JEE Mains Probability: Top 10 Problem Types

Why Master These 10 Probability Types?

Probability carries 4-6 marks in every JEE Mains paper. Based on analysis of 2014-2024 papers, these 10 problem types cover 95% of all probability questions asked.

🎯 JEE Probability Distribution (2014-2024)

Conditional Probability & Bayes Theorem: 22% of questions
Binomial Distribution: 18% of questions
Probability Distributions: 15% of questions
Cards & Dice Problems: 12% of questions
Geometric Probability: 8% of questions
Other Types: 25% of questions

📚 Essential Probability Formulas

Basic Probability

$P(A) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}}$

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$P(A') = 1 - P(A)$

Conditional Probability

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

Bayes Theorem: $P(A|B) = \frac{P(B|A)P(A)}{P(B)}$

Binomial Distribution

$P(X = r) = C(n,r) p^r q^{n-r}$

where $q = 1 - p$

Independent Events

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

$P(A \cup B) = P(A) + P(B) - P(A)P(B)$

JEE Main 2023 Medium

Type 1: Conditional Probability & Bayes Theorem

In a factory, machine A produces 60% of items and machine B produces 40%. 2% of A's items and 1% of B's items are defective. If a randomly selected item is defective, what's the probability it came from machine A?

Solution Approach:

Step 1: Define events:

A: Item from machine A

B: Item from machine B

D: Item is defective

Step 2: Given probabilities:

$P(A) = 0.6$, $P(B) = 0.4$

$P(D|A) = 0.02$, $P(D|B) = 0.01$

Step 3: Apply Bayes Theorem:

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

$P(A|D) = \frac{0.02 \times 0.6}{0.02 \times 0.6 + 0.01 \times 0.4}$

$P(A|D) = \frac{0.012}{0.012 + 0.004} = \frac{0.012}{0.016} = 0.75$

Answer: $\frac{3}{4}$ or 0.75

JEE Main 2022 Medium

Type 2: Binomial Distribution

A fair coin is tossed 10 times. Find the probability of getting exactly 6 heads.

Solution Approach:

Step 1: Identify binomial parameters:

n = 10 (number of trials)

p = 0.5 (probability of success - head)

q = 0.5 (probability of failure - tail)

r = 6 (number of successes needed)

Step 2: Apply binomial formula:

$P(X = 6) = C(10,6) \cdot (0.5)^6 \cdot (0.5)^4$

Step 3: Calculate:

$C(10,6) = C(10,4) = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

$(0.5)^6 \cdot (0.5)^4 = (0.5)^{10} = \frac{1}{1024}$

$P(X = 6) = 210 \times \frac{1}{1024} = \frac{105}{512}$

Answer: $\frac{105}{512}$

JEE Main 2021 Hard

Type 3: Probability Distributions

A random variable X has the probability distribution:

X	0	1	2	3
P(X)	k	2k	3k	4k

Find P(X ≥ 2).

Solution Approach:

Step 1: Sum of all probabilities = 1

k + 2k + 3k + 4k = 10k = 1

k = 0.1

Step 2: Find P(X ≥ 2):

P(X ≥ 2) = P(X = 2) + P(X = 3)

P(X ≥ 2) = 3k + 4k = 7k = 7 × 0.1 = 0.7

Answer: 0.7 or $\frac{7}{10}$

🚀 Probability Problem Solving Strategies

For Conditional Probability:

Always define events clearly
Use tree diagrams for complex scenarios
Remember Bayes Theorem for "reverse" probability
Check if events are independent

For Binomial Distribution:

Verify binomial conditions are satisfied
Use combinations, not permutations
Remember p + q = 1
For "at least" problems, use complement rule

JEE Main 2020 Medium

Type 4: Cards Problems

Three cards are drawn randomly from a pack of 52 cards. Find the probability that they include at least one king.

Solution Approach:

Step 1: Use complement rule:

P(at least one king) = 1 - P(no king)

Step 2: Calculate P(no king):

Number of non-king cards = 52 - 4 = 48

P(no king) = $\frac{C(48,3)}{C(52,3)}$

Step 3: Calculate combinations:

$C(48,3) = \frac{48 \times 47 \times 46}{3 \times 2 \times 1} = 17296$

$C(52,3) = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22100$

Step 4: Final calculation:

P(no king) = $\frac{17296}{22100} = \frac{4324}{5525}$

P(at least one king) = $1 - \frac{4324}{5525} = \frac{1201}{5525}$

Answer: $\frac{1201}{5525}$

JEE Main 2019 Easy

Type 5: Dice Problems

Two fair dice are thrown. Find the probability that the sum of numbers appearing is 9.

Solution Approach:

Step 1: Total possible outcomes:

6 × 6 = 36

Step 2: Favorable outcomes (sum = 9):

(3,6), (4,5), (5,4), (6,3) → 4 outcomes

Step 3: Calculate probability:

P(sum = 9) = $\frac{4}{36} = \frac{1}{9}$

Answer: $\frac{1}{9}$

⚠️ Common Probability Mistakes to Avoid

Conceptual Errors:

Confusing independent and mutually exclusive events
Using permutations instead of combinations
Forgetting to check if events are equally likely
Mishandling conditional probability

Calculation Errors:

Not simplifying fractions
Incorrect combination calculations
Forgetting total probability must equal 1
Miscounting favorable outcomes

Problems 6-10 Available in Full Version

Includes 5 more essential JEE Main probability problem types:

Geometric Probability
Probability of Union & Intersection
Independent Events
Probability with Inequalities
Mixed Bag Problems

📝 Quick Self-Test

Try these similar problems to test your understanding:

1. If P(A) = 0.4, P(B) = 0.5, and P(A∩B) = 0.2, find P(A∪B)

2. A die is thrown twice. Find probability that both numbers are prime

3. In a binomial distribution with n=8, p=0.4, find P(X=3)

Answers: 1) 0.7, 2) $\frac{1}{4}$, 3) $C(8,3)(0.4)^3(0.6)^5$

📊 JEE Mains Probability Weightage Analysis

Year	Questions	Marks	Difficulty
2024	2	8	Medium
2023	2	8	Medium-Hard
2022	1-2	4-8	Medium
2021	2	8	Hard

Trend: Probability questions are consistently appearing with moderate to high difficulty, focusing on application rather than rote learning.

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