Probability and Two-Way Tables

Calculate probabilities from two-way tables and counting principles.

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Probability and Two-Way Tables on the SAT

Basic Probability

$P(\text{event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Probability is always between 0 and 1 (or 0% and 100%)
$P = 0$ : impossible
$P = 1$ : certain

Two-Way Tables (Contingency Tables)

A two-way table organizes data by two categorical variables.

Example:

| | Likes Pizza | Doesn't Like Pizza | Total | |---|---|---|---| | Students | 45 | 15 | 60 | | Teachers | 20 | 10 | 30 | | Total | 65 | 25 | 90 |

Reading the Table

Row totals are on the right
Column totals are on the bottom
Grand total is bottom-right

Types of Probability from Two-Way Tables

Joint Probability

The probability of two specific categories together.

$P(\text{Student AND Likes Pizza}) = \frac{45}{90} = \frac{1}{2}$

The denominator is the grand total.

Marginal Probability

The probability of just one category.

$P(\text{Student}) = \frac{60}{90} = \frac{2}{3}$

$P(\text{Likes Pizza}) = \frac{65}{90} = \frac{13}{18}$

Conditional Probability

The probability of one event GIVEN another has occurred.

$P(\text{Likes Pizza} \mid \text{Student}) = \frac{45}{60} = \frac{3}{4}$

Key: The denominator is the subtotal of the given condition, not the grand total!

$P(\text{Student} \mid \text{Likes Pizza}) = \frac{45}{65} = \frac{9}{13}$

Conditional Probability Formula

$P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)}$

The SAT usually tests this with two-way tables rather than the formula directly.

Complement Rule

$P(\text{NOT } A) = 1 - P(A)$

If the probability of rain is 0.3, the probability of no rain is $1 - 0.3 = 0.7$ .

SAT Question Types

Type 1: "What is the probability that a randomly selected person...?"

Identify the numerator (favorable outcomes) and denominator (total)
Watch for whether it's conditional ("...given that they are a student")

Type 2: "What fraction of [group] are [category]?"

This is conditional probability. The denominator is the SIZE of the given group.

Type 3: "Which group has a higher proportion of...?"

Compare conditional probabilities between groups.

Type 4: Complete a Two-Way Table

Fill in missing values using row/column totals. Every row and column must add up.

Common SAT Mistakes

Using the wrong denominator — the #1 mistake! For conditional probability, use the row/column total, NOT the grand total
Confusing "and" with "given" — $P(A \text{ and } B) \neq P(A \mid B)$
Misreading which row/column represents which category
Not simplifying fractions when answer choices are simplified
Forgetting the complement — sometimes it's easier to calculate $1 - P(\text{not happening})$

📚 Practice Problems

1Problem 1easy

❓ Question:

A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles. What is the probability of randomly selecting a blue marble?

💡 Show Solution

Step 1: Count total marbles: $5 + 3 + 2 = 10$

Step 2: Apply the probability formula: $P(\text{blue}) = \frac{\text{blue marbles}}{\text{total marbles}} = \frac{3}{10}$

Answer: $\frac{3}{10}$ or $0.3$ or $30\%$

2Problem 2easy

❓ Question:

A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles. What is the probability of randomly selecting a blue marble?

💡 Show Solution

Step 1: Count total marbles: $5 + 3 + 2 = 10$

Step 2: Apply the probability formula: $P(\text{blue}) = \frac{\text{blue marbles}}{\text{total marbles}} = \frac{3}{10}$

Answer: $\frac{3}{10}$ or $0.3$ or $30\%$

3Problem 3medium

❓ Question:

Use the table below:

| | Passed | Failed | Total | |---|---|---|---| | Studied | 42 | 8 | 50 | | Did Not Study | 18 | 32 | 50 | | Total | 60 | 40 | 100 |

What is the probability that a student passed, given that they studied?

💡 Show Solution

Key: This is a conditional probability question because of the phrase "given that they studied."

Step 1: Identify the condition: "given that they studied" means we only look at the "Studied" row.

Step 2: The denominator is the total who studied: 50 The numerator is those who studied AND passed: 42

$P(\text{Passed} \mid \text{Studied}) = \frac{42}{50} = \frac{21}{25} = 0.84$

Answer: $\frac{21}{25}$ or $84\%$

Common mistake: Using 100 as the denominator (that would give the joint probability, not the conditional probability).

4Problem 4medium

❓ Question:

Use the table below:

| | Passed | Failed | Total | |---|---|---|---| | Studied | 42 | 8 | 50 | | Did Not Study | 18 | 32 | 50 | | Total | 60 | 40 | 100 |

What is the probability that a student passed, given that they studied?

💡 Show Solution

Key: This is a conditional probability question because of the phrase "given that they studied."

Step 1: Identify the condition: "given that they studied" means we only look at the "Studied" row.

Step 2: The denominator is the total who studied: 50 The numerator is those who studied AND passed: 42

$P(\text{Passed} \mid \text{Studied}) = \frac{42}{50} = \frac{21}{25} = 0.84$

Answer: $\frac{21}{25}$ or $84\%$

Common mistake: Using 100 as the denominator (that would give the joint probability, not the conditional probability).

