Prove A and B1$\cup$B2 are independent

Question

Suppose events A and $B_1$, A and $B_2$ are independent, the events $B_1$ and $B_2$ are disjoint.Prove that the events A and $B$$ \cup$$B_2$ are independent.

I know $B_1\cap B_2=0$, $P(A\cap B_1)=P(A)*P(B_1)$, $P(A\cap B_2)=P(A)*P(B_2)$ now. My attempt:

I try to prove that $P(A) * P(B_1\cup B_2)=P(A \cap (B_1\cup B_2))$

$P(A) * P(B_1 \cup B_2)$

$=P(A)*(P(B_1)+P(B_2)-0)$

$=P(A\cap B_1)+P(A\cap B_2)$ ...

Thank you.

Answer 1

$$P(A\cap(B_1\cup B_2))=P((A\cap B_1)\cup (A\cap B_2))=$$ $$=P(A\cap B_1)+P(A\cap B_2)$$

because if $B_1$ and $B_2$ are disjoint then $A\cap B_1$ and $A\cap B_2$ are also disjoint. Then

$$P(A\cap B_1)+P(A\cap B_2)=P(A)(P(B_1)+P(B_2))$$ because of the hypothetical independence. And finally

$$P(B_1)+P(B_2)=P(B_1\cup B_2)$$

because $B_1$ and $B_2$ are disjoint.

SO we proved that

$$P(A\cap(B_1\cup B_2))=P(A)P(B_1\cup B_2).$$

Answer 2

You are very close to a solution here. Since $B_1$ and $B_2$ are disjoint, we also have that $A\cap B_1$ and $A\cap B_2$ are disjoint, so your last line is equal to $$ P\Big((A\cap B_1)\cup(A\cap B_2)\Big) $$ Now all that's left is knowing (or proving, if you want) that it is true in general for intersections and unions that $$ (X\cap Y)\cup(X\cap Z) = X\cap(Y\cup Z) $$ and you can write your final, desired equality.