GMAT Focus Edition — Data Insights: Table Analysis · Graphics Interpretation · Multi-Source Reasoning · Two-Part Analysis
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Data Insights Lesson 16 of 20

Lesson 16:
Conditional Probability & Subset Logic

Apply conditional probability in DI contexts: given one condition, what is the probability of another? Master Bayes' intuition.

⏱ 50 mins
🎯 DI 75 to 88
📚 Prereq: Lesson 15 (Tree Diagrams)
Note: Lesson 16 of 20 — Conditional Probability & Subset Logic.
1

Core Concepts: Conditional Probability & Subset Logic

Apply conditional probability in DI contexts: given one condition, what is the probability of another? Master Bayes' intuition.

Key Framework
P(A|B) = P(A∩B) / P(B)
Conditional probability: probability of A given B has occurred.
Reduced sample space
Condition B restricts you to the subset where B is true. Compute within that subset.
Bayes intuition: prior × likelihood / total probability
Used to update beliefs given new evidence.
Joint probability tables: read the right row
Given one row's total, compute within-row fractions for conditional questions.
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 16
📊
P(A|B) = P(A∩B) / P(B)
📈
Reduced sample space
🔍
Bayes intuition: prior × likelihood / total probability
📋
Joint probability tables: read the right row
4

Quick Reference Rules

▸
P(A|B) = P(A∩B) / P(B): Conditional probability: probability of A given B has occurred.
▸
Reduced sample space: Condition B restricts you to the subset where B is true. Compute within that subset.
▸
Bayes intuition: prior × likelihood / total probability: Used to update beliefs given new evidence.
▸
Joint probability tables: read the right row: Given one row's total, compute within-row fractions for conditional questions.
5

10 Traps for Lesson 16

⚠ Confusing the key formula for this topic

Review the core rule: P(A|B) = P(A∩B) / P(B)

⚠ 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

Reduced sample space — 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~650

P(A∩B) = 0.12, P(B) = 0.4. P(A|B) = ?

Explanation: 0.3. P(A|B) = P(A∩B)/P(B) = 0.12/0.4 = 0.3.
Q2 of 10
GI~600

Of 200 survey respondents, 80 are women and 60 of those women prefer Brand X. P(Brand X | Woman) = ?

Explanation: 75%. P(Brand X | Woman) = 60/80 = 75%.
Q3 of 10
GI~650

A table shows: 40 men like product A, 60 men don't; 50 women like A, 30 don't. Total sample = 180. P(like A | female) = ?

Explanation: 62.5%. Women total = 50+30 = 80. P(like A | female) = 50/80 = 62.5%.
Q4 of 10
GI~700

P(A) = 0.5, P(B) = 0.4, P(A∩B) = 0.2. Are A and B independent?

Explanation: Yes. For independence: P(A∩B) must equal P(A)×P(B) = 0.5×0.4 = 0.2. Since P(A∩B) = 0.2, the events are independent.
Q5 of 10
GI~700

A doctor's test for a disease has: sensitivity = 90% (P(positive|disease)) and the disease prevalence = 10%. P(disease AND positive test) = ?

Explanation: 9%. P(disease AND positive) = P(positive|disease) × P(disease) = 0.9 × 0.1 = 0.09 = 9%.
Q6 of 10
GI~650

From a class of 30 students, 18 passed Math, 15 passed English, 9 passed both. Given a student passed Math, P(also passed English) = ?

Explanation: 9/18 = 50%. P(English | Math) = P(both)/P(Math) = 9/18 = 0.5.
Q7 of 10
GI~700

P(A|B) = P(A) implies:

Explanation: A and B are independent. If knowing B occurred gives no information about A — P(A|B) = P(A) — the events are statistically independent.
Q8 of 10
GI~750

In a company: 40% are senior, 60% junior. 30% of seniors and 20% of juniors are in management. P(management) = ?

Explanation: 24%. P(mgmt) = P(senior)×P(mgmt|senior) + P(junior)×P(mgmt|junior) = 0.4×0.3 + 0.6×0.2 = 0.12+0.12 = 0.24 = 24%.
Q9 of 10
GI~700

A bag has 3 red, 7 blue balls. Two drawn without replacement. P(second is red | first is blue) = ?

Explanation: 3/9 = 1/3. After drawing 1 blue (from 7 blue, 3 red = 10 total), 9 remain: still 3 red. P(second red | first blue) = 3/9 = 1/3.
Q10 of 10
GI~650

P(A|B) can be greater than P(A) when:

Explanation: B makes A more likely. P(A|B) > P(A) when events A and B are positively associated — knowing B occurred increases the likelihood of A. This is common in correlated or causally linked events.
Lesson Summary
P(A|B) = P(A∩B) / P(B)

Conditional probability: probability of A given B has occurred.

Reduced sample space

Condition B restricts you to the subset where B is true. Compute within that subset.

Bayes intuition: prior × likelihood / total probability

Used to update beliefs given new evidence.

Joint probability tables: read the right row

Given one row's total, compute within-row fractions for conditional questions.

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