Find $P\{ (A \, \text{or} \, B) \, \text{and} \, (A_1 \, \text{or} \, B_1) \}$ or a lower bound in this specific case Define $X$, $Y$, $X_1$, $Y_1$, and $Z$ to be some positive random variables, for each of which we know the distribution. Note that these variables are independent of each other.
Let $t, a$ two positive constants and $f(X,Y)$ a function of $X$ and $Y$.
Also, define the following events:
Event $A: X \ge t$. Event $B: (X <t) \, \text{and} \, (f(X,Y)  \ge t) \, \text{and} \, (Z \ge a)$.
Event $A_1: X_1 \ge t $.  Event $B_1: (X_1 <t) \, \text{and} \, (f(X_1,Y_1)  \ge t) \, \text{and} \, (Z \ge a)$.
Event $E= (A \, \text{or} \, B) \, \text{and} \, (A_1 \, \text{or} \, B_1)$.
Let $C: (Z \ge a)$, $D:(X <t) \, \text{and} \, (f(X,Y) \ge t) $. Also, let $D_1:(X_1 <t) \, \text{and} \, (f(X_1,Y_1) \ge t) $. Thus, we have $B=(D \, \text{and} \, C)$,  $B_1=(D_1 \, \text{and} \, C)$.
-One can notice that $A$ and $B$ are mutually exclusive; the same can be noticed for $A_1$ and $B_1$.
-I think that the two events  $(A \, \text{or} \, B)$ and $(A_1 \, \text{or} \, B_1)$ are not independent; am I correct ?
So $P\{ (A \, \text{or} \, B)\}=P\{ A \} +P\{ B\}$, $P\{ (A_1 \, \text{or} \, B_1)\}=P\{ A_1 \} +P\{ B_1\}$.
-The probabilities: $P\{A\}$; $P\{B\}=P\{(X <t) \, \text{and} \, (f(X,Y)  \ge t)\} P\{(Z \ge a) \}$; $P\{A_1\}$, $P\{B_1\}=P\{(X_1 <t) \, \text{and} \, (f(X_1,Y_1)  \ge t)\} P\{(Z \ge a) \}$; ... are all known.   
Question: My goal is to derive the probability $P\{E\}=$ $$P\{  (A \, \text{or} \, B) \, \text{and} \, (A_1 \, \text{or} \, B_1)  \}=P\{  (A \, \text{or} \, (D \, \text{and} \, C)) \, \text{and} \, (A_1 \, \text{or} \, (D_1 \, \text{and} \, C) ) \}.$$ The problem is that, as I mentioned earlier, $(A \, \text{or} \, B)$ and $(A_1 \, \text{or} \, B_1)$ are not independent. Based on the information given above, is there any apporach I can adopt in order to derive the probability or a lower bound on this probability?
Lower bound ex: suppose that we can prove that $P\{E\} \ge P\{  (A \, \text{or} \, (D)) \, \text{and} \, (A_1 \, \text{or} \, (D_1 \, \text{and} \, C) ) \}.$ Hence, since $(A \, \text{or} \, (D))$ and $(A_1 \, \text{or} \, (D_1 \, \text{and} \, C) ) $ are independent, and using the disjoint property, we get $(P\{ A\}+P\{ D\})(P\{ A_1\}+P\{ D_1 \text{and} \, C\})$. This is the lower bound in this case, since all the probabilities in this bound are known.
Second question: Define function $g(X_1,Y_1,Z)$. Now event $B_1$ is defined as $B_1: (X_1<t) \, \text{and} \, g(X_1,Y_1,Z)$.   Events $A,A_1, B$ are defined as before. 
Let Event $E= (A \, \text{or} \, B) \, \text{and} \, (A_1 \, \text{and} \, B_1)$.
Also, I am interested in computing $P\{E\}$ or a lower bound on this probability.
Any hint for one or both of these questions ? Thank you!
