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KVPY Sample Paper KVPY Stream-SX Model Paper-3

  • question_answer
    There are three bags \[{{\text{B}}_{\text{1}}}\text{,}\]\[{{\text{B}}_{\text{2}}}\] and \[{{\text{B}}_{\text{3}}}\text{.}\]The bag \[{{\text{B}}_{\text{1}}}\] contains 5 red and 5 green balls,  contains 3 red and 5 green balls, and \[{{\text{B}}_{\text{2}}}\] contains 5 red \[{{\text{B}}_{3}}\] and 3 green balls. Bags \[{{\text{B}}_{\text{1}}}\text{,}\]\[{{\text{B}}_{\text{2}}}\] and \[{{\text{B}}_{\text{3}}}\] have probabilities \[\frac{3}{10},\,\frac{3}{10}\] and \[\frac{4}{10}\] respectively of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?

    A) Probability that the chosen ball is green, given that the selected bag is \[{{\text{B}}_{\text{3}}}\] equals \[\frac{3}{8}\]

    B) Probability that the selected bag is \[{{\text{B}}_{\text{3}}}\] and the chosen ball is green equals \[\frac{3}{10}\]

    C) Probability that the selected bag is \[{{\text{B}}_{\text{3}}}\] and the chosen ball is green equals \[\frac{5}{13}\]

    D) Probability that the chosen ball is green equals \[\frac{39}{80}\]

    Correct Answer: A , D

    Solution :

    \[\,\,\,\,\]
    \[\text{P}({{\text{B}}_{1}})\,=\,\frac{3}{10}\left| \text{P}({{\text{B}}_{2}})=\frac{3}{10} \right|\text{P}({{\text{B}}_{3}})=\frac{4}{10}\]
    [a]        \[\text{P}({{\text{G}}_{1}}\left| {{\text{B}}_{3}} \right.)\,=\,\frac{3}{8}=\]
    [b]        \[\text{P}({{\text{B}}_{3}}\left| \text{G} \right.)=\]
    [c]         \[\text{P}({{\text{B}}_{3}}\left| \text{G} \right.)=\frac{12}{39}=\]
    [d]        \[\text{P}(\text{G})=\frac{3}{10}.\frac{5}{10}+\frac{3}{10}.\frac{5}{8}+\frac{4}{10}.\frac{3}{8}\]
                \[=\frac{12+15+12}{80}=\]

    Solution :

    Same as above


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