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 $H^{\text{C}}$ denotes its complement, then which of the following statements is(are) TRUE?

$P(E)=\dfrac{1}{8},\quad P(F)=\dfrac{1}{6},\quad P(G)=\dfrac{1}{4}$

$P(E \cap F \cap G)=\dfrac{1}{10}$

$\text{Check: } (1)\ P(E\cap F\cap G^{C})\le\dfrac{1}{40}\quad (2)\ P(E^{C}\cap F\cap G)\le\dfrac{1}{15}$

$(3)\ P(E\cup F\cup G)\le\dfrac{13}{24}\quad (4)\ P(E^{C}\cap F^{C}\cap G^{C})\le\dfrac{5}{12}$

$\text{1. } A \subseteq B \Rightarrow P(A) \le P(B)$

$\text{2. } P(E\cup F\cup G) \le P(E)+P(F)+P(G)$

$E\cap F\cap G^{C} = (E\cap F) - (E\cap F\cap G)$

$E^{C}\cap F\cap G = (F\cap G) - (E\cap F\cap G)$

$P(E\cap F) \le P(E) = \dfrac{1}{8}$

$P(E\cap F\cap G^{C}) \le \dfrac{1}{8}-\dfrac{1}{10}=\dfrac{1}{40}\ \checkmark$

$P(F\cap G) \le P(F) = \dfrac{1}{6}$

$P(E^{C}\cap F\cap G) \le \dfrac{1}{6}-\dfrac{1}{10}=\dfrac{1}{15}\ \checkmark$

$P(E\cup F\cup G) \le P(E)+P(F)+P(G)$

$= \dfrac{1}{8}+\dfrac{1}{6}+\dfrac{1}{4} = \dfrac{13}{24}\ \checkmark$

$P(E^{C}\cap F^{C}\cap G^{C}) = 1 - P(E\cup F\cup G)$

$\ge 1-\dfrac{13}{24} = \dfrac{11}{24}$

$\dfrac{11}{24} > \dfrac{5}{12} \Rightarrow (4)\ \text{is FALSE}$

$\boxed{\text{Answer: (1), (2) and (3)}}$