∪ (A ∩ En)
Also A ∩ Ei and A ∩ Ej are respectively the subsets of Ei and Ej We know that
Ei  and Ej are disjoint, for i
≠j
, therefore, A  ∩ Ei and A  ∩ Ej  are also disjoint for all
i ≠ j,  i, j = 1, 2, , n Thus,
P(A) = P [(A ∩ E1) ∪ (A ∩ E2)∪