JEE Mathematics • Official PYQs

Probability & Statistics

Practice JEE Main Probability & Statistics PYQs online. Master Bayes’ Theorem, variance shift formula, binomial distribution & step-by-step solutions.

📝 15 Solved PYQs
📅 Exam Years: 2019–2024
📊 Weightage: 8%–10% (2 Questions per shift)
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🎯 High-Yield Syllabus Breakdown

These high-frequency concepts account for over 90% of all questions asked in Probability & Statistics:

⚡ Essential Formulas & Speed Shortcuts

Formulas and time-saving shortcuts to solve Probability & Stats questions in under 45 seconds:

Speed Trick #1:

Shift Invariance: Adding or subtracting constant c does not change Variance: Var(X + c) = Var(X).

Speed Trick #2:

Scaling: Var(k·X) = k² · Var(X) and SD(k·X) = |k| · SD(X).

Speed Trick #3:

Independent Events: P(A ∩ B) = P(A) · P(B).

📖 Sample Solved Exam Questions

Preview authentic questions from this chapter with full step-by-step solutions:

Question 1 of 3 JEE Main 2024 (Online) 27th Jan Shift-1
If A and B are two events such that P(A) = 0.4, P(B) = 0.8, and P(B|A) = 0.6, then P(A' ∩ B') is:
(A) 0.04
(B) 0.24
(C) 0.16
(D) 0.36
💡 View Step-by-Step Solution & Answer

Correct Answer: 0.04

  1. Step 1: By definition of conditional probability: P(B|A) = P(A ∩ B) / P(A) ⇒ P(A ∩ B) = P(A) · P(B|A).
  2. Step 2: Calculate P(A ∩ B) = 0.4 × 0.6 = 0.24.
  3. Step 3: By the union formula: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.4 + 0.8 - 0.24 = 0.96.
  4. Step 4: By De Morgan's Law: P(A' ∩ B') = P((A ∪ B)') = 1 - P(A ∪ B) = 1 - 0.96 = 0.04.

⚡ Speed Tip: De Morgan's Shortcut: P(A' ∩ B') = 1 - P(A ∪ B) = 1 - (0.4 + 0.8 - 0.24) = 0.04.

Question 2 of 3 JEE Main 2023 (Online) 24th Jan Shift-1
A bag contains 4 red and 6 black balls. Three balls are drawn at random without replacement. The probability that exactly two of them are red is:
(A) 3/10
(B) 1/5
(C) 3/8
(D) 1/2
💡 View Step-by-Step Solution & Answer

Correct Answer: 3/10

  1. Step 1: Total balls = 4 (red) + 6 (black) = 10 balls.
  2. Step 2: Total number of ways to draw 3 balls from 10 is ¹⁰C₃ = (10 × 9 × 8) / (3 × 2 × 1) = 120.
  3. Step 3: Favorable ways: Choose 2 red balls from 4 and 1 black ball from 6: ⁴C₂ · ⁶C₁ = 6 × 6 = 36.
  4. Step 4: Probability = 36 / 120 = 3 / 10.

⚡ Speed Tip: Hypergeometric Formula: P = (⁴C₂ · ⁶C₁) / ¹⁰C₃ = (6 × 6) / 120 = 36 / 120 = 3/10.

Question 3 of 3 JEE Main 2024 (Online) 29th Jan Shift-2
The mean and variance of 7 observations are 8 and 16 respectively. If five of the observations are 2, 4, 10, 12, 14, then the absolute difference of the remaining two observations is:
(A) 2
(B) 4
(C) 6
(D) 1
💡 View Step-by-Step Solution & Answer

Correct Answer: 2

  1. Step 1: Let the remaining two observations be x and y. Total observations N = 7.
  2. Step 2: Mean = 8 ⇒ (2 + 4 + 10 + 12 + 14 + x + y)/7 = 8 ⇒ 42 + x + y = 56 ⇒ x + y = 14.
  3. Step 3: Variance = 16 ⇒ (∑ xᵢ²)/7 - (Mean)² = 16 ⇒ (∑ xᵢ²)/7 - 64 = 16 ⇒ (∑ xᵢ²)/7 = 80 ⇒ ∑ xᵢ² = 560.
  4. Step 4: 2² + 4² + 10² + 12² + 14² + x² + y² = 4 + 16 + 100 + 144 + 196 + x² + y² = 460 + x² + y² = 560 ⇒ x² + y² = 100. With x + y = 14 and x² + y² = 100, the numbers are 8 and 6 (8² + 6² = 100). Absolute difference = |8 - 6| = 2.

⚡ Speed Tip: Pythagorean pair: x + y = 14 and x² + y² = 100 immediately yields {6, 8}, so |x - y| = 2.

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❓ Frequently Asked Questions (FAQs)

Does change of origin affect variance in JEE Main questions?

No! Adding or subtracting any constant from all data points leaves variance and standard deviation completely unchanged.

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