Independent Events (Union)

Deep Breakdown

What it tells me (Situation Awareness): When $A$ and $B$ are independent, the intersection term in the union formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ simplifies: $P(A \cap B) = P(A)P(B)$. This gives the union probability for independent events without needing to observe the joint directly.

Flowchart

Real-world use

Where it is used

System reliability: probability that at least one redundant component works. A/B test multi-metric: probability of detecting a significant result on at least one independent metric. Ensemble model: probability that at least one model flags an anomaly.

Problem it solves

Computes the union probability when events are known to be independent, without needing empirical joint observations.

How it solves the problem

Substitutes $P(A \cap B) = P(A) P(B)$ (from independence) into the inclusion-exclusion formula.

Example in a business

Two independent fraud signals each detect fraud with probability 0.70. Combined probability of detection: $0.70 + 0.70 - 0.70 \times 0.70 = 0.91$. Ensemble models use this logic to increase recall.

The formula, piece by piece

$P(A \cup B) = P(A) + P(B) - P(A) \cdot P(B)$

• Valid ONLY when $A$ and $B$ are independent.
• Equivalent to: $1 - P(A^c) P(B^c) = 1 - (1-P(A))(1-P(B))$ — probability neither fails.

The formula in words

"For independent A and B, the probability that at least one occurs equals A plus B minus their product." $P = 0.7 + 0.7 - 0.49 = 0.91$.

Critical Concepts

• The "neither occurs" form is sometimes cleaner: $P(A \cup B) = 1 - (1-p_A)(1-p_B)$.
• Generalizes: $P(\bigcup_i A_i) = 1 - \prod_i (1 - P(A_i))$ for mutually independent events.
• Used in ensemble methods: $P(\text{ensemble correct}) = 1 - \prod_i P(\text{model}_i \text{ wrong})$.

Step-by-step calculation (tiny example)

Step 1: Two independent classifiers, each 70% accurate $P(A) = P(B) = 0.70$. Independent.

Step 2: P(at least one correct) $P(A \cup B) = 0.70 + 0.70 - 0.70 \times 0.70 = 1.40 - 0.49 = 0.91$

Step 3: Verify using complement $1 - (1 - 0.70)^2 = 1 - 0.09 = 0.91$. Consistent.

Code examples

Beginner example (Python)

# Ensemble: P(at least one of 3 classifiers is correct)

def p_union_independent(probabilities: list[float]) -> float:
    """P(at least one occurs) for independent events."""
    p_none = 1.0
    for p in probabilities:
        p_none *= (1 - p)
    return 1 - p_none   # 3 classifiers with different individual accuracies
acc = [0.70, 0.75, 0.80]
ensemble_prob = p_union_independent(acc)
print(f"P(at least one correct) = {ensemble_prob:.4f}")

Beginner example (R)

p_union_independent <- function(probabilities) { 1 - prod(1 - probabilities)
}
acc <- c(0.70, 0.75, 0.80)
cat("P(at least one correct) =", round(p_union_independent(acc), 4), "\n")

Professional example (Python)

import numpy as np

def ensemble_detection_probability( individual_recalls: np.ndarray, assume_independent: bool = True ) -> float:
    """Compute P(at least one model detects event) for an ensemble. Under independence: P(union) = 1 - prod(1 - recall_i). Args: individual_recalls: Array of per-model recall values. assume_independent: If True, use independence product formula. Returns: Ensemble-level detection probability. """
    if not assume_independent:
        raise NotImplementedError("Dependent case requires joint distribution data.")
    p_no_detection = np.prod(1 - individual_recalls)
    return float(1 - p_no_detection)   # Analyze ensemble recall vs number of independent models
baseline_recall = 0.70
for n_models in [1, 2, 3, 5, 10]:
    recalls = np.full(n_models, baseline_recall)
ensemble_recall = ensemble_detection_probability(recalls)
print(f"n_models={n_models:>3}: ensemble recall = {ensemble_recall:.4f}")

Source: domain knowledge

#probability-statistics#probability-theory#conditional-probability-and-independence