Conditional Probability and Two-Way Tables

Master conditional probability and two-way tables for the Digital SAT. Calculate P(A|B) and interpret frequency data.

Conditional probability answers the question: "Given that one thing has already happened, what is the probability of another?" Two-way tables (also called contingency tables) organise data by two categories and are the primary tool for conditional probability on the Digital SAT.

Core Concepts

Two-Way Tables

A two-way table shows frequencies for two categorical variables.

Sports No Sports Total
Male 45 30 75
Female 35 40 75
Total 80 70 150

Conditional Probability

P(AB)=P(A and B)P(B)=number in both A and Bnumber in BP(A | B) = \frac{P(A \text{ and } B)}{P(B)} = \frac{\text{number in both A and B}}{\text{number in B}}

The key: the denominator is the given condition's total, not the grand total.

Example: P(SportsMale)=4575=0.6P(\text{Sports} | \text{Male}) = \frac{45}{75} = 0.6

Example: P(MaleSports)=4580=0.5625P(\text{Male} | \text{Sports}) = \frac{45}{80} = 0.5625

Notice these are different! The condition changes the denominator.

Relative Frequency

Joint relative frequency: each cell divided by the grand total. P(Male and Sports)=45150=0.3P(\text{Male and Sports}) = \frac{45}{150} = 0.3

Marginal relative frequency: row/column totals divided by the grand total. P(Male)=75150=0.5P(\text{Male}) = \frac{75}{150} = 0.5

Independence Test

Two events are independent if P(AB)=P(A)P(A|B) = P(A).

Is playing sports independent of gender? P(Sports)=80150=0.533P(\text{Sports}) = \frac{80}{150} = 0.533. P(SportsMale)=0.6P(\text{Sports}|\text{Male}) = 0.6.

Since 0.60.5330.6 \neq 0.533, sports and gender are NOT independent.

Strategy Tips

Tip 1: Identify the Denominator

The word after "given" (or the condition) determines the denominator. "Given male" → denominator is total males.

Tip 2: Read the Table Row/Column Carefully

Make sure you're reading the correct row and column.

Tip 3: "Among" or "Of" = Conditional

"Among males, what fraction play sports?" = P(SportsMale)P(\text{Sports}|\text{Male}).

Worked Example: Example 1

Problem

Using the table above: What is the probability that a randomly selected person is female and plays no sports?

P=40150=415P = \frac{40}{150} = \frac{4}{15}

Solution

Worked Example: Example 2

Problem

Among females, what proportion plays sports?

P(SportsFemale)=3575=715P(\text{Sports}|\text{Female}) = \frac{35}{75} = \frac{7}{15}

Solution

Worked Example: SAT-Style

Problem

A survey found: 60 people prefer coffee, 40 prefer tea. Of coffee drinkers, 36 are adults. Of tea drinkers, 20 are adults. If a person is selected at random from adults, what is the probability they prefer coffee?

Total adults: 36+20=5636 + 20 = 56. P(coffeeadult)=3656=914P(\text{coffee}|\text{adult}) = \frac{36}{56} = \frac{9}{14}.

Solution

Practice Problems

  1. Problem 1

    Using a two-way table with 200 students: 120 passed, 80 failed. Of those who studied, 100 passed and 20 failed. P(passedstudied)P(\text{passed}|\text{studied})?

    Problem 2

    P(A)=0.4P(A) = 0.4, P(B)=0.5P(B) = 0.5, P(A and B)=0.2P(A \text{ and } B) = 0.2. Find P(AB)P(A|B).

    Problem 3

    Are A and B from Problem 2 independent?

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Common Mistakes

  • Using the wrong denominator. The condition (given) determines the denominator.
  • Confusing P(AB)P(A|B) with P(BA)P(B|A). They are usually different.
  • Using the grand total as denominator for conditional probability. Only use the grand total for unconditional (joint or marginal) probability.

Key Takeaways

  • Conditional probability: P(AB)=A and BB totalP(A|B) = \frac{\text{A and B}}{\text{B total}}.

  • The denominator is the total of the given condition.

  • Two-way tables are the primary tool — read rows and columns carefully.

  • "Among", "of", "given" all signal conditional probability.

  • Check independence by comparing P(AB)P(A|B) to P(A)P(A).

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