TITLE: Are the two events $(A \cup (B\cap C))$ and $(A_1 \cup (B_1 \cap C))$ independent
QUESTION [0 upvotes]: Let $A$, $A_1$, $B$, $B_1$, and $C$ be some events.
$B$ can be written as $B=S \cap R$, where $S=\bar{A}$ (i.e. complementary event of $A$). What define the event $A$ is to have $x \ge \tau$, where $x$ is a random variable; thus, $S=\bar{A}$ is defined as $x < \tau$. Note that $R$ depends on $x$.
$B_1$ can be written as $B_1=S_1 \cap R_1$, where $S_1=\bar{A_1}$ (i.e. complementary event of $A_1$). What define the event $A_1$ is to have $x_1\ge \tau$, where $x_1$ is a random variable; thus, $S_1=\bar{A}_1$ is defined as $x_1 < \tau$. Note that $R_1$ depends on $x_1$.
In addition, we have: $A$ and $A_1$ are independent, $A$ and $C$ are independent, $A$ and $B_1$ are independent, $B$ and $B_1$ are independent, $B$ and $C$ are independent, $B$ and $A_1$ are independent, $A_1$ and $C$ are independent, $B_1$ and $C$ are independent.
I have the following combination: $(A \cup (B\cap C)) \cap (A_1 \cup (B_1 \cap C)) .$
Questions: 1) are the two events $(A \cup (B\cap C))$ and $(A_1 \cup (B_1 \cap C))$ independent ? if so, why ?
2) Suppose we have two events $D$ and $D_1$ that are disjoint. For an event $E$ independent of $D$ and $D_1$,  are $D$ and $(D_1 \cap E)$ also disjoint ?

REPLY [1 votes]: 2) Suppose $D$ and $D_1$ are disjoint, i.e., $D\cap D_1=\emptyset,$ and let $E$ be any event; then $D$ and $D_1\cap E$ are disjoint, because $D\cap(D_1\cap E)\subseteq D\cap D_1=\emptyset.$
1) The fact that $B$ can be written as $B=S\cap R$ is irrelevant, because any event $B$ can be so written. Namely, let $R=B,$ and define $\tau$ and $x$ so that $x\lt\tau$ always holds. So we can forget about $R,R_1,S,S_1,x,x_1$ and $\tau.$ I will give a counterexample where the events $A\cup(B\cap C)$ and $A_1\cup(B_1\cap C)$ are not independent.
Consider a random experiment consisting of five independent coin tosses. Let $A=$ "heads on first toss", $A_1=$ "heads on second toss", $B=$ "heads on third toss", $B_1=$ "heads on fourth toss", $C=$ "heads on fifth toss". The events $A,A_1,B,B_1,C$ are mutually independent.
$$P(A\cup(B\cap C))=P(A)+P(B\cap C)-P(A\cap B\cap C)=\frac12+\frac14-\frac18=\frac58=P(A_1\cup(B_1\cap C)).$$
$$P((A\cup(B\cap C))\cap(A_1\cup(B_1\cap C)))=P(A\cap A_1)+P(A\cap\overline{A_1}\cap B_1\cap C)+P(\overline A\cap A_1\cap B\cap C)+P(\overline A\cap\overline{A_1}\cap B\cap B_1\cap C)=\frac14+\frac1{16}+\frac1{16}+\frac1{32}=\frac{13}{32}\ne\frac58\cdot\frac58$$
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