Match the LIST-I with LIST-II | | LIST-I | | LIST-II | |---|---|---|---| | A. | If A and B are mutually exclusive events then P(A B) = | I. | P(A B)/P(B), P(B) ≠ 0 | | B. | If A and B are independent events then P(A B) = | II. | P(A B)/P(A), P(A) ≠ 0 | | C. | If A and B are two events of a sample space of an experiment, then P(A/B) = | III. | P(A) * P(B) | | D. | If A and B are two events of a sample space of an experiment, then P(B/A) = | IV. | P(A) + P(B) | Choose the correct answer from the options given below:

Question

Option 1: A-IV, B-III, C-I, D-II

Option 2: A-II, B-I, C-III, D-IV

Option 3: A-I, B-II, C-IV, D-III

Option 4: A-III, B-IV, C-I, D-II

AfterBoards · Accepted Answer

The answer is 'A-IV, B-III, C-I, D-II'. View full solution:

List - I	List - II
(A) Two events E and F will be independent if $P(E'F')$ is equal to	(I) $1 - P(E/F)$
(B) If $P(F) \neq 0$ , then $P(E'/F)$ is equal to	(II) $P(E) = P(F)$
(C) If E and F are independent events, then	(III) $P(E \cap F') = P(E) \cdot P(F')$
(D) If $P(E \cap F) \neq 0$ and $P(E/F) = P(F/E)$ , then	(IV) $[1-P(E)][1-P(F)]$

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

CUET Mathematics 2026 31st May Shift 2 PYQs

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