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Match List-I with List-II

List-IList-II
(A) P(A∩B)=P(A).P(B)P(A \cap B) = P(A).P(B)(I) P(B∩A)P(A),P(A)≠0\frac{P(B \cap A)}{P(A)}, P(A) \ne 0
(B) P(S∩B)P(B)\frac{P(S \cap B)}{P(B)}, S = Sample space(II) A, B are independent events
(C) P(B/A)=P(B/A) =(III) P(Aˉ∩Bˉ)P(\bar{A} \cap \bar{B})
(D) P(A∪B‾)P(\overline{A \cup B})(IV) 1

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 2

The product rule P(A∩B)=P(A)P(B)P(A\cap B) = P(A)P(B) defines independence. P(S∩B)/P(B)=P(B)/P(B)=1P(S\cap B)/P(B) = P(B)/P(B) = 1. Conditional probability P(B/A)=P(B∩A)/P(A)P(B/A) = P(B\cap A)/P(A). By De Morgan's law the complement of A∪BA\cup B is Aˉ∩Bˉ\bar{A}\cap\bar{B}.

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