Total Probability Rule

Tags: #◇Computer Science #◇STT101A #◇Conceptual #◇Total Probability Rule

◇Overview

The Total Probability Rule (Law of Total Probability) computes the unconditional (marginal) probability $P(A)$ of a target event $A$ by conditioning on a collection of mutually exclusive and collectively exhaustive scenarios $\{B_1, B_2, \dots, B_k\}$ that form a partition of the sample space $S$ .

◇Key Definitions

General Formula (for Partition $\{B_1, \dots, B_k\}$ ):
$P(A) = \sum_{i=1}^k P(A \cap B_i) = \sum_{i=1}^k P(A \mid B_i) P(B_i)$
Binary Partition Form ( $\{B, B^c\}$ ):
$P(A) = P(A \mid B)P(B) + P(A \mid B^c)P(B^c)$
Role in Bayes' Theorem: The Law of Total Probability provides the normalizing denominator $P(A)$ in Bayes' Rule: $P(B_j \mid A) = \frac{P(A \mid B_j)P(B_j)}{\sum_{i=1}^k P(A \mid B_i)P(B_i)}$ .

◇Worked Example

Problem: An e-commerce cluster routes $50\%$ of requests to Server $B_1$ , $30\%$ to Server $B_2$ , and $20\%$ to Server $B_3$ . Timeout rates on the three servers are $P(T \mid B_1) = 0.01$ , $P(T \mid B_2) = 0.02$ , and $P(T \mid B_3) = 0.05$ . What is the overall probability $P(T)$ that a randomly arriving request times out?

Solution:
$P(T) = P(T \mid B_1)P(B_1) + P(T \mid B_2)P(B_2) + P(T \mid B_3)P(B_3)$
$P(T) = (0.01)(0.50) + (0.02)(0.30) + (0.05)(0.20) = 0.005 + 0.006 + 0.010 = 0.021 \quad (2.1\%)$

Previous Topic: ◇Partition of Sample Space
Next Topic: ◇Bayes's Rules
Related Notes: ◇Bayes's Formula, ◇Contingency Table

CONFIRMATION // 理解度の確認

ACTIVE RECALL VERIFICATION

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

What condition MUST the events $B_1, B_2, \dots, B_k$ satisfy for the Law of Total Probability $P(A) = \sum_{i=1}^k P(A \mid B_i)P(B_i)$ to hold?

Suppose $70\%$ of emails arriving at a gateway are legitimate ( $L$ ) and $30\%$ are spam ( $S$ ). A keyword filter flags $80\%$ of spam emails ( $P(F \mid S) = 0.80$ ) and falsely flags $5\%$ of legitimate emails ( $P(F \mid L) = 0.05$ ). What is the total probability $P(F)$ that an incoming email is flagged?

In a probability tree diagram where the first stage branches into $B_1, \dots, B_k$ and the second stage branches into $A$ and $A^c$ , how does the Law of Total Probability compute $P(A)$ ?

Why is $P(A)$ in the Law of Total Probability interpretable as a weighted average of the conditional probabilities $P(A \mid B_i)$ ?

Box 1 has $3$ red and $1$ blue ball; Box 2 has $1$ red and $3$ blue balls. A fair coin is tossed to pick a box ( $P(B_1) = P(B_2) = 0.5$ ), and one ball is drawn. What is the total probability of drawing a red ball?