 A: If you visualized the sample space being partitioned according to the events you defined, a vertical line for the $t$ threshold on the $X$ axis, and a horizontal bisecting line for the $t$ threshold on the $Y$ axis, you would realize you actually have 4 events in the joint probability space: A not A1, A1 not A, A1 A, not A1 not A. If you do not have the joint probability for $X$ and $X_1$, you cannot solve the problem.
A: Answer to the first question:
$P\{  (A \cup B) \cap  (A_1 \cup B_1)  \}=P\{ (A \cap (A_1 \cup B_1)) \cup (B \cap (A_1 \cup B_1))\} $. Since $A$ and $B$ are disjoint, we have $(A \cap (A_1 \cup B_1))$ and $(B \cap (A_1 \cup B_1))$ are also disjoint. Thus $$ P\{ (A \cap (A_1 \cup B_1)) \cup (B \cap (A_1 \cup B_1))\} = P\{ A \cap (A_1 \cup B_1) \}+P\{ B \cap (A_1 \cup B_1) \} .$$
A is independent of both $A_1$ and $B_1$, so $P\{ A \cap (A_1 \cup B_1) \}= P\{ A\} P\{ A_1 \cup B_1 \}$. Events $A_1$ and $B_1$ are disjoint, so $P\{ A_1 \cup B_1 \}=P\{ A_1\}+P\{ B_1 \}$. Hence $$ P\{ A \cap (A_1 \cup B_1) \} = P\{ A\} (P\{ A_1\}+P\{ B_1 \}).$$
Regarding $P\{ B \cap (A_1 \cup B_1) \}$: 
we have $P\{ B \cap (A_1 \cup B_1) \}=P\{ (B \cap A_1) \cup (B \cap B_1) \}$.
Since $A_1$ and $B_1$ are disjoint, we get
$$ P\{ (B \cap A_1) \cup (B \cap B_1) \}=P\{ B \cap A_1\}+P\{ B \cap B_1\} $$
Events $B$ and $A_1$ are independent, so $P\{ B \cap A_1\}=P\{ B\}P\{ A_1\}$.
Recall that $B= D \cap C$ and $B_1=D_1 \cap C$. Thus, event $B \cap B_1$ can be written as: $$ B \cap B_1 = D \cap C \cap D_1 \cap C =D \cap D_1 \cap C.$$
Therefore, we get $$ P\{ B \cap B_1\} = P\{ D \cap D_1 \cap C\}=P\{D\} P\{D_1\}P\{C\},$$ which results from the fact that $D$, $D_1$, and $C$ are all independent of each other.
Finally, we get $$ P \{E\}=P\{ A\} (P\{ A_1\}+P\{ B_1 \})+ (P\{ B\}P\{ A_1\}+P\{D\} P\{D_1\}P\{C\} ) .$$
Answer to the second question:
$P\{  (A \cup B) \cap  (A_1 \cap B_1) \} = P\{ (A \cap (A_1 \cap B_1)) \cup (B \cap (A_1 \cap B_1))\} $.
Since $A$ and $B$ are disjoint, we have  $$  P\{ (A \cap (A_1 \cap B_1)) \cup (B \cap (A_1 \cap B_1))\} =  P\{ A \cap (A_1 \cap B_1)\}+ P\{ B \cap (A_1 \cap B_1) \}. $$
Event $A$ is independent of both $A_1$ and $B_1$, so $ P\{ A \cap (A_1 \cap B_1)\}= P\{ A \} P\{ A_1 \cap B_1 \}$.
Recall that $B= D \cap C$, $D$ is independent of $C$, and $D$ is independent of $A_1$ and $B_1$. Thus  $$ P\{ B \cap (A_1 \cap B_1) \} = P\{ C \cap D \cap (A_1 \cap B_1) \} = P\{ D \} P\{ C \cap A_1 \cap B_1 \}.$$
Therefore, we obtain  $$ P\{E\}= P\{ A \} P\{ A_1 \cap B_1 \}+ P\{ C \cap D \cap (A_1 \cap B_1) \} = P\{ D \} P\{ C \cap A_1 \cap B_1 \}.$$ 
