A1, A2, and B1 are independent; A1, A2, and B2 are independent; B1 and B2 are disjoint. Are A1 and A2 and $B1 \cup B2$ independent?

Question

I started by applying definitions. $P(A \cap B) = P(A)*P(B)$ for independent events. $P(A \cup B) = P(A) + P(B)$ for disjoint events. I then tried to apply Demorgan's Laws $$(A1 \cap A2 \cap (B1 \cup B2)) = (A1 \cap (A2 \cap B1) \cup (A2 \cap B2))$$ However, I don't know where to go from here and I don't know if/when the probability rules will come into play.

Answer 1

$$P(A_1\cap A_2\cap (B_1\cup B_2))=P((A_1\cap A_2 \cap B_1)\cup(A_1\cap A_2\cap B_2))$$ Since $(A_1\cap A_2 \cap B_1)$ and $(A_1\cap A_2\cap B_2)$ are disjoint, $$ P((A_1\cap A_2 \cap B_1)\cup(A_1\cap A_2\cap B_2))=P(A_1\cap A_2\cap B_1)+P(A_1\cap A_2 \cap B_2) $$ Using independence, $$ \begin{align} P(A_1\cap A_2\cap B_1)+P(A_1\cap A_2 \cap B_2) &=P(A_1)P(A_2)P(B_1)+P(A_1)P(A_2)P(B_2)\\ &=P(A_1)P(A_2)(P(B_1)+P(B_2)) \end{align} $$ Again, since $B_1$ and $B_2$ are disjoint, $$ P(A_1)P(A_2)(P(B_1)+P(B_2))= P(A_1)P(A_2)P(B_1\cup B_2) $$ Now chain those all together.

By the way, to prove that three events $E,F,G$ are independent, not only do you need to check $P(E\cap F\cap G)=P(E)P(F)P(G)$, but you also need to check $P(E\cap F)=P(E)P(F)$, and same for the other two pairs.

Answer 2

You can apply probability rules any time and for this problem we can apply them directly to the left hand side of your equation.

The event $A1\cap A2\cap(B1\cup B2)=(A1\cap A2\cap B1)\cup (A1\cap A2\cap B2)$.

Since $A1\cap A2\cap B1\subset B1$ and $A1\cap A2\cap B2\subset B2$ and $B1$ and $B2$ are disjoint, these two intersections are also disjoint.

Therefore, $P(A1\cap A2\cap(B1\cup B2))=P((A1\cap A2\cap B1)\cup (A1\cap A2\cap B2))=P(A1\cap A2\cap B1)) + P(A1\cap A2\cap B2)$.

Since $A1$, $A2$ and $B1$ as well as $A1$, $A2$ and $B2$ are independent triples, $P(A1\cap A2\cap B1)) + P(A1\cap A2\cap B2)=P(A1)P(A2)P(B1) + P(A1)P(A2)P(B2) =P(A1)P(A2)(P(B1) + P(B2))$.

Therefore, $A1$, $A2$ and $B1\cup B2$ are independent.