Given a set of events \( E_1 \ldots \ldots . . E_n \) , defined on the sample space \( S \) such that:
(i) \( \forall i \) and \( \mathrm{j}, \mathrm{i} \neq \mathrm{j}, \mathrm{E}_{\mathrm{i}} \cap \mathrm{E}_{\mathrm{j}}=\phi \)
(ii) \( \bigcup_{i=1}^n E_i=S \)
(iii) \( \quad \mathrm{P}\left(\mathrm{E}_{\mathrm{i}}\right)>0, \forall \mathrm{i}=1, \mathrm{n} \)
Then the events are:
Step-by-step Solution:
We are given a set of events \( E_1, E_2, \dots, E_n \) defined on the sample space \( S \) satisfying the following conditions: 1. Pairwise disjoint condition: \[ \forall i, j, \quad i \neq j, \quad E_i \cap E_j = \phi \] This means that no two events share common outcomes, making them mutually exclusive. 2. Exhaustive condition: \[ \bigcup_{i=1}^{n} E_i = S \] This implies that the union of all the events covers the entire sample space, meaning the events are exhaustive. 3. Non-zero probability condition: \[ P(E_i) > 0, \quad \forall i = 1, n \] Each event has a positive probability. \[Step 1:\] Evaluating the given conditions - Since the events are pairwise disjoint, they cannot be independent. - Independence requires \( P(E_i \cap E_j) = P(E_i) P(E_j) \), but here \( P(E_i \cap E_j) = 0 \), which contradicts the independence condition unless \( P(E_i) \) or \( P(E_j) \) is zero. - Thus, Option B and Option D are incorrect. - The events are mutually exclusive (pairwise disjoint) and exhaustive. - Option A correctly describes this. - Option C (Dependent and mutually exclusive) is incorrect because mutual exclusivity does not imply dependence in the general probability sense. Final Answer: \[ \boxed{\text{Pairwise disjoint and exhaustive.}} \]