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

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