Probability Series Reading Time: 15 min Part 1 of 3

The Law of Total Probability & Bayes' Theorem - Part 1 (The Setup)

Build strong foundations in probability theory with intuitive explanations and real-world examples. Essential for JEE probability problems.

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Key Concepts

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Why This Foundation Matters

The Law of Total Probability and Bayes' Theorem are among the most powerful tools in probability theory. They appear in every JEE paper and form the basis for solving complex real-world probability problems.

🎯 JEE Strategic Importance

Direct questions worth 3-4 marks in every paper
Essential for solving conditional probability problems
Foundation for probability distributions in advanced topics
Used in real-world applications from medicine to machine learning

🧭 Navigation Guide

Prerequisites Partition Concept Total Probability Law Bayes' Theorem

1. Essential Prerequisites

Conditional Probability Refresher

Definition

The probability of event A given that event B has occurred is:

$$ P(A|B) = \frac{P(A \cap B)}{P(B)} $$

Provided $P(B) > 0$

📚 Example: Card Drawing

From a standard deck of 52 cards, if we know a card is red, what's the probability it's a heart?

Given: Total red cards = 26, Hearts (red) = 13

Conditional Probability: $P(\text{Heart}|\text{Red}) = \frac{13}{26} = \frac{1}{2}$

Multiplication Rule

Formula

From conditional probability, we get:

$$ P(A \cap B) = P(A|B) \cdot P(B) = P(B|A) \cdot P(A) $$

💡 Intuitive Understanding

Think of probability as "fraction of possibilities":

$P(B)$ = Fraction of cases where B happens
$P(A|B)$ = Fraction of B-cases where A also happens
$P(A \cap B)$ = Fraction of total cases where both happen

2. The Concept of Partition

What is a Partition?

Mathematical Definition

Events $B_1, B_2, \ldots, B_n$ form a partition of sample space S if:

$B_i \cap B_j = \emptyset$ for $i \neq j$ (Mutually Exclusive)
$B_1 \cup B_2 \cup \cdots \cup B_n = S$ (Collectively Exhaustive)
$P(B_i) > 0$ for all $i$ (Non-zero Probability)

🎯 Real-world Example: Weather Partition

Consider weather conditions for a day:

B₁
Sunny

B₂
Cloudy

B₃
Rainy

These form a partition because:

A day can't be both sunny and rainy (mutually exclusive)
Every day must be one of these (collectively exhaustive)
Each has non-zero probability

Visualizing Partitions

Sample Space S partitioned into B₁, B₂, B₃

B₁

B₂

B₃

Mutually Exclusive + Collectively Exhaustive

💡 Why Partitions Matter

Partitions allow us to break complex probability problems into simpler pieces. Instead of calculating probability directly, we calculate it through different "pathways" provided by the partition.

3. Law of Total Probability

The Fundamental Idea

Theorem Statement

If $B_1, B_2, \ldots, B_n$ form a partition of sample space S, then for any event A:

$$ P(A) = \sum_{i=1}^{n} P(A|B_i) \cdot P(B_i) $$

Step-by-step Derivation

Step 1: Since $B_i$'s cover entire sample space:

$$ A = A \cap S = A \cap (B_1 \cup B_2 \cup \cdots \cup B_n) $$

Step 2: Using distributive property:

$$ A = (A \cap B_1) \cup (A \cap B_2) \cup \cdots \cup (A \cap B_n) $$

Step 3: Since $B_i$'s are mutually exclusive, so are $A \cap B_i$'s:

$$ P(A) = \sum_{i=1}^{n} P(A \cap B_i) $$

Step 4: Using multiplication rule:

$$ P(A) = \sum_{i=1}^{n} P(A|B_i) \cdot P(B_i) $$

Real-world Application

🏥 Medical Testing Example

A disease affects 2% of population. Test is 95% accurate for sick people and 90% accurate for healthy people. What's the probability of testing positive?

Define Partition:

$B_1$: Person has disease → $P(B_1) = 0.02$
$B_2$: Person is healthy → $P(B_2) = 0.98$

Define Conditional Probabilities:

$P(\text{Positive}|B_1) = 0.95$ (True Positive)
$P(\text{Positive}|B_2) = 0.10$ (False Positive)

Apply Law of Total Probability:

$$ P(\text{Positive}) = 0.95 \times 0.02 + 0.10 \times 0.98 = 0.117 $$

About 11.7% of people will test positive, even though only 2% are actually sick!

4. Introduction to Bayes' Theorem

The "Reverse" Probability

Bayes' Theorem Formula

For events A and B with $P(A) > 0$, $P(B) > 0$:

$$ P(B|A) = \frac{P(A|B) \cdot P(B)}{P(A)} $$

🔄 What Makes Bayes' Theorem Special?

Bayes' Theorem allows us to update our beliefs when we get new evidence:

Prior Probability $P(B)$: What we believe before evidence
Likelihood $P(A|B)$: How likely evidence is if our belief is true
Posterior Probability $P(B|A)$: Updated belief after seeing evidence

Bayes' Theorem with Partitions

Extended Form

When we have a partition $B_1, B_2, \ldots, B_n$:

$$ P(B_i|A) = \frac{P(A|B_i) \cdot P(B_i)}{\sum_{j=1}^{n} P(A|B_j) \cdot P(B_j)} $$

Notice the denominator is exactly $P(A)$ from Law of Total Probability!

🔍 Continuing Medical Example

If someone tests positive, what's the probability they actually have the disease?

We want: $P(\text{Disease}|\text{Positive})$

Using Bayes' Theorem:

$$ P(\text{Disease}|\text{Positive}) = \frac{0.95 \times 0.02}{0.117} \approx 0.162 $$

Only about 16.2% chance of actually having disease despite positive test!

📝 Quick Practice Problems

Test your understanding of the foundational concepts:

1. A factory has 3 machines producing items. Machine A makes 50% with 2% defect rate, Machine B makes 30% with 3% defect rate, Machine C makes 20% with 5% defect rate. What's the probability a randomly chosen item is defective?

2. In the factory above, if an item is defective, what's the probability it came from Machine C?

3. Do these events form a partition: {Hearts, Diamonds, Clubs, Spades} in a card deck? Why or why not?

💎 Key Takeaways - Part 1

Conceptual Foundations

Partitions break sample space into mutually exclusive, collectively exhaustive pieces
Law of Total Probability calculates probabilities through different pathways
Bayes' Theorem reverses conditional probabilities using evidence
These tools work together to solve complex probability problems

Problem-Solving Strategy

Identify if you have a natural partition in the problem
Use Law of Total Probability when you need $P(A)$ but have conditional information
Use Bayes' Theorem when you want to "reverse" conditional probabilities
Always verify your partition meets all three conditions

Ready for Part 2?

In the next part, we'll dive into advanced applications and JEE-level problems

Continue to Part 2 →

🔍 Related JEE Probability Topics

Conditional Probability Independent Events Probability Distributions Random Variables