Discover the power of conditional probability! Learn to use two-way tables to analyze relationships between events and understand when events are independent or dependent.
• A table that shows the relationship between two categorical variables
• Rows represent one variable, columns represent another
• Each cell shows the count (frequency) of observations
• Totals are shown in margins (row totals, column totals, grand total)
Survey: 100 students were asked about their favorite subject and whether they play sports.
| Math | Science | English | Total | |
|---|---|---|---|---|
| Play Sports | 15 | 20 | 10 | 45 |
| Don't Play Sports | 25 | 15 | 15 | 55 |
| Total | 40 | 35 | 25 | 100 |
P(A|B) = P(A and B) / P(B)
• P(A|B) = "Probability of A given B"
• P(A and B) = "Probability of both A and B occurring"
• P(B) = "Probability of B occurring"
Question: What is the probability that a student likes Math, given that they play sports?
Step 1: Identify the given condition
Given: Student plays sports (45 students total)
Step 2: Find the intersection
Students who play sports AND like Math = 15
Step 3: Calculate conditional probability
P(Math | Sports) = 15/45 = 1/3 ≈ 0.333
Interpretation: 33.3% of students who play sports like Math
Question: What is the probability that a student plays sports, given that they like Science?
Step 1: Identify the given condition
Given: Student likes Science (35 students total)
Step 2: Find the intersection
Students who like Science AND play sports = 20
Step 3: Calculate conditional probability
P(Sports | Science) = 20/35 = 4/7 ≈ 0.571
Interpretation: 57.1% of students who like Science play sports
Two events A and B are independent if:
P(A|B) = P(A) or P(B|A) = P(B)
This means that knowing one event occurred doesn't change the probability of the other event.
Question: Are "liking Math" and "playing sports" independent events?
Step 1: Calculate P(Math)
P(Math) = 40/100 = 0.4
Step 2: Calculate P(Math | Sports)
P(Math | Sports) = 15/45 = 1/3 ≈ 0.333
Step 3: Compare the probabilities
P(Math) = 0.4 ≠ P(Math | Sports) = 0.333
Conclusion: The events are NOT independent
Knowing that a student plays sports changes the probability that they like Math.
Scenario: A medical test for a disease has 95% accuracy. In a population where 2% have the disease, what's the probability someone has the disease given a positive test result?
Given Information:
• Disease prevalence: 2%
• Test accuracy: 95%
• False positive rate: 5%
Two-Way Table (per 1000 people):
| Has Disease | No Disease | Total | |
|---|---|---|---|
| Positive Test | 19 | 49 | 68 |
| Negative Test | 1 | 931 | 932 |
| Total | 20 | 980 | 1000 |
Solution:
P(Disease | Positive Test) = 19/68 ≈ 0.279
Only 27.9% of people with positive test results actually have the disease!
P(A|B) ≠ P(B|A). The order matters! P(A|B) means "A given B", not "B given A"
For P(A|B), use the total for B as the denominator, not the grand total
Don't assume events are independent. Always test using P(A|B) = P(A)
Problem 1:
From the table above, find P(English | Don't Play Sports)
P(English | Don't Play Sports) = 15/55 = 3/11 ≈ 0.273
Problem 2:
Are "liking Science" and "playing sports" independent events?
P(Science) = 35/100 = 0.35
P(Science | Sports) = 20/45 ≈ 0.444
Since 0.35 ≠ 0.444, the events are NOT independent
Problem 3:
Find P(Don't Play Sports | Math)
P(Don't Play Sports | Math) = 25/40 = 5/8 = 0.625
Formula:
P(A|B) = P(B|A) × P(A) / P(B)
Use: Updating probabilities with new information
Formula:
P(A) = Σ P(A|Bᵢ) × P(Bᵢ)
Use: Finding total probability across all scenarios
Disease Testing
False positive/negative rates
Treatment effectiveness and side effects
Manufacturing
Defect detection and prevention
Process optimization and monitoring
Insurance
Premium calculation and claims
Investment portfolio management