PROJECTENGRAM

MULTIPLICATIVE RULE

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


◇Overview

In probability theory, the Multiplicative Rule of Probability determines the probability of the intersection of two or more events, P(A∩B)P(A \cap B). It comes in two forms: the General Multiplicative Rule (for any events, using conditional probability) and the Special Multiplicative Rule (for independent events).

◇Key Definitions

  • General Multiplicative Rule (Dependent or Independent Events):
    P(A∩B)=P(A)P(B∣A)=P(B)P(A∣B)P(A \cap B) = P(A) P(B \mid A) = P(B) P(A \mid B)
  • Special Multiplicative Rule (Independent Events Only): If and only if AA and BB are independent:
    P(A∩B)=P(A)P(B)P(A \cap B) = P(A) P(B)
  • Extension to kk Independent Events: If A1,A2,…,AkA_1, A_2, \dots, A_k are mutually independent, P(A1∩A2∩⋯∩Ak)=∏i=1kP(Ai)P(A_1 \cap A_2 \cap \cdots \cap A_k) = \prod_{i=1}^k P(A_i).

◇Worked Example

Problem: Contrast drawing 22 Aces from a 5252-card deck (a) with replacement vs. (b) without replacement.

Solution:

  • (a) With Replacement (Independent): Special Multiplicative Rule applies: P(A1∩A2)=P(A1)P(A2)=452×452=1169≈0.00592P(A_1 \cap A_2) = P(A_1)P(A_2) = \frac{4}{52} \times \frac{4}{52} = \frac{1}{169} \approx 0.00592.
  • (b) Without Replacement (Dependent): General Multiplicative Rule applies: P(A1∩A2)=P(A1)P(A2∣A1)=452×351=1221≈0.00452P(A_1 \cap A_2) = P(A_1)P(A_2 \mid A_1) = \frac{4}{52} \times \frac{3}{51} = \frac{1}{221} \approx 0.00452.

◇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

When is it valid to use the Special Multiplicative Rule P(A∩B)=P(A)P(B)P(A \cap B) = P(A)P(B) instead of P(A∩B)=P(A)P(B∣A)P(A \cap B) = P(A)P(B \mid A)?

Q2

A box contains 66 working cables and 44 broken cables. Two cables are selected without replacement. What is the probability that both cables are broken?

Q3

Using the same box (66 working, 44 broken, drawn without replacement), what is the probability that the first is working and the second is broken?

Q4

Three independent components in a series system have reliability probabilities 0.90,0.90,0.90, 0.90, and 0.800.80. Because the system functions only if all three components function, what is the system reliability?

Q5

Why does a parallel redundant system of those same three independent components fail only with probability (0.10)(0.10)(0.20)=0.002(0.10)(0.10)(0.20) = 0.002?

Multiplicative Rule | PROJECT ENGRAM