Question

I f F and F are independent events such that $O < P(E) < 1$ and $O < P (F ) < 1$, then

• A. E and F are mutually exclusive
• B. E and F $^c$ (the complement of the event F) are independent
• C. E $^c$ and F$^c$ are independen
• D. $P (E /F )+ P(E^c /F) = 1$

Answer 1

Answer: (D)

$P (E /F )+ P(E^c /F) = 1$

Answer 2

Since, E and F are independent events. Therefore,
$P (E \cap F ) = P(E) . P(F) \ne 0,$ so $E$ and $F$ are not mutually exclusive events.
Now, $P(E \cap \overline{F})=P(E)-P(E \cap F)=P(E)-P(E).P(F)$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =P(E)[1-P(F)]=P(E)-P(\overline{F})$
and $\, \, \, P(\overline{E} \cap \overline{F})=P(\overline{E \cup F})=1-P(E \cup F)$
\hspace20mm \, \, \, \, \, =1-[1-P(\overline{E}),P(\overline{F})]
\hspace30mm \, \, \, [\because \, E and F are independent]
\hspace20mm \, \, P(\overline{E}).P(\overline{F})
So, $E$ and$\overline{F}$ as well as $\overline{E}$ and $\overline{F}$ are independent events.
Now, $P (E / F) + P(\overline{E}$ / $\overline{F}$)=$\frac{P(E \cap F)+P(\overline{E} \cap \overline{F})}{P(F)}$
\hspace20mm \frac{P(F)}{P(F)}=1