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) \[P\,(E/\bar{F})+P\,(\bar{E}/\bar{F})=1\]
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. \[P\left( \frac{E}{{\bar{F}}} \right)+P\left( \frac{{\bar{E}}}{{\bar{F}}} \right)=\frac{P(E\cap \bar{F})}{P(F)}+\frac{P(\bar{E}\cap \bar{F})}{P(F)}=\frac{P(\bar{F})}{P(\bar{F})}=1\]You need to login to perform this action.
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