PROVE:
P(( union )[i=1]^(k))=(∑)P(A[i]^ ) ∀ A[i]^ such that (A[i]^ intersect A[]j = ∅ ∀i <>j) ∀k
To prove the given statement, let's break it down step by step:
First, let's define some terms:
- P(A[i]) represents the probability of event A[i] occurring.
- ∑ denotes the sum of probabilities.
- A[i]∧ represents the intersection of event A[i].
- A[i]∧A[j] represents the intersection of events A[i] and A[j].
Now, let's prove the statement:
1. We are given that for any two different events A[i] and A[j], their intersection A[i]∧A[j] is empty (denoted as ∅). This means that the events A[i] are mutually exclusive or disjoint from each other.
2. Since the events A[i] are mutually exclusive, the union of these events (denoted as ∪) represents the occurrence of at least one of the events A[i]. The probability of this union is given by P(∪[i=1]^[k]).
3. According to the inclusion-exclusion principle, the probability of the union of events is the sum of individual probabilities minus the sum of probabilities of intersections of events, then adding the probabilities of intersections of three events, and so on.
4. In this case, we have no intersections of events since the given condition states that any two different events have no intersection (∅). Therefore, all terms involving the intersection of events become zero.
5. As a result, the probability of the union of mutually exclusive events simplifies to the sum of their individual probabilities.
6. Hence, ∑P(A[i]) represents the sum of probabilities of events A[i] for all i from 1 to k, where k is the number of events.
Therefore, we can conclude that P(∪[i=1]^[k]) = ∑P(A[i]) for all different events A[i] such that their intersections are empty (∅) for i ≠ j, for all k.