5Problem 5medium

❓ Question:

Using the same table above, what fraction of students who passed had studied?

💡 Show Solution

Key: This question asks "of those who passed" — so the condition is passing.

Step 1: The denominator is the total who passed: 60 The numerator is those who passed AND studied: 42

$P(\text{Studied} \mid \text{Passed}) = \frac{42}{60} = \frac{7}{10}$

Answer: $\frac{7}{10}$

Important: Notice this is DIFFERENT from the previous question! $P(\text{Passed} | \text{Studied}) = \frac{42}{50}$ but $P(\text{Studied} | \text{Passed}) = \frac{42}{60}$ . The order matters in conditional probability!

6Problem 6medium

❓ Question:

Using the same table above, what fraction of students who passed had studied?

💡 Show Solution

Key: This question asks "of those who passed" — so the condition is passing.

Step 1: The denominator is the total who passed: 60 The numerator is those who passed AND studied: 42

$P(\text{Studied} \mid \text{Passed}) = \frac{42}{60} = \frac{7}{10}$

Answer: $\frac{7}{10}$

7Problem 7hard

❓ Question:

A survey asked 200 people about their exercise habits and diet:

| | Exercises Regularly | Does Not Exercise | Total | |---|---|---|---| | Healthy Diet | 65 | ? | 100 | | Unhealthy Diet | ? | 60 | ? | | Total | ? | ? | 200 |

Complete the table and find the probability that a randomly selected person exercises regularly OR has a healthy diet.

💡 Show Solution

Step 1: Fill in the table.

Healthy Diet row: Does Not Exercise = $100 - 65 = 35$ Unhealthy Diet total = $200 - 100 = 100$ Unhealthy Diet, Exercises = $100 - 60 = 40$ Exercises total = $65 + 40 = 105$ Does Not Exercise total = $35 + 60 = 95$

Completed table: | | Exercises | Doesn't | Total | |---|---|---|---| | Healthy | 65 | 35 | 100 | | Unhealthy | 40 | 60 | 100 | | Total | 105 | 95 | 200 |

Step 2: Find $P(\text{Exercises OR Healthy Diet})$

Use the inclusion-exclusion principle: $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$ $= \frac{105}{200} + \frac{100}{200} - \frac{65}{200} = \frac{140}{200} = \frac{7}{10}$

Answer: $\frac{7}{10}$ or $70\%$

8Problem 8hard

❓ Question:

A survey asked 200 people about their exercise habits and diet:

| | Exercises Regularly | Does Not Exercise | Total | |---|---|---|---| | Healthy Diet | 65 | ? | 100 | | Unhealthy Diet | ? | 60 | ? | | Total | ? | ? | 200 |

Complete the table and find the probability that a randomly selected person exercises regularly OR has a healthy diet.

💡 Show Solution

Step 1: Fill in the table.

Completed table: | | Exercises | Doesn't | Total | |---|---|---|---| | Healthy | 65 | 35 | 100 | | Unhealthy | 40 | 60 | 100 | | Total | 105 | 95 | 200 |

Step 2: Find $P(\text{Exercises OR Healthy Diet})$

Use the inclusion-exclusion principle: $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$ $= \frac{105}{200} + \frac{100}{200} - \frac{65}{200} = \frac{140}{200} = \frac{7}{10}$

Answer: $\frac{7}{10}$ or $70\%$

9Problem 9expert

❓ Question:

In a class, the probability of a student playing basketball is 0.4, the probability of playing soccer is 0.3, and the probability of playing both is 0.1. What is the probability that a randomly chosen student plays basketball but NOT soccer?

💡 Show Solution

Step 1: Use the relationship: $P(\text{Basketball only}) = P(\text{Basketball}) - P(\text{Basketball AND Soccer})$ $= 0.4 - 0.1 = 0.3$

Step 2: Verify with a Venn diagram mental model:

Basketball only: 0.3
Soccer only: $0.3 - 0.1 = 0.2$
Both: 0.1
Neither: $1 - (0.3 + 0.2 + 0.1) = 0.4$

All probabilities sum to 1: $0.3 + 0.2 + 0.1 + 0.4 = 1$ ✓

Answer: $0.3$ or $30\%$

SAT Tip: "A but NOT B" means subtract the overlap from A's probability.

10Problem 10expert

❓ Question:

💡 Show Solution

Step 1: Use the relationship: $P(\text{Basketball only}) = P(\text{Basketball}) - P(\text{Basketball AND Soccer})$ $= 0.4 - 0.1 = 0.3$

Step 2: Verify with a Venn diagram mental model:

Basketball only: 0.3
Soccer only: $0.3 - 0.1 = 0.2$
Both: 0.1
Neither: $1 - (0.3 + 0.2 + 0.1) = 0.4$

All probabilities sum to 1: $0.3 + 0.2 + 0.1 + 0.4 = 1$ ✓

Answer: $0.3$ or $30\%$

SAT Tip: "A but NOT B" means subtract the overlap from A's probability.

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