Probability Question 188
Question: If $ \bar{E} $ and $ \bar{F} $ are the complementary events of events E and F respectively and if $ 0<P(F)<1, $ then
[IIT 1998]
Options:
A) $ P(E/F)+P(\bar{E}/F)=1 $
B) $ P(E/F)+P(E/\bar{F})=1 $
C) $ P(\bar{E}/F)+P(E/\bar{F})=1 $
D) $ P(E/\bar{F})+P(\bar{E}/\bar{F})=1 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ P(E/F)+P(\bar{E}/F)=\frac{P(E\cap F)+P(\bar{E}\cap F)}{P(F)} $
$ =\frac{P{(E\cap F)\cup (\bar{E}\cap F)}}{P(F)} $
$ [\because \ E\cap F $ and $ \bar{E}\cap F $ are disjoint] $ =\frac{P{(E\cup \bar{E})\cap F}}{P(F)}=\frac{P(F)}{P(F)}=1 $
Similarly we can show that and are not true while is true.
$ P( \frac{E}{{\bar{F}}} )+P( \frac{{\bar{E}}}{{\bar{F}}} )=\frac{P(E\cap \bar{F})}{P(F)}+\frac{P(\bar{E}\cap \bar{F})}{P(F)}=\frac{P(\bar{F})}{P(\bar{F})}=1 $
BETA
