Calculate probabilities from two-way tables and counting principles.
How can I study Probability and Two-Way Tables effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 10 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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What course covers Probability and Two-Way Tables?▾
Probability and Two-Way Tables is part of the SAT Prep course on Study Mondo, specifically in the Problem Solving and Data Analysis section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Probability and Two-Way Tables?
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(Student AND Likes Pizza)=9045=21
The denominator is the grand total.
Marginal Probability
The probability of just one category.
P(Student)=9060=32
P(Likes Pizza)=9065=1813
Conditional Probability
The probability of one event GIVEN another has occurred.
P(Likes Pizza∣Student)=6045=43
Key: The denominator is the subtotal of the given condition, not the grand total!
P(Student∣Likes Pizza)=6545=139
Conditional Probability Formula
P(A∣B)=P(B)P(A and B)
The SAT usually tests this with two-way tables rather than the formula directly.
Complement Rule
P(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 and B)=P(A∣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(not happening)
=
10
Step 2: Apply the probability formula:
P(blue)=total marblesblue marbles=103
Answer:103 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(blue)=total marblesblue marbles=10
Answer:103 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(Passed∣Studied)=
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(Passed∣Studied)=
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(Studied∣Passed)=6042=107
Answer:107
Important: Notice this is DIFFERENT from the previous question! P(Passed∣Studied)=5042 but . 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(Studied∣Passed)=6042=107
Answer:107
Important: Notice this is DIFFERENT from the previous question! P(Passed∣Studied)=5042 but . The order matters in conditional probability!
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 =
Exercises total =
Does Not Exercise total =
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.
Healthy Diet row: Does Not Exercise = 100−65=35
Unhealthy Diet total = 200−100=100
Unhealthy Diet, Exercises =
Exercises total =
Does Not Exercise total =
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(Basketball only)=P(Basketball)−P(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)
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:
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(Basketball only)=P(Basketball)−P(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)
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.
▾
Yes, this page includes 10 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.
3
5042
=
2521=
0.84
Answer:2521 or 84%
Common mistake: Using 100 as the denominator (that would give the joint probability, not the conditional probability).
5042
=
2521=
0.84
Answer:2521 or 84%
Common mistake: Using 100 as the denominator (that would give the joint probability, not the conditional probability).
P(Studied∣Passed)=
6042
P(Studied∣Passed)=
6042
100−60=40
65+40=105
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(Exercises OR Healthy Diet)
Use the inclusion-exclusion principle:
P(A or B)=P(A)+P(B)−P(A and B)=200105+200100−
Answer:107 or 70%
100−60=40
65+40=105
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(Exercises OR Healthy Diet)
Use the inclusion-exclusion principle:
P(A or B)=P(A)+P(B)−P(A and B)=200105+200100−