: Especially useful for sequential events with changing probabilities.
5

10 Traps for Lesson 15

⚠ Confusing the key formula for this topic

Review the core rule: Branch probability = product along the path

⚠ Reading the wrong data series

Always verify axis labels and legends before extracting values.

⚠ Ignoring units or scale

Check units on every axis before computing ratios or changes.

⚠ Treating estimates as exact

GI data requires interpolation — use approximate values and pick the closest answer.

⚠ Forgetting the denominator

All leaf probabilities sum to 1 — always identify what you're dividing by.

⚠ Applying simple average when weighted is needed

If group sizes differ, compute weighted average, not simple mean.

⚠ "All" statements need every row verified

One counterexample makes a universal statement False.

⚠ Scope: data sample ≠ full population

Conclusions are limited to the data range provided, not broader populations.

⚠ Confusing absolute and relative measures

Absolute change ($) and relative change (%) answer different questions.

⚠ Not using the complement when helpful

P(at least one) = 1 − P(none). In set logic, use complement for "not" conditions.

✦

10 Practice Questions

Q1 of 10
GI~550

A tree diagram shows two coin flips. What is the probability of getting Head then Tail?

Explanation: 1/4. P(H) × P(T) = 1/2 × 1/2 = 1/4.
Q2 of 10
GI~600

A tree has three branches: Rain (60%), Cloudy (30%), Sunny (10%). Each day is independent. P(Rain both days) = ?

Explanation: 0.36. P(Rain) × P(Rain) = 0.6 × 0.6 = 0.36.
Q3 of 10
GI~650

P(at least one head in 4 coin flips) = ?

Explanation: 15/16. P(all tails) = (1/2)^4 = 1/16. P(at least one head) = 1 − 1/16 = 15/16.
Q4 of 10
GI~700

A tree diagram for drawing 2 cards (no replacement) from a deck of 4 red, 6 blue cards. P(both red) = ?

Explanation: 2/15. P(first red) = 4/10. P(second red | first red) = 3/9. P(both red) = 4/10 × 3/9 = 12/90 = 2/15.
Q5 of 10
GI~550

A tree diagram has 3 levels. Each level has 2 branches with equal probability. The total number of end branches (leaves) is:

Explanation: 8. 2 branches × 2 branches × 2 branches = 2³ = 8 leaf nodes.
Q6 of 10
GI~650

In a decision tree, Path A-B-C has probabilities 0.6, 0.4, 0.5. Probability of this specific path = ?

Explanation: 0.12. 0.6 × 0.4 × 0.5 = 0.12.
Q7 of 10
GI~550

Sum of all leaf probabilities in a complete probability tree must be:

Explanation: 1. All possible outcomes are exhaustive and mutually exclusive. Their probabilities sum to 1 — this is a basic axiom of probability.
Q8 of 10
GI~700

A tree shows: P(A) = 0.7, P(B|A) = 0.6, P(B|not A) = 0.3. P(A and B) = ?

Explanation: 0.42. P(A and B) = P(A) × P(B|A) = 0.7 × 0.6 = 0.42.
Q9 of 10
GI~750

P(B) using the total probability theorem, with P(A)=0.7, P(B|A)=0.6, P(B|not A)=0.3:

Explanation: 0.51. P(B) = P(A)×P(B|A) + P(not A)×P(B|not A) = 0.7×0.6 + 0.3×0.3 = 0.42+0.09 = 0.51.
Q10 of 10
GI~650

A tree diagram shows 3 paths to success. Their probabilities are 0.12, 0.08, and 0.15. P(success via any path) = ?

Explanation: 0.35. Since the paths are mutually exclusive routes to success: P(any) = 0.12+0.08+0.15 = 0.35.
Lesson Summary
Branch probability = product along the path

Multiply conditional probabilities from root to leaf.

All leaf probabilities sum to 1

Use this as a sanity check when building trees.

At least one = 1 − P(none)

The complement is always faster for "at least one" scenarios.

Tree diagrams make conditional logic visual

Especially useful for sequential events with changing probabilities.