Probability with Two-Way Tables: Union of Events

What This Problem Teaches

• How to apply the inclusion-exclusion principle to find P(A or B)
• Reading and interpreting two-way frequency tables for probability calculations
• Understanding why you subtract the intersection to avoid double-counting
• Using the complement strategy as an alternative solution method
• Converting between counts and probabilities in real-world contexts

Visualizing the Data

Let's create a visual representation to understand the relationships between the events:

This Venn diagram shows how the 641 people are distributed across the four possible categories. Notice that 19 people appear in both circles—they drive red cars and got tickets.

Solution: Method 1 — The Inclusion-Exclusion Approach

Step 1 — Identify the events and their counts

From the table, we need:

• People with red cars: 457
• People with speeding tickets: 60
• People with both red cars and tickets: 19
• Total people surveyed: 641

Step 2 — Apply the inclusion-exclusion principle

For "A or B" events, we use: P(A or B) = P(A) + P(B) - P(A and B)

Step 3 — Convert counts to probabilities

Divide each count by the total sample size:

• P(red car) = 457/641
• P(ticket) = 60/641
• P(both) = 19/641

Step 4 — Perform the calculation

Step 5 — Simplify if possible

To check if this fraction can be simplified, we need the GCD of 498 and 641. Since 641 is prime, and 498 = 2 × 3 × 83, these numbers share no common factors. The fraction 498/641 is already in lowest terms.

Solution: Method 2 — The Complement Strategy

Step 1 — Identify the complement

Instead of finding P(red car or ticket), we'll find P(NOT red car AND NOT ticket), then subtract from 1.

Step 2 — Find people with neither characteristic

From the table, people who drive non-red cars AND have no ticket: 143 people

Step 3 — Calculate the complement probability

Step 4 — Use the complement rule

This confirms our first method's answer!

Verification

Let's verify our answer using a direct count approach:

People who drive red cars OR got tickets include:

• Red car drivers with tickets: 19
• Red car drivers without tickets: 438
• Non-red car drivers with tickets: 41

Total: 19 + 438 + 41 = 498 people

Probability: 498/641 ✓

As a decimal: 498 ÷ 641 ≈ 0.7769 or about 77.7% ✓

Watch Out For These

Does This Seem Reasonable?

Our answer of about 77.7% makes intuitive sense. Here's why:

• Red car drivers alone make up 457/641 ≈ 71.3% of the sample
• Ticket recipients add another 60/641 ≈ 9.4%
• But there's overlap of 19/641 ≈ 3.0% that we must subtract
• Net result: 71.3% + 9.4% - 3.0% = 77.7% ✓

The majority of people (71.3%) already drive red cars, so adding the ticket holders only increases this to about 78%. The small overlap prevents it from reaching the naive sum of 81%.

The Pattern Behind This

This problem demonstrates the fundamental inclusion-exclusion principle for the union of two events:

In words: "The probability of A or B equals the probability of A plus the probability of B minus the probability of both A and B."

This principle extends to any two-way table problem where you need to find the probability of belonging to at least one category. The key insight is that simple addition double-counts the intersection, so you must subtract it once to get the correct total.

Real Applications

This type of "union probability" calculation appears frequently in:

• Medical testing: Finding the probability that a patient has condition A or condition B when conditions sometimes co-occur
• Marketing analysis: Calculating the reach of advertising campaigns that target overlapping demographics
• Quality control: Determining the probability that a product fails for reason X or reason Y, where both failure modes can happen simultaneously

What If?