Q. If $E$ and $F\$are two independent events such that $0 < P(E) < 1\ and\ 0 < P(F) < 1,\$then
A) $E$ and $F^{c}$ are independent
B) $E^{c}andF^{c}$ are independent
C) $P\left( \frac{E}{F} \right) + P\left( \frac{E^{c}}{F^{c}} \right) = 1$
D) None of these
Explanation
(c)
$P(E \cap F) = P(E).P(F)$
Now, $P(E \cap F) = P(E) - P(E \cap F) = P(E)\lbrack 1 - P(F)\rbrack$
$= P(E).P(F^{c})$
and $P\left( E^{c} \cap F^{c} \right) = 1 - P(E \cup F)$
$= 1 - \lbrack P(E) + P(F) - P(E \cap F)$
$= \left\lbrack 1 - P(E) \right\rbrack\left\lbrack 1 - P(F) \right\rbrack = P\left( E^{c} \right)P(F^{c})$
Also $P(E/F) = P(E)\ and\ P\left( \frac{E^{c}}{F^{c}} \right) = P(E^{c})$
$\Rightarrow \ P(E/F) + P\left( \frac{E^{c}}{F^{c}} \right) = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.