• # 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

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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