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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