Probability Tree & Contingency Table Calculator

Calculate multi-stage probability experiments, path probabilities (1st path rule), total probability (2nd path rule), Bayes conditional probabilities and synchronized contingency tables.

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Klassische Praxisbeispiele (Presets)
10.0% (0.100)
0%P(Gesund (G)) = 90.0%100%
95.0% (0.950)
0%P(Test Negativ (-)|Krank (K)) = 5.0%100%
5.0% (0.050)
0%P(Test Negativ (-)|Gesund (G)) = 95.0%100%
P(B) - Totale Wk.14.00%0.1400
P(Krank (K) ∩ Test Positiv (+))9.50%0.0950
P(Krank (K)|Test Positiv (+)) (Bayes)67.86%0.6786
UnabhängigkeitAbhängig

Baumdiagramm (2-Stufig)

Stochastisch abhängig
0.1000.9000.9500.0500.0500.950ΩKrank (K)Gesund (G)Test Positiv (+)Test Negativ (-)Test Positiv (+)Test Negativ (-)P(Krank (K)∩Test Positiv (+)) = 0.0950P(Krank (K)∩Test Negativ (-)) = 0.0050P(Gesund (G)∩Test Positiv (+)) = 0.0450P(Gesund (G)∩Test Negativ (-)) = 0.8550

Vierfeldertafel (Wahrscheinlichkeiten)Summe Zeilen & Spalten = Randwahrscheinlichkeiten

EreignisKrank (K)Gesund (G)Σ (Gesamt)
Test Positiv (+)0.09509.50%0.04504.50%0.1400P(Test Positiv (+))
Test Negativ (-)0.00500.50%0.855085.50%0.8600P(Test Negativ (-))
Σ (Gesamt)0.1000P(Krank (K))0.9000P(Gesund (G))1.0000 (100%)

📚Schritt-für-Schritt Rechenweg

1. Pfadregel (Multiplikationssatz fu¨r Pfade):\text{\textbf{1. Pfadregel (Multiplikationssatz für Pfade):}}
P(Krank(K)∩TestPositiv(+))=P(Krank(K))⋅P(TestPositiv(+)∣Krank(K))=0.1000⋅0.9500=0.0950P(Krank (K) \cap Test Positiv (+)) = P(Krank (K)) \cdot P(Test Positiv (+)|Krank (K)) = 0.1000 \cdot 0.9500 = 0.0950
P(Krank(K)∩TestNegativ(−))=P(Krank(K))⋅P(TestNegativ(−)∣Krank(K))=0.1000⋅0.0500=0.0050P(Krank (K) \cap Test Negativ (-)) = P(Krank (K)) \cdot P(Test Negativ (-)|Krank (K)) = 0.1000 \cdot 0.0500 = 0.0050
P(Gesund(G)∩TestPositiv(+))=P(Gesund(G))⋅P(TestNegativ(−)∣Gesund(G))=0.9000⋅0.0500=0.0450P(Gesund (G) \cap Test Positiv (+)) = P(Gesund (G)) \cdot P(Test Negativ (-)|Gesund (G)) = 0.9000 \cdot 0.0500 = 0.0450
P(Gesund(G)∩TestNegativ(−))=P(Gesund(G))⋅P(TestNegativ(−)∣Gesund(G))=0.9000⋅0.9500=0.8550P(Gesund (G) \cap Test Negativ (-)) = P(Gesund (G)) \cdot P(Test Negativ (-)|Gesund (G)) = 0.9000 \cdot 0.9500 = 0.8550
2. Pfadregel (Totale Wahrscheinlichkeit):\text{\textbf{2. Pfadregel (Totale Wahrscheinlichkeit):}}
P(TestPositiv(+))=P(Krank(K)∩TestPositiv(+))+P(Gesund(G)∩TestPositiv(+))=0.0950+0.0450=0.1400P(Test Positiv (+)) = P(Krank (K) \cap Test Positiv (+)) + P(Gesund (G) \cap Test Positiv (+)) = 0.0950 + 0.0450 = 0.1400
P(TestNegativ(−))=1−P(TestPositiv(+))=0.8600P(Test Negativ (-)) = 1 - P(Test Positiv (+)) = 0.8600
Satz von Bayes (Bedingte Wahrscheinlichkeit):\text{\textbf{Satz von Bayes (Bedingte Wahrscheinlichkeit):}}
P(Krank(K)∣TestPositiv(+))=P(Krank(K)∩TestPositiv(+))P(TestPositiv(+))=0.09500.1400=0.6786  (67.86%)P(Krank (K)|Test Positiv (+)) = \frac{P(Krank (K) \cap Test Positiv (+))}{P(Test Positiv (+))} = \frac{0.0950}{0.1400} = 0.6786 \; (67.86\%)
Stochastische Unabha¨ngigkeit: Da P(Krank(K)∩TestPositiv(+))=0.0950≠P(Krank(K))⋅P(TestPositiv(+))=0.0140, sind die Ereignisse A und B stochastisch abha¨ngig.\text{\textbf{Stochastische Unabhängigkeit:}}\text{ Da } P(Krank (K) \cap Test Positiv (+)) = 0.0950 \neq P(Krank (K)) \cdot P(Test Positiv (+)) = 0.0140, \text{ sind die Ereignisse } A \text{ und } B \text{ \textbf{stochastisch abhängig}.}

Historical Origin & The Challenge Solved

Why was this mathematics developed, where does it come from, and what real-world problem did it solve?

