JEE Main & Advanced Sample Paper JEE Main Sample Paper-19

  • question_answer Three persons \[A,\,\,\,B,\,\,\,C\] throw a die in succession. The one getting 'six' wins. If \[A\] starts then the probability of \[B\] winning is

    A) \[\frac{36}{91}\]                                              

    B) \[\frac{25}{91}\]

    C) \[\frac{41}{91}\]                                              

    D) \[\frac{30}{91}\]

    Correct Answer: D

    Solution :

    \[P(\bar{E}E)+P(\bar{E}\bar{E}\bar{E}\bar{E}\bar{E})\]                 \[=\frac{5}{6}\times \frac{1}{6}+{{\left( \frac{5}{6} \right)}^{4}}\times \frac{1}{6}+{{\left( \frac{5}{6} \right)}^{8}}\frac{1}{6}...\infty \]                 \[=\frac{5}{36}\left[ 1+{{\left( \frac{5}{6} \right)}^{3}}... \right]=\frac{30}{91}\]

adversite


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