Contingency Tables (Vierfeldertafel)

15 min · 2 drills

Why do we need this table?

Imagine this problem: "70% are vaccinated, 5% of the vaccinated got sick, 20% of the unvaccinated got sick." Your head is spinning from all those percentages!

A Vierfeldertafel (German for "four-field table") organizes everything for you. Instead of juggling percentages in your head, you put everything into a table and see the complete picture.

On the EPM exam, these problems appear constantly. Good news: once you master the table, solving them becomes almost automatic.

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How the table works

The idea is simple: we have people (or objects) and divide them by two characteristics. For example:

• Characteristic A: vaccinated / not vaccinated
• Characteristic B: got sick / stayed healthy

This gives us 4 groups, hence the name "four-field":

The golden rule: numbers in a row add up to the row total, numbers in a column add up to the column total. It's like Sudoku: if you know some numbers, you can figure out the rest.

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How to fill in the table: step-by-step

Let's work through a concrete example.

Problem: 70% of the population is vaccinated. Among the vaccinated, 5% got sick. Among the unvaccinated, 20% got sick.

Step 1: Choose a convenient total number

Percentages are fractions of a whole. To avoid messy decimals, let's pretend we have exactly 1000 people (you could use 100, but 1000 is easier when percentages give decimals).

Step 2: Fill in what we know directly

• Vaccinated: 70% of 1000 = 700 people
• Unvaccinated: the rest = 300 people

Step 3: Translate "of them X% got sick"

Here's where people mess up! "Of the vaccinated, 5% got sick" means:

• Take the 700 vaccinated people
• Calculate 5% of 700 (not of 1000!)
• We get: $700\times 0.05=35$ people

Same for the unvaccinated:

• 20% of 300 = $300\times 0.2=60$ people

Step 4: Fill in the rest

Now it's just arithmetic:

• Healthy vaccinated: $700-35=665$
• Healthy unvaccinated: $300-60=240$
• Total sick: $35+60=95$
• Total healthy: $665+240=905$

Step 5: Check your work

Add up the totals: $95+905=1000$ ✓

Now we can answer any question from this table!

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How to calculate probabilities from the table

Regular probability: "what's the probability that a random person is sick?"

This is the easiest. We look: 1000 people total, 95 are sick.

$$P(\text{sick})==\frac{\text{how many are sick}}{\text{how many total}}==\frac{95}{1000}==0.095$$

The intuition: if you randomly pick someone from a crowd of 1000, about 95 of them would be sick. So your chance of "landing on" a sick person is 95 out of 1000.

Conditional probability: "if someone is sick, what's the probability they were vaccinated?"

This is trickier. We're told: "the person is already sick." This means we're not picking from all 1000; we're picking only from the 95 sick people.

Among these 95 sick people, 35 were vaccinated.

$$P(\text{vaccinated}\mid \mid \text{sick})==\frac{35}{95}==\frac{7}{19}\approx \approx 0.37$$

The intuition: imagine only sick people are standing in front of you (95 of them). How many were vaccinated? 35. That's your answer: 35 out of 95.

The key difference

• Regular probability: divide by the total count (all 1000)
• Conditional probability: divide by the count of those we already know something about (only the sick, only the vaccinated, etc.)

The bar "|" is read as "given that." $P(A|B)$ means the probability of A, knowing that B has already happened.

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Watch out: common trap!

$P(\text{vaccinated}|\text{sick})$ and $P(\text{sick}|\text{vaccinated})$ are different numbers!

From our table:

• $P(\text{vaccinated}|\text{sick})=\frac{35}{95}\approx 0.37$ (among sick people, 37% are vaccinated)
• $P(\text{sick}|\text{vaccinated})=\frac{35}{700}=0.05$ (among vaccinated people, 5% are sick)

It's like "all cats are animals" doesn't mean "all animals are cats."

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Example with an incomplete table

On the exam, you'll often get a table with missing values. Don't panic: use the sum rule.

Problem:

Solve it like a puzzle:

1. How many who studied passed? The "Studied" row totals 60, and 15 failed. So passed: $60-15=45$

2. How many didn't study? Total 100, studied 60. So: $100-60=40$

3. How many who didn't study failed? The "Didn't study" row totals 40, and 20 passed. So: $40-20=20$

4. Column totals: passed $45+20=65$, failed $15+20=35$

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Independence of events

Sometimes they ask: "Does sickness depend on vaccination?" or "Are events A and B independent?"

