independence of sets of a sigma algebra

Question

I have been trying to solve this question all day, i am not good at probability theory so i am not able to solve this.

Let (Ω, A, P) be a probability space and A, B, C ∈ A and A, B, C are independent of each other. \ Show that A and B ∪ C are independent as well.

My approaches are:

1. using properties of a probability space: the intersection of the disjoint sequence of events is an empty set.
   P(A).P(BuC) = P(A).(P(B)+ P(C) - P(BnC))
   but P(BnC) = P(empty set) = 0
   therefore it remains = P(A).(P(B) + P(C)) \

I don't think this is right. \

2. looking at properties of a sigma algebra: if A,B,C ∈ A, this means that complements of A, B, C ∈ A, and so are their unions.
   P(A) = P(A\B) + P(B\C) + P(AnBnC)
   P(BuC)= P(B\C) + P(C\B) + P(BnC)
   P(BuC) = P((B\C) + P(BnC)) + (P(C\B) + P(BnC)) - P(BnC)
   after combining this, i don't know how to solve it further.

You don't have to give the answer. An idea of the solution or what properties one should follow is all i am asking.

Answer 1

Let $I \perp \!\!\! \perp J$, $I \neq J, I,J \in \{ A , B, C\}$

We want to show $A \perp \!\!\!\perp B \cup C$.

Hints:

1. $P(I \cap J) = P(I)P(J)$. This comes from independence
2. $P(I \cup J) = P(I) + P(J) - P(I \cap J) = P(I) + P(J) - P(I)P(J)$

The spoiler below is the solution

$$P(A \cap (B \cup C))\\ = P((A\cap B) \cup (A \cap C)) \\ = P(A\cap B)+P(A\cap C) - P((A\cap B) \cap (A\cap C)))\\ = P(A)P(B) + P(A)P(C) - P(A\cap B \cap C)\\= P(A)P(B) + P(A)P(C) -P(A)P(B)P(C)\\ = P(A)[P(B) + P(C) -P(B)P(C)]\\ = P(A)P(B \cup C)$$Hence, $A \perp \!\!\! \perp B \cup C$.