KVPY Sample Paper KVPY Stream-SX Model Paper-25

  • question_answer
    A, B are C are contesting the election for the post of secretary of a club which does not allow ladies to become members. The probabilities of A, B and C winning the election are \[\frac{1}{3},\frac{2}{9}\] and \[\frac{4}{9}\] respectively. The probabilities of introducing the clause of admitting lady member to the club by A, B and C are 0.6, 0.7 and 0.5, respectively. The probability that ladies will be taken as members in the club after the election is

    A) \[\frac{26}{45}\]

    B) \[\frac{5}{9}\]

    C) \[\frac{19}{45}\]           

    D) none of these

    Correct Answer: A

    Solution :

    Let \[{{\operatorname{E}}_{A}}\]= The event of A becoming secretary. Similarly, \[{{\operatorname{E}}_{\operatorname{B}}}\] and \[{{\operatorname{E}}_{C}}\]
    \[{{\operatorname{E}}_{L}}\]= The event of admitting lady members.
    Here, \[\operatorname{P}\left( {{E}_{A}} \right)=\frac{1}{3},P\left( {{E}_{B}} \right)=\frac{2}{9}\]and \[\operatorname{P}\left( {{E}_{C}} \right)=\frac{4}{9}\]Clearly, \[{{\operatorname{E}}_{A}},{{E}_{B}}\]and \[{{\operatorname{E}}_{C}}\] are mutually exclusive and exhaustive.
    Also, \[\operatorname{P}\left( \frac{{{E}_{L}}}{{{E}_{A}}} \right)=0.6,P\left( \frac{{{E}_{L}}}{{{E}_{B}}} \right)=0.7\]and \[P\left( \frac{{{E}_{L}}}{{{E}_{C}}} \right)=0.5\]
    \[\therefore \]Required probability
    \[=P\left( {{E}_{A}} \right).P\left( \frac{{{E}_{L}}}{{{E}_{A}}} \right)+P\left( {{E}_{B}} \right).P\left( \frac{{{E}_{L}}}{{{E}_{B}}} \right)+P\left( {{E}_{C}} \right).P\left( \frac{{{E}_{L}}}{{{E}_{C}}} \right)\]\[=\frac{1}{3}\times \frac{3}{5}\times \frac{2}{9}\times \frac{7}{10}+\frac{4}{9}\times \frac{5}{10}=\frac{26}{45}\]

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