If A is any event associated with sample space and if \(E_1, E_2, E_3\) are mutually exclusive and exhaustive events. Then which of the following are true?
(A) \(P(A) = P(E_1)P(E_1|A) + P(E_2)P(E_2|A) + P(E_3)P(E_3|A)\)
(B) \(P(A) = P(A|E_1)P(E_1) + P(A|E_2)P(E_2) + P(A|E_3)P(E_3)\)
(C) \(P(E_i|A) = \displaystyle\frac{P(A|E_i)P(E_i)}{\sum_{i=1}^{3}P(A|E_i)P(E_i)}, i = 1, 2, 3\)
(D) \(P(A|E_i) = \displaystyle\frac{P(E_i|A)P(E_i)}{\sum_{i=1}^{3}P(E_i|A)P(E_i)}, i = 1, 2, 3\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (B) and (C) only
Since \(E_1, E_2, E_3\) are mutually exclusive and exhaustive, A splits into the disjoint pieces \(A \cap E_1, A \cap E_2, A \cap E_3\):
\(P(A) = P(A \cap E_1) + P(A \cap E_2) + P(A \cap E_3)\)
\(P(A \cap E_j) = P(A|E_j)P(E_j)\)
This gives the theorem of total probability. Then \(P(E_i|A) = \frac{P(A \cap E_i)}{P(A)}\) gives Bayes’ theorem (the sum in the denominator runs over j = 1, 2, 3):
\(\displaystyle P(E_i|A) = \frac{P(A|E_i)P(E_i)}{\sum_{j=1}^{3}P(A|E_j)P(E_j)}\)
Checking the statements
(A) \(P(A) = P(E_1)P(E_1|A) + P(E_2)P(E_2|A) + P(E_3)P(E_3|A)\) — Incorrect. Each P(Ej) must be multiplied by \(P(A|E_j)\), not by \(P(E_j|A)\).
(B) \(P(A) = P(A|E_1)P(E_1) + P(A|E_2)P(E_2) + P(A|E_3)P(E_3)\) — Correct. This is the theorem of total probability.
(C) \(P(E_i|A) = \displaystyle\frac{P(A|E_i)P(E_i)}{\sum_{i=1}^{3}P(A|E_i)P(E_i)}, i = 1, 2, 3\) — Correct. This is Bayes’ theorem.
(D) \(P(A|E_i) = \displaystyle\frac{P(E_i|A)P(E_i)}{\sum_{i=1}^{3}P(E_i|A)P(E_i)}, i = 1, 2, 3\) — Incorrect. The conditionals are reversed: Bayes’ theorem gives \(P(E_i|A)\) in terms of \(P(A|E_i)\), not the other way round.
NCERT (2026-27 edition):
Probability
