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JEE Main & Advanced Mathematics Probability Question Bank Conditional probability bayes theorem

  • question_answer
    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]

    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.                 \[P\left( \frac{E}{{\bar{F}}} \right)+P\left( \frac{{\bar{E}}}{{\bar{F}}} \right)=\frac{P(E\cap \bar{F})}{P(F)}+\frac{P(\bar{E}\cap \bar{F})}{P(F)}=\frac{P(\bar{F})}{P(\bar{F})}=1\]


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