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KVPY Sample Paper KVPY Stream-SX Model Paper-12

  • question_answer
    A person goes to office either by car, scooter, bus or train whose probabilities are respectively, \[\frac{1}{7},\frac{3}{7},\frac{2}{7}\] and\[\,\frac{1}{7}\]. The probability that he travelled by car is

    A) \[\frac{1}{7}\]

    B) \[\frac{2}{7}\]

    C) \[\frac{6}{7}\]

    D) \[\frac{5}{7}\]

    Correct Answer: A

    Solution :

    Let \[{{E}_{1}},{{E}_{2}},{{E}_{3}}\] and \[{{E}_{4}}\] denote the events of the person travelling by car, scooter, bus or train, respectively. Let E denote the event of his reaching office in time. By hypothesis
    \[P({{E}_{1}})=\frac{1}{7},P({{E}_{2}})\]\[=\frac{3}{7},P({{E}_{3}})=\frac{2}{7},P({{E}_{4}})=\frac{1}{7}\]
    \[\operatorname{P}\left( \frac{E}{{{E}_{1}}} \right)=\frac{7}{9},P\left( \frac{E}{{{E}_{2}}} \right)=\frac{8}{9}\]
    \[\operatorname{P}\left( \frac{E}{{{E}_{3}}} \right)=\frac{5}{9},P\left( \frac{E}{{{E}_{4}}} \right)=\frac{8}{9}\]
    By Bayes? theorem
    \[\operatorname{P}\left( \frac{{{E}_{1}}}{E} \right)=\frac{P\left( {{E}_{1}} \right)P\left( E/{{E}_{1}} \right)}{\sum\limits_{j=1}^{4}{P\left( {{E}_{j}} \right)P\left( E/{{E}_{j}} \right)}}\]\[=\frac{\frac{1}{7}\times \frac{7}{9}}{\frac{1}{7}\times \frac{7}{9}+\frac{3}{7}\times \frac{8}{9}+\frac{2}{7}\times \frac{5}{9}+\frac{1}{7}\times \frac{8}{9}}=\frac{7}{49}=\frac{1}{7}\]


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