📜
Epoche & ZeitalterClassical Probability (1654 – 1763)
Pioniere: Blaise Pascal, Pierre de Fermat, Thomas Bayes

1. Unfinished Games of Chance & Diagnostic Fallacies

In 1654, Chevalier de Méré asked Pascal: how should the prize money be divided fairly if a dice game is interrupted prematurely? Furthermore, how does one assess diagnostic test accuracy when a disease is exceptionally rare?

2. Tree Diagram Path Rules & Bayes' Theorem

Fermat and Pascal developed multiplication and addition path rules. Thomas Bayes formulated Bayes' theorem for conditional probability, demonstrating how prior beliefs (prior) are updated by new evidence (posterior).

3. Spam Filters, Medical Diagnostics & Forensic DNA

Naive Bayes email spam filters, AI diagnostic assistants in radiology, and legal forensic evidence evaluation rely on Bayesian conditional probability.

„Probability theory is nothing but common sense reduced to calculation.“ — Pierre-Simon Laplace

Real-World Applications of Probability Trees & Bayes

Explore how multi-stage probability trees empower data-driven decisions in medicine, engineering, and daily risk management.

Healthcare & Diagnostics

Medical Screening & Bayes Paradox

Calculating positive predictive value under low population disease prevalence.

Praxis-Beispiel:

Even with 98% test accuracy, a rare 1% disease yields true positive probability of around 16%.

Quality Engineering

Manufacturing Defect Trees

Multi-stage production line failure rate analysis and defect origin tracking.

Praxis-Beispiel:

Determining total defect rate across multiple automated robotic assembly lines.

Logistics & Supply Chain

Supply Chain Risk Modeling

Combined stochastic delivery delays across multi-stop customs and weather hubs.

Praxis-Beispiel:

Predicting on-time delivery percentages for complex international logistics chains.

Guide: Probability Trees, Path Rules & Bayes Theorem

A probability tree diagram systematically structures sequential random events. It enables straightforward computation of intersection and marginal probabilities alongside contingency tables.

1st Path Rule (Multiplication Rule)

The probability of an entire branch path is the product of its individual probabilities: P(A ∩ B) = P(A) · P(B|A).

2nd Path Rule (Law of Total Probability)

To compute the overall probability of event B, add the probabilities of all mutually exclusive paths leading to B: P(B) = P(A ∩ B) + P(not A ∩ B).

Bayes Theorem

Calculates inverted conditional probability: P(A|B) = P(A ∩ B) / P(B) = (P(A) · P(B|A)) / P(B).

Stochastic Independence

Events A and B are stochastically independent if and only if P(A ∩ B) = P(A) · P(B).

Frequently Asked Questions about Probability Trees

What is the difference between replacement and without replacement?▼

With replacement, branch probabilities remain constant at every stage (independent draws). Without replacement, available item counts decrease, updating conditional probabilities dynamically.

When should I use a contingency table versus a tree diagram?▼

Contingency tables are optimal for static 2x2 categorical comparisons and fast marginal sum checks. Tree diagrams are best for chronological, sequential decision processes.

Why is Bayes Theorem crucial in medical diagnostics?▼

Because in low prevalence scenarios, false positives frequently outnumber true positives. Bayes Theorem computes the actual probability that a tested individual has the condition.

What is the intuitive meaning of stochastic independence?▼

Occurring of event A does not alter the likelihood of event B happening, meaning P(B|A) = P(B).

Wie hängen Baumdiagramm und Vierfeldertafel (2x2-Tabelle) zusammen?▼

Ein zweistufiges Baumdiagramm und eine Vierfeldertafel stellen dieselben stochastischen Zusammenhänge dar. Die Pfade des Baumdiagramms entsprechen den Schnittmengen P(A ∩ B) in den vier inneren Feldern der Vierfeldertafel. Aus der Vierfeldertafel lassen sich die Randwahrscheinlichkeiten P(A) und P(B) sowie die bedingten Wahrscheinlichkeiten P(B|A) = P(A ∩ B) / P(A) (Satz von Bayes) direkt ablesen und berechnen.