PROJECTENGRAM

ADDITIVE RULE

Tags: #◇Computer Science #◇STT101A #◇Conceptual #◇Additive Rule


◇Overview

The Additive Rule of Probability (also known as the Principle of Inclusion-Exclusion for two events) calculates the probability that at least one of two events AA or BB occurs, denoted P(A∪B)P(A \cup B). Because summing P(A)P(A) and P(B)P(B) counts outcomes in the overlap A∩BA \cap B twice, the intersection probability P(A∩B)P(A \cap B) must be subtracted once.

◇Key Definitions

  • Union (A∪BA \cup B): The event that AA occurs, BB occurs, or both occur simultaneously.
  • General Additive Rule: P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B) for any two events AA and BB.
  • Special Additive Rule (Mutually Exclusive Events): When A∩B=∅A \cap B = \emptyset, P(A∩B)=0P(A \cap B) = 0, simplifying the rule to P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B).
  • Three-Event Inclusion-Exclusion: P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C).

◇Worked Example

Problem: In a university server cluster, P(High CPU)=0.35P(\text{High CPU}) = 0.35, P(High Memory)=0.25P(\text{High Memory}) = 0.25, and the probability of both occurring simultaneously is P(High CPU∩High Memory)=0.12P(\text{High CPU} \cap \text{High Memory}) = 0.12. What is the probability that a server experiences at least one of these resource spikes?

Solution:
P(CPU∪Mem)=0.35+0.25−0.12=0.48P(\text{CPU} \cup \text{Mem}) = 0.35 + 0.25 - 0.12 = 0.48

◇References & Navigation

CONFIRMATION // 理解度の確認

ACTIVE RECALL VERIFICATION

Select an answer to verify conceptual comprehension and reveal the technical rationale.

Q1

For any two events AA and BB in a sample space SS, which formula gives the general Additive Rule?

Q2

Why must P(A∩B)P(A \cap B) be subtracted in the general Additive Rule?

Q3

Given P(A)=0.45P(A) = 0.45, P(B)=0.30P(B) = 0.30, and P(A∪B)=0.60P(A \cup B) = 0.60, what is P(A∩B)P(A \cap B)?

Q4

Under what condition does P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B) hold exactly?

Q5

In the three-event Additive Rule for P(A∪B∪C)P(A \cup B \cup C), why is +P(A∩B∩C)+P(A \cap B \cap C) added back at the end?