If sets $A$ and $B$ are defined as follows:
$A = \{(x, y) : y = \frac{1}{x}, 0 \neq x \in R\}$
$B = \{(x, y) : y = -x, x \in R\}$,then

Let $E, F$ and $G$ be three events having probabilities $P(E) = \frac{1}{8}, P(F) = \frac{1}{6}$ and $P(G) = \frac{1}{4}$,and let $P(E \cap F \cap G) = \frac{1}{10}$. For any event $H$,if $H^C$ denotes its complement,then which of the following statements is(are) $TRUE$?
$(A) P(E \cap F \cap G^C) \leq \frac{1}{40}$
$(B) P(E^C \cap F \cap G) \leq \frac{1}{15}$
$(C) P(E \cup F \cup G) \leq \frac{13}{24}$
$(D) P(E^C \cap F^C \cap G^C) \leq \frac{5}{12}$

Consider the sets $A$ and $B$ where $A = \{2, 4, 6, 8\}$ and $B = \{6, 8, 10, 12\}$. Find $A \cap B$.

If $A$ and $B$ are two sets,then $A \cap (B - A)$ is equal to:

If $A = \{ x : x \text{ is a natural number} \}$,$B = \{ x : x \text{ is an even natural number} \}$,$C = \{ x : x \text{ is an odd natural number} \}$,and $D = \{ x : x \text{ is a prime number} \}$,find $A \cap C$.

If $A, B, C$ are three sets,then $A \cap (B \cup C)$ is equal to