Question

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. $P\left(E \cap F \cap G^{C}\right) \leq \frac{1}{40}$
• B. $P\left(E^{C} \cap F \cap G\right) \leq \frac{1}{15}$
• C. $P(E \cup F \cup G) \leq \frac{13}{24}$
• D. $P\left(E^{C} \cap F^{C} \cap G^{C}\right) \leq \frac{5}{12}$

Answer 1

Answer: (A), (B), (C)

$P(E)=\frac{1}{8}, P(F)=\frac{1}{6}, P(G)=\frac{1}{4}, P(E \cap F \cap G)=\frac{1}{10}$
(A) $P\left(E \cap F \cap G^{C}\right)=P(E \cap F)-P(E \cap F \cap G)$
$\leq P ( E )- P ( E \cap F \cap G )$
$\leq \frac{1}{8}-\frac{1}{10}$
$\leq \frac{5-4}{40} $
$\leq \frac{1}{40}$
(B) $P\left(E^{C} \cap F \cap G\right)=P(F \cap G)-P(E \cap F \cap G)$
$\leq P(F)-P(E \cap F \cap G)$
$\leq \frac{1}{6}-\frac{1}{10}$
$\leq \frac{10-6}{60}$
$\leq \frac{4}{60}$
$\leq \frac{1}{15}$
(C) $P ( E \cup F \cup G ) \leq P ( E )+ P ( E )+ P ( G )$
$\leq \frac{1}{8}+\frac{1}{6}+\frac{1}{4} $
$\leq \frac{13}{24}$
(D) $P\left(E^{C} \cap F^{C} \cap G^{C}\right)=1-P(E \cup F \cup G)$
$\geq 1-\frac{13}{24}$
$\geq \frac{11}{24}>\frac{10}{24}>\frac{5}{12}$