The Law of Total Probability & Bayes' Theorem - Part 2 (The Revelation)
Master advanced applications with step-by-step problem solving, probability trees, and real JEE examples.
From Basics to Advanced Applications
In Part 1, we covered the fundamentals. Now, we dive into advanced problem-solving techniques that appear in JEE Mains and Advanced. These concepts help solve complex probability problems by breaking them into manageable parts.
Core Formulas Recap
Law of Total Probability
$$P(A) = \sum_{i=1}^{n} P(A|B_i)P(B_i)$$
where $B_1, B_2, ..., B_n$ form a partition of sample space
Bayes' Theorem
$$P(B_i|A) = \frac{P(A|B_i)P(B_i)}{\sum_{j=1}^{n} P(A|B_j)P(B_j)}$$
Updates probability based on new evidence
Concept 1: Multi-Stage Probability Trees
A bag contains 4 red and 6 blue balls. Two balls are drawn successively without replacement. What is the probability that the second ball is red?
Probability Tree Visualization
Start → First Draw:
• Red (4/10) → Second Draw: Red (3/9) or Blue (6/9)
• Blue (6/10) → Second Draw: Red (4/9) or Blue (5/9)
Solution Approach:
Step 1: Define events:
• $R_1$: First ball is red
• $B_1$: First ball is blue
• $R_2$: Second ball is red
Step 2: Apply Law of Total Probability:
$P(R_2) = P(R_2|R_1)P(R_1) + P(R_2|B_1)P(B_1)$
Step 3: Substitute values:
$P(R_2) = \frac{3}{9} \times \frac{4}{10} + \frac{4}{9} \times \frac{6}{10}$
Step 4: Calculate:
$P(R_2) = \frac{12}{90} + \frac{24}{90} = \frac{36}{90} = \frac{2}{5}$
Key Insight: The probability remains the same as drawing a red ball first!
Concept 2: Medical Testing & False Positives
A disease affects 1% of population. A test is 99% accurate (99% true positive, 99% true negative). If a person tests positive, what is the probability they actually have the disease?
Bayesian Reasoning:
Step 1: Define events:
• $D$: Has disease, $P(D) = 0.01$
• $D^c$: No disease, $P(D^c) = 0.99$
• $T^+$: Tests positive
Step 2: Given probabilities:
• $P(T^+|D) = 0.99$ (True positive)
• $P(T^+|D^c) = 0.01$ (False positive)
Step 3: Apply Bayes' Theorem:
$P(D|T^+) = \frac{P(T^+|D)P(D)}{P(T^+|D)P(D) + P(T^+|D^c)P(D^c)}$
Step 4: Substitute values:
$P(D|T^+) = \frac{0.99 \times 0.01}{0.99 \times 0.01 + 0.01 \times 0.99}$
Step 5: Calculate:
$P(D|T^+) = \frac{0.0099}{0.0099 + 0.0099} = \frac{1}{2} = 0.5$
Revelation: Even with 99% accuracy, a positive test means only 50% chance of actually having the disease!
Concept 3: Multiple Choice & Elimination
In a multiple choice question with 4 options, a student knows the answer with probability 0.7. If they don't know, they guess randomly. Given they answered correctly, what's the probability they actually knew the answer?
Bayesian Analysis:
Step 1: Define events:
• $K$: Knows answer, $P(K) = 0.7$
• $K^c$: Doesn't know, $P(K^c) = 0.3$
• $C$: Answers correctly
Step 2: Conditional probabilities:
• $P(C|K) = 1$ (If they know, always correct)
• $P(C|K^c) = 0.25$ (Guessing from 4 options)
Step 3: Apply Bayes' Theorem:
$P(K|C) = \frac{P(C|K)P(K)}{P(C|K)P(K) + P(C|K^c)P(K^c)}$
Step 4: Substitute values:
$P(K|C) = \frac{1 \times 0.7}{1 \times 0.7 + 0.25 \times 0.3}$
Step 5: Calculate:
$P(K|C) = \frac{0.7}{0.7 + 0.075} = \frac{0.7}{0.775} \approx 0.903$
Interpretation: If they answer correctly, there's ~90% chance they actually knew the answer.
🎯 4-Step Framework for Total Probability Problems
Step 1: Identify the Partition
Find mutually exclusive and exhaustive events $B_1, B_2, ..., B_n$ that cover all possibilities.
Step 2: Set Up the Tree
Create probability tree with branches for each $B_i$ and conditional probabilities.
Step 3: Apply the Formula
Use $P(A) = \sum P(A|B_i)P(B_i)$ for total probability or Bayes' for reverse probability.
Step 4: Interpret the Result
Understand what the calculated probability means in the context of the problem.
Concepts 4-6 Available in Full Version
Includes 3 more advanced applications: Multi-source Information, Sequential Testing, and Real-world Case Studies
📝 Quick Self-Test
Apply what you've learned to these JEE-level problems:
1. There are 3 boxes: Box I (2 gold coins), Box II (1 gold, 1 silver), Box III (2 silver). A box is selected randomly and a coin drawn is gold. What's the probability the other coin in the box is also gold?
2. A factory has 3 machines: M1 (30% output, 2% defective), M2 (50% output, 1% defective), M3 (20% output, 3% defective). A randomly selected item is defective. What's the probability it came from M1?
3. In a tournament, Player A has 60% chance of beating Player B. They play until someone wins 2 games. What's the probability A wins the match?
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