CONDITIONAL PROBABILITY AND MULTIPLICATIVE RULES

Tags: #◇Computer Science #◇STT101A #◇Conceptual #◇Conditional Probability and Multiplicative Rules

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◇Overview

Conditional Probability and the Multiplicative Rule link the probability of joint intersections $P(A \cap B)$ with sequential conditioning $P(A)P(B \mid A)$. Together, they enable multi-stage probability tree analysis, sampling without replacement, and Bayesian inference.

◇Key Definitions

• Conditional Probability: $P(B \mid A) = \frac{P(A \cap B)}{P(A)}$ for $P(A) > 0$.
• General Multiplicative Rule: $P(A \cap B) = P(A)P(B \mid A) = P(B)P(A \mid B)$.
• Chain Rule for $n$ Events: $P(A_1 \cap A_2 \cap \cdots \cap A_n) = P(A_1) P(A_2 \mid A_1) P(A_3 \mid A_1 \cap A_2) \cdots P(A_n \mid A_1 \cap \cdots \cap A_{n-1})$.
• Law of Total Probability (2 Branches): $P(B) = P(A)P(B \mid A) + P(A^c)P(B \mid A^c)$.

◇Worked Example

Problem: A box of $20$ microprocessors contains $4$ defective units. Two processors are selected at random without replacement. Find (a) the probability that both are defective, and (b) the probability that the second is defective given the first was non-defective.

Solution:

• Let $D_1, D_2$ be the events that the 1st and 2nd processors are defective.
• (a) By the Multiplicative Rule: $P(D_1 \cap D_2) = P(D_1) P(D_2 \mid D_1) = \frac{4}{20} \times \frac{3}{19} = \frac{12}{380} = \frac{3}{95} \approx 0.0316$.
• (b) Given $D_1^c$ occurred ($1$ good unit removed), $4$ defective units remain among $19$ total units: $P(D_2 \mid D_1^c) = \frac{4}{19} \approx 0.2105$.

◇References & Navigation

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• Related Notes: ◇Multiplicative Rule, ◇Additive Rule