# Are the two events $(A \cup (B\cap C))$ and $(A_1 \cup (B_1 \cap C))$ independent

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 ?

• Are they pairwise independent? – user64066 Dec 29 '16 at 10:26
• @user64066 I have edited the question. Hope it's clearer now. – din Dec 29 '16 at 10:41
• Are you sure you stated the second question correctly? Because it seems really trivial. If $D$ is disjoint from $D_1$ then $D$ is also disjoint from $D_1\cap E,$ because $D\cap(D_1\cap E)\subseteq D\cap D_1.$ It doesn't matter if $E$ is independent of $D$ and $D_1$ or not. – bof Dec 29 '16 at 11:11
• @bof Yes it's stated correctly. I thought that this additional information might have an impact on the answer. Btw thank you! – din Dec 29 '16 at 11:17

HINT

Just to give an example of events $A,B,C,A_1,B_1$ being pairwise independent and yet $A \cup (B\cap C)$ and $A_1 \cup (B_1 \cap C)$ are not independent.

Consider the following diagram. Here the probabilities of the little squares are equal, $\frac14.$ I hope that based on the color code every events can be identified. For example

$$B_1=\{5,9,6,10,7,11,8,12\}.$$

It is easy to check that all possible pairs of distinct events will consist of four little squares. This proves pairwise independence.

However

$$P[A\cup(B\cap C)]=P[\{1,2,3,5,6,7,9,10,11,13,14,15\}]=\frac{12}{16}$$

and

$$P[A_1\cup(B_1\cap C)]=P[\{7,8,9,10,11,12,13,14,15,16\}]=\frac{10}{16}$$

and

$$P[A\cup(B\cap C)\cap A_1\cup(B_1\cap C)]=P[\{7,9,10,11,13,14,15\}]=\frac7{16}.$$

So, the two combinations are not independent.

• Based on your answer and on all the information given in the question, could you please tell me if the following is correct: $$P\{ (A \cup (B \cap C)) \mid (A_1 \cup (B_1 \cap C)) \}=P\{ (A \cup (B \cap C)) \mid C \}$$ – din Dec 29 '16 at 20:25

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. $$P(A\cup(B\cap C))=P(A)+P(B\cap C)-P(A\cap B\cap C)=\frac12+\frac14-\frac18=\frac58=P(A_1\cup(B_1\cap C)).$$ $$P((A\cup(B\cap C))\cap(A_1\cup(B_1\cap C)))=P(A\cap A_1)+P(A\cap\overline{A_1}\cap B_1\cap C)+P(\overline A\cap A_1\cap B\cap C)+P(\overline A\cap\overline{A_1}\cap B\cap B_1\cap C)=\frac14+\frac1{16}+\frac1{16}+\frac1{32}=\frac{13}{32}\ne\frac58\cdot\frac58$$

• Based on your answer and on all the information given in the question, could you please tell me if the following is correct: $$P\{(A\cup(B \cap C)) \mid (A_1 \cup(B_1 \cap C))\}=P\{(A\cup(B \cap C)) \mid C\}$$ – din Dec 29 '16 at 20:09
• In the case of my example your statement isn't correct $$P[A\cup (B\cap C)\mid A_1\cup (B_1\cap C)]=\frac{P[(A\cup (B\cap C))\cap (A_1\cup (B_1\cap C))]}{P[A_1\cup (B_1\cap C)]}=$$ $$=\frac{P[\{1,2,3,5,6,7,9,10,11,13,14,15\}\cap\{7,8,9,10,11,12,13,14,15,16\} ]}{P[\{7,8,9,10,11,12,13,14,15,16\}]}=$$ $$=\frac{P[\{7,9,10,11,13,14,15\}]}{P[\{7,8,9,10,11,12,13,14,15,16\}]}=\frac{7}{10}$$ and $$P[A\cup(B\cap C)\mid C]=\frac{P[\{1,2,3,5,6,7,9,10,11,13,14,15\}\cap \{3,4,7,8,11,12,15,16\}]}{P[\{3,4,7,8,11,12,15,16\}]}=$$ $$=\frac{P[\{3,7,11,15\}]}{P[\{3,4,7,8,11,12,15,16\}]}=\frac{1}{2}.$$ – zoli Dec 29 '16 at 23:25
• Referring to the example in my answer (i.e., $C,A,B,A_1,B_1$ are five mutually independent events, each with probability $\frac12,$ routine calculation shows that $P\{(A\cup(B \cap C)) \mid (A_1 \cup(B_1 \cap C))\}=\frac{13}{20}$ while $P\{(A\cup(B \cap C)) \mid C\}=\frac34.$ – bof Dec 30 '16 at 2:07