The intuition: events are independent if knowing about one tells you nothing about the other. For example, the color of your socks doesn't affect your exam grade; these are independent events.

How to check using the table:

If events are independent, then the proportion of B among A is the same as the proportion of B among everyone:

$$P(B\mid \mid A)==P(B)$$

Or put differently: the probability of "both happening" equals the product of individual probabilities:

$$P(A\cap \cap B)==P(A)\times \times P(B)$$

Example:

Let's check:

• $P(A)=50/100=0.5$
• $P(B)=40/100=0.4$
• $P(A\cap B)=20/100=0.2$
• $P(A)\times P(B)=0.5\times 0.4=0.2$ ✓

They're equal! So A and B are independent.

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Worked Examples

Example 1: Completing a table

In a school with 400 students, 180 play sports. 240 are girls, of whom 100 play sports.

Calculation: Total boys: $400-240=160$. Boys playing sports: $180-100=80$. Boys not playing sports: $160-80=80$.

$P(\text{Sports}|\text{Girls})=\frac{100}{240}\approx 0.417$ $P(\text{Sports}|\text{Boys})=\frac{80}{160}=0.5$

Answer: Boys play sports more often (50% vs. 41.7%).

Example 2: Checking independence

From the table above: Are "Sports" and "Gender" independent?

$P(\text{Sports})=\frac{180}{400}=0.45$ $P(\text{Sports}|\text{Girls})=\frac{100}{240}\approx 0.417\ne 0.45$

Since $P(\text{Sports}|\text{Girls})\ne P(\text{Sports})$, the events are not independent.

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Common exam mistakes

Mistake 1: Confusing "of them" with "of all"

"Of the vaccinated, 5% got sick" means 5% of the vaccinated, not of everyone!

Wrong: $1000\times 0.05=50$

Right: $700\times 0.05=35$

Mistake 2: Mixing up P(A|B) and P(B|A)

Read the question carefully:

• "If sick, what's the probability of being vaccinated?" → look among the sick
• "If vaccinated, what's the probability of getting sick?" → look among the vaccinated

Mistake 3: Forgetting to check the sums

After filling in the table, make sure:

• Each row adds up to its total
• Each column adds up to its total
• The grand total = N

If something doesn't add up, you made an arithmetic error somewhere.

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Cheat sheet

Regular probability:

$$P(A)==\frac{\text{how many have trait A}}{\text{how many total}}$$

Conditional probability (A given B):

$$P(A\mid \mid B)==\frac{\text{how many have both A and B}}{\text{how many have B}}$$

Independence: $P(A|B)=P(A)$ or $P(A\cap B)=P(A)\times P(B)$

Pro tip: always build the table. Even if you think you can solve it "in your head," the table will save you from mistakes.

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Check Your Understanding

Your turn1 of 2

In a company 25% of the employees work part-time. Of the part-time employees, 14% commute by train, and of the full-time employees, 2%. What share of the train commuters works part-time?

A87.5% of the train commutersB14% of the train commutersC70% of the train commutersD56% of the train commuters

Check

Your turn2 of 2

Of 250 respondents, 150 do sport regularly and 125 are vegetarian. 50 respondents do sport but are not vegetarian. What share of those who do sport is vegetarian?

A$\frac{2}{3}$ of those who do sportB$\frac{3}{5}$ of those who do sportC$\frac{4}{5}$ of those who do sportD$\frac{1}{2}$ of those who do sport

Check

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Formulas to Remember

Sum rule: Every row and column of the contingency table adds up to its marginal total.

Conditional probability: $P(A|B)=\frac{P(A\cap B)}{P(B)}$

Independence: $A$ and $B$ are independent $\iff$ $P(A\cap B)=P(A)\cdot P(B)$

Complement: $P(\bar{A})=1-P(A)$

Union: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, equivalently $P(A\cup B)=1-P(\bar{A}\cap \bar{B})$

Exam tip: Always fill in the marginal totals first, then the inner cells; this way you avoid arithmetic errors.