Bayes' Theorem
Probability
9
JEE Qs
8%
Hard
75
min
Master the art of correctly identifying the 'causes' (partitions) and the 'effect' in a problem, then systematically apply the Total Probability Theorem to compute the denominator before using the Bayes' formula.
🧮 Key Formulas
✅ Key Points for JEE
- 1Bayes' Theorem calculates the probability of a 'cause' (Ei) given an 'effect' (A) has occurred (posterior probability).
- 2It requires identifying the set of mutually exclusive and exhaustive 'causes' (partitions of the sample space, E1, E2,..., En) and the 'effect' (A).
- 3The numerator involves the prior probability of the cause P(Ei) and the likelihood of the effect given the cause P(A|Ei).
- 4The denominator, P(A), is typically calculated using the Total Probability Theorem, summing up P(A|Ej)P(Ej) for all possible causes Ej.
- 5Tree diagrams are often highly effective for visualizing the problem structure and correctly identifying all probabilities needed.
⚠️ Common Mistakes
- ✕Confusing P(A|B) with P(B|A); Bayes' Theorem precisely helps to find one from the other.
- ✕Incorrectly identifying the 'causes' (Ei) and the 'effect' (A), leading to swapped probabilities.
- ✕Errors in calculating the denominator using the Total Probability Theorem, especially in multi-stage problems.
- ✕Not ensuring that the events Ei are mutually exclusive and exhaustive, which is crucial for the application of the Total Probability Theorem.
📝 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 12 Mathematics Ch 13: Probability