Question

Let $E,F$ and $G$ be three events having probabilities $P\left(E\right)=\frac{1}{8},P\left(F\right)=\frac{1}{6}$ and $P\left(G\right)=\frac{1}{4},$ 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. $P(E\cap F\cap G^c)\le \frac{1}{40}$
• B. $P(E^c\cap F\cap G)\le \frac{1}{15}$
• C. $P(E\cup F\cup G)\le \frac{13}{24}$
• D. $P(E^c\cap F^c\cap G^c)\le \frac{5}{12}$

Answer 1

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

$P(E\cup F\cup G)\le \frac{13}{24}$

$P(E\cap F\cap G^c)$
$=P(E\cap F)-P(E\cap F\cap G)$
$\le P\left(E\right)-P(E\cap F\cap G)$
$\le \frac{1}{8}-\frac{1}{10}$
$\le \frac{1}{40}$
$\therefore P(E\cap F\cap G^c)\le \frac{1}{40}$


$P(E^c\cap F\cap G)$
$=P(F\cap G)-P(E\cap F\cap G)$
$\le P\left(F\right)-P(E\cap F\cap G)$
$\le \frac{1}{6}-\frac{1}{10}$
$\le \frac{1}{15}$
$\therefore P(E^c\cap F\cap G)\le \frac{1}{15}$


$P(E\cup F\cup G)$
$\le P\left(E\right)+P\left(F\right)+P\left(G\right)$
$\le \frac{1}{8}+\frac{1}{6}+\frac{1}{4}$
$\le \frac{3+4+6}{24}$
$\therefore P(E\cup F\cup G)\le \frac{13}{24}$


$P(E^c\cap F^c\cap G^c)$
$=1-P(E\cup F\cup G)$
$\ge 1-\frac{13}{24}$
$\ge \frac{11}{24}\ge \frac{5}{12}$