How to Solve Probability (Addition Theorem, Conditional, Bayes’ Theorem, Binomial Distribution) – IIT JEE Guide

How to Solve Probability (Addition Theorem, Conditional, Bayes’ Theorem, Binomial Distribution) – IIT JEE Guide

Introduction

"Mastering probability in IIT JEE can add 12-15 marks to your score—enough to jump from a 90 to a 105+ percentile. Whether it’s predicting exam outcomes, medical test accuracy, or stock market trends, probability is the hidden language of real-world decisions. Today, you’ll learn the exact steps to solve any probability problem in JEE Main or Advanced—no guesswork, just precision."

What You Need To Know First

Before diving in, ensure you understand: 1. Basic Probability Rules – Sample space, events, P(A) = (Favorable outcomes)/(Total outcomes). 2. Set Theory – Union (A ∪ B), Intersection (A ∩ B), Complement (A’). 3. Permutations & Combinations – Counting arrangements and selections (nCr, nPr).

If any of these feel shaky, pause and review them first.

Key Vocabulary

Formulas To Know

1. Addition Theorem (General Case)

Formula: P(A ∪ B) = P(A) + P(B) – P(A ∩ B) - P(A ∪ B): Probability of A or B happening. - P(A ∩ B): Probability of A and B happening. MEMORISE THIS (Not always given in JEE sheet).

Special Case (Mutually Exclusive): P(A ∪ B) = P(A) + P(B) (since P(A ∩ B) = 0)

2. Conditional Probability

Formula: P(A|B) = P(A ∩ B) / P(B) - P(A|B): Probability of A given B has occurred. - P(B) ≠ 0 (B must have a chance of happening). MEMORISE THIS (Given in JEE sheet, but understand it deeply).

3. Bayes’ Theorem

Formula: P(A|B) = [P(B|A) × P(A)] / P(B) - P(B): Total probability of B = P(B|A)P(A) + P(B|A’)P(A’). MEMORISE THIS (Not always given, but derivable from conditional probability).

4. Binomial Distribution

Formula: P(X = k) = nCk × p^k × (1–p)^(n–k) - n: Number of trials. - k: Number of successes. - p: Probability of success in one trial. - nCk: Combination (n choose k). MEMORISE THIS (Given in JEE sheet, but know when to use it).

Step-by-Step Method

Step 1: Identify the Problem Type

• Addition Theorem? → "OR" probability (A or B).
• Conditional? → "Given that" or "if" language.
• Bayes’? → "Updated probability" or "test accuracy" problems.
• Binomial? → Fixed trials, two outcomes, independent.

Step 2: Define Events Clearly

• Write down what A and B represent in plain English.
• Example: A = "Rains today", B = "Car breaks down".

Step 3: Check for Independence or Mutual Exclusivity

• Independent? P(A ∩ B) = P(A) × P(B).
• Mutually Exclusive? P(A ∩ B) = 0.

Step 4: Apply the Correct Formula

• Addition: P(A ∪ B) = P(A) + P(B) – P(A ∩ B).
• Conditional: P(A|B) = P(A ∩ B) / P(B).
• Bayes’: Plug into P(A|B) = [P(B|A)P(A)] / P(B).
• Binomial: P(X=k) = nCk × p^k × (1–p)^(n–k).

Step 5: Calculate and Simplify

• Use nCk = n! / [k!(n–k)!] for combinations.
• Simplify fractions early to avoid errors.

Step 6: Verify the Answer

• Probability must be between 0 and 1.
• Check if the answer makes logical sense.

Worked Examples

Example 1 – Basic (Addition Theorem)

Problem: A die is rolled. Find P(Even or Prime). - Even numbers: 2, 4, 6. - Prime numbers: 2, 3, 5.

Solution: 1. Define Events:
- A = "Even" = {2, 4, 6} → P(A) = 3/6 = 1/2.
- B = "Prime" = {2, 3, 5} → P(B) = 3/6 = 1/2. 2. Check Overlap:
- A ∩ B = {2} → P(A ∩ B) = 1/6. 3. Apply Addition Theorem:
P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 1/2 + 1/2 – 1/6 = 5/6.

What we did and why: We used the general addition rule because "Even" and "Prime" can both occur (number 2). If we’d ignored the overlap, we’d have gotten 1 (wrong!).

Example 2 – Medium (Conditional Probability)

Problem: In a class, 60% are boys, 40% are girls. 30% of boys and 20% of girls play cricket. If a student is selected at random and plays cricket, find the probability they are a girl.

Solution: 1. Define Events:
- B = "Boy", G = "Girl", C = "Plays Cricket".
- P(B) = 0.6, P(G) = 0.4.
- P(C|B) = 0.3, P(C|G) = 0.2. 2. Find P(C):
P(C) = P(C|B)P(B) + P(C|G)P(G) = (0.3 × 0.6) + (0.2 × 0.4) = 0.18 + 0.08 = 0.26. 3. Apply Conditional Probability:
P(G|C) = P(C|G)P(G) / P(C) = (0.2 × 0.4) / 0.26 = 0.08 / 0.26 = 4/13.

What we did and why: We used conditional probability because we needed the probability of "Girl" given "Plays Cricket". The denominator P(C) was found using the total probability rule.

Example 3 – Exam-Style (Bayes’ Theorem)

Problem: A medical test for a disease is 95% accurate. 1% of the population has the disease. If a person tests positive, what’s the probability they actually have the disease?

Solution: 1. Define Events:
- D = "Has Disease", T = "Tests Positive".
- P(D) = 0.01, P(T|D) = 0.95 (True Positive Rate).
- P(T|D’) = 0.05 (False Positive Rate). 2. Find P(T):
P(T) = P(T|D)P(D) + P(T|D’)P(D’) = (0.95 × 0.01) + (0.05 × 0.99) = 0.0095 + 0.0495 = 0.059. 3. Apply Bayes’ Theorem:
P(D|T) = [P(T|D)P(D)] / P(T) = (0.95 × 0.01) / 0.059 = 0.0095 / 0.059 ≈ 0.161 (16.1%).

What we did and why: This is a classic Bayes’ problem. Even with a 95% accurate test, the low disease prevalence (1%) means most positive results are false positives. Always calculate P(T) first!

Common Mistakes

Exam Traps

1-Minute Recap

"Listen up—this is your last-minute cheat sheet for probability in JEE: 1. Addition Rule: P(A or B) = P(A) + P(B) – P(A and B). Never forget the overlap! 2. Conditional Probability: P(A|B) = P(A and B) / P(B). Denominator is the "given" event. 3. Bayes’ Theorem: Flip the condition—use when you have new info and need to update probability. 4. Binomial: Fixed trials, two outcomes, independent. Formula: nCk × p^k × (1–p)^(n–k). 5. Always define events first. Write down P(A), P(B), P(A ∩ B) before plugging into formulas. 6. Check for independence. If unsure, assume they’re not independent and calculate P(A ∩ B) separately. Now go crush that exam—you’ve got this!