GENERALIZED MULTIPLICATION

Tags: #◇Computer Science #◇STT101A #◇Conceptual #◇Generalized Multiplication

---

◇Overview

The Generalized Multiplication Rule (both in combinatorics and in probability) extends two-stage product rules to an arbitrary finite sequence of $k$ stages or $k$ events:

• Combinatorial Form: $N = \prod_{i=1}^k n_i = n_1 \cdot n_2 \cdots n_k$.
• Probabilistic Chain Rule: For events $A_1, A_2, \dots, A_k$ with $P(A_1 \cap \cdots \cap A_{k-1}) > 0$:
  $P\left(\bigcap_{i=1}^k A_i\right) = P(A_1) \cdot P(A_2 \mid A_1) \cdot P(A_3 \mid A_1 \cap A_2) \cdots P\left(A_k \;\middle|\; \bigcap_{j=1}^{k-1} A_j\right)$

◇Key Definitions

• Sequential Conditioning (Chain Rule): At step $i$, the probability of $A_i$ is conditioned on the joint occurrence of all preceding events $A_1 \cap A_2 \cap \cdots \cap A_{i-1}$.
• Mutual Independence Simplification: If $A_1, \dots, A_k$ are mutually independent, every conditional probability $P(A_i \mid A_1 \cap \cdots \cap A_{i-1})$ simplifies to $P(A_i)$, yielding $P(\bigcap_{i=1}^k A_i) = \prod_{i=1}^k P(A_i)$.

◇Worked Example

Problem: A batch of $12$ hard drives contains $9$ healthy drives and $3$ failing drives. Three drives are selected sequentially without replacement. What is the probability that all $3$ selected drives are healthy?

Solution:
Let $H_1, H_2, H_3$ be the events that the 1st, 2nd, and 3rd drives are healthy:
$P(H_1 \cap H_2 \cap H_3) = P(H_1) \cdot P(H_2 \mid H_1) \cdot P(H_3 \mid H_1 \cap H_2) = \frac{9}{12} \times \frac{8}{11} \times \frac{7}{10} = \frac{504}{1320} = \frac{21}{55} \approx 0.3818$

◇References & Navigation

• Previous Topic: ◇Multiplication Rule
• Next Topic: ◇Independent Events
• Related Notes: ◇Multiplicative Rule, ◇Contingency Table