Two-Way Tables and Conditional Probability

Analyze two-way tables and calculate conditional probabilities

Two-Way Tables and Conditional Probability

What is a Two-Way Table?

A two-way table (also called a contingency table) shows the relationship between two categorical variables.

Example: Survey of 100 students about pets:

| | Dog | Cat | Total | |-----------|-----|-----|-------| | Male | 24 | 16 | 40 | | Female | 36 | 24 | 60 | | Total | 60 | 40 | 100 |

Reading Two-Way Tables

Key vocabulary:

Joint frequency: Cell value (e.g., 24 males have dogs)
Marginal frequency: Row/column total (e.g., 40 males total)
Total: Overall sum (100 students)

Conditional Probability

Formula: $P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$

Read as: "Probability of A given B"

Example Questions

Q1: What's the probability a randomly selected student is female? $P(\text{Female}) = \frac{60}{100} = 0.6$

Q2: What's the probability a student has a dog, given they're male? $P(\text{Dog}|\text{Male}) = \frac{24}{40} = 0.6$

Q3: What's the probability a student is male, given they have a cat? $P(\text{Male}|\text{Cat}) = \frac{16}{40} = 0.4$

SAT Strategy

Step 1: Identify What You're Finding

Look for "given" or "if" → use conditional probability

"Given that a student is female, what's the probability they have a cat?"

Condition: Female (use Female row only)
Want: Cat
Answer: $\frac{24}{60} = 0.4$

Step 2: Find the Right Row or Column

The condition determines your denominator:

"Given male" → Use male row (denominator = 40)
"Given has dog" → Use dog column (denominator = 60)

Step 3: Calculate

$\text{Conditional Probability} = \frac{\text{Intersection}}{\text{Condition Total}}$

Common SAT Mistakes

❌ Using the wrong denominator (using 100 instead of the condition total)
❌ Confusing $P(A|B)$ with $P(B|A)$ — order matters!
❌ Not identifying the condition properly
❌ Adding when you should divide

Practice Tip

Always ask: "Out of WHAT?" This tells you the denominator.

"Out of all males" → denominator is 40
"Out of all students" → denominator is 100
"Out of all dog owners" → denominator is 60

📚 Practice Problems

1Problem 1easy

❓ Question:

A survey of 200 students asked about their preferred lunch option. The results are shown in the table below:

            Pizza   Salad   Sandwich   Total

Freshman 30 10 20 60 Sophomore 25 15 30 70 Junior 20 20 30 70 Total 75 45 80 200

What is the probability that a randomly selected student is a sophomore who prefers pizza?

A) 25/200 B) 25/70 C) 25/75 D) 70/200

💡 Show Solution

We need to find the probability of selecting a sophomore who prefers pizza.

From the table: • Sophomores who prefer pizza = 25 • Total students = 200

Probability = (Number of sophomores who prefer pizza) / (Total students) P(Sophomore AND Pizza) = 25/200

Simplify: 25/200 = 1/8

Answer: A) 25/200

Why not the others? B) 25/70 - This would be if we asked "Given it's a sophomore, what's probability they like pizza?" (conditional) C) 25/75 - This would be if we asked "Given they like pizza, what's probability they're a sophomore?" (conditional) D) 70/200 - This is just probability of being a sophomore

Key: The question asks for joint probability (sophomore AND pizza), not conditional probability.

2Problem 2medium

❓ Question:

Using the same table from the previous question, what is the probability that a student prefers salad, given that the student is a junior?

            Pizza   Salad   Sandwich   Total

Freshman 30 10 20 60 Sophomore 25 15 30 70 Junior 20 20 30 70 Total 75 45 80 200

A) 20/200 B) 20/45 C) 20/70 D) 45/70

💡 Show Solution

This is asking for CONDITIONAL PROBABILITY: P(Salad | Junior)

Read as: "Probability of salad GIVEN that student is junior"

When we know the student is a junior, we only look at the junior row.

Junior row: • Pizza: 20 • Salad: 20 • Sandwich: 30 • Total juniors: 70

P(Salad | Junior) = (Juniors who prefer salad) / (Total juniors) = 20/70 = 2/7

Answer: C) 20/70

Conditional Probability Formula: P(A | B) = (Number in both A and B) / (Number in B)

In words: Given that condition B is met, what fraction also satisfies A?

Why not the others? A) 20/200 - Joint probability, not conditional B) 20/45 - Reversed: P(Junior | Salad) D) 45/70 - This makes no sense (45 > number of juniors who prefer salad)

SAT Tip: "Given that" signals conditional probability. Focus only on the row or column specified after "given that."

3Problem 3hard

❓ Question:

A company surveyed 500 employees about remote work preference and productivity:

                High Productivity   Low Productivity   Total

Prefers Remote 180 70 250 Prefers Office 120 130 250 Total 300 200 500

Based on this data, which statement is most supported?

A) Preferring remote work causes high productivity B) Employees who prefer remote work are more likely to have high productivity than employees who prefer office work C) 180 employees have high productivity because they prefer remote work D) The probability of an employee preferring remote work is equal to the probability of high productivity

💡 Show Solution

Let's analyze each statement:

A) "Causes" - correlation vs causation issue: • This is observational data, not an experiment • Cannot establish causation • Wrong ✗

B) "More likely" - compare conditional probabilities: • P(High Prod | Prefers Remote) = 180/250 = 0.72 = 72% • P(High Prod | Prefers Office) = 120/250 = 0.48 = 48% • 72% > 48%, so yes, those who prefer remote are more likely to have high productivity • This is a valid comparison of the data ✓ • Correct! ✓

C) "Because they prefer remote work": • Claims causation again • We only know association • Wrong ✗

D) Check if probabilities are equal: • P(Prefers Remote) = 250/500 = 0.50 = 50% • P(High Productivity) = 300/500 = 0.60 = 60% • 50% ≠ 60% • Wrong ✗

Answer: B) Employees who prefer remote work are more likely to have high productivity than employees who prefer office work

Key SAT Concept: • Two-way tables show ASSOCIATIONS (correlations) • Can compare likelihoods using conditional probabilities • Cannot claim causation without experimental design • "More likely" = comparing conditional probabilities

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