Probability 3 Question 28
28. If $E$ and $F$ are independent events such that $0<P(E)<1$ and $0<P(F)<1$, then
$(1989,2 M)$
(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 independent
(d) $P(E / F)+P\left(E^{c} / F\right)=1$
Fill in the Blanks
Show Answer
Solution:
- Since, $E$ and $F$ are independent events. Therefore, $P(E \cap F)=P(E) \cdot P(F) \neq 0$, so $E$ and $F$ are not mutually exclusive events.
Now, $P(E \cap \bar{F})=P(E)-P(E \cap F)=P(E)-P(E) \cdot P(F)$
$$ =P(E)[1-P(F)]=P(E) \cdot P(\bar{F}) $$
and $P(\bar{E} \cap \bar{F})=P(\overline{E \cup F})=1-P(E \cup F)$
$$ \begin{aligned} & =1-[1-P(\bar{E}) \cdot P(\bar{F})] \\ & =P(\bar{E}) \cdot P(\bar{F}) \end{aligned} $$
So, $E$ and $\bar{F}$ as well as $\bar{E}$ and $\bar{F}$ are independent events.
Now, $P(E / F)+P(\bar{E} / F)=\frac{P(E \cap F)+P(\bar{E} \cap F)}{P(F)}$
$$ =\frac{P(F)}{P(F)}=1 $$
