Classical Probability — Equally likely outcomes
Probability
9
JEE Qs
8%
Hard
75
min
Master counting techniques from Permutations and Combinations, as accurate enumeration of outcomes is the biggest hurdle in classical probability problems.
🧮 Key Formulas
✅ Key Points for JEE
- 1The classical definition of probability is applicable ONLY when all outcomes in the sample space are equally likely.
- 2The most crucial step is to correctly define the sample space (S) and the event (E) as a subset of S, ensuring all outcomes in S are distinct and equally probable.
- 3Proficiency in Permutations and Combinations is essential for accurately counting the number of favorable outcomes (n(E)) and the total number of outcomes (n(S)).
- 4Always verify that the calculated probability P(E) lies between 0 and 1, inclusive (0 <= P(E) <= 1).
⚠️ Common Mistakes
- ✕Incorrectly assuming outcomes are equally likely when they are not (e.g., sums when rolling two dice are not equally likely).
- ✕Errors in defining or listing the sample space, leading to an incorrect n(S).
- ✕Mistakes in applying counting principles (Permutations and Combinations) to find n(E) or n(S), especially in problems involving identical objects or complex arrangements.
📝 Practice Questions
See allQ5. If A and B are two events such that P(A ∩B) = 0.1, and P(A ∣B) and P(B ∣A) are the roots of the equation – 12x2 −7x + 1 = 0, then the value of P(A∪B) is : P(A∩B) (1) 4 (2) 7 3 4 (3) 5 (4) 9 3 4
Q10.Let A = [aij] be a square matrix of order 2 with entries either 0 or 1 . Let E be the event that A is an invertible matrix. Then the probability P(E) is : 2025 (24 Jan Shift 2) JEE Main Previous Year Paper (1) 3 (2) 5 16 8 (3) 3 (4) 1 8 8
Q3. Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected ball is black, given that the second selected ball is also black, is m , n where gcd(m, n) = 1, then m + n is equal to : (1) 4 (2) 14 (3) 13 (4) 11
Q16.A coin is tossed three times. Let X denote the number of times a tail follows a head. If μ and σ2 denote the mean and variance of X , then the value of 64 (μ + σ2) is : (1) 51 (2) 64 (3) 32 (4) 48
Q2. One die has two faces marked 1 , two faces marked 2 , one face marked 3 and one face marked 4 . Another die has one face marked 1 , two faces marked 2 , two faces marked 3 and one face marked 4. The probability of getting the sum of numbers to be 4 or 5 , when both the dice are thrown together, is (1) 2 (2) 1 3 2 (3) 4 (4) 3 9 5
Q8. Two number k1 and k2 are randomly chosen from the set of natural numbers. Then, the probability that the value of ik1 + ik2, (i = √−1) is non-zero, equals ⎪ ⎪ 2025 (28 Jan Shift 1) JEE Main Previous Year Paper (1) 1 (2) 3 2 4 (3) 1 (4) 2 4 3
NCERT Chapters
- Class 11 Maths Ch 7: Permutations and Combinations
- Class 11 Maths Ch 16: Probability