• # question_answer There are 6 boxes labelled ${{B}_{1}},{{B}_{2}},{{B}_{3}},...,{{B}_{6}}.$ By In each trial two fair dice ${{D}_{1}},{{D}_{2}}$ are thrown. If ${{D}_{1}}$ shows j and ${{D}_{2}}$shows k, then j balls are put into the box ${{B}_{K}},$ After n trials, what is the probability that ${{B}_{1}}$contains at most one ball? A) $\left( \frac{{{5}^{n-1}}}{{{6}^{n-1}}} \right)+\left( \frac{{{5}^{n}}}{{{6}^{n}}} \right)\left( \frac{1}{6} \right)$ B) $\left( \frac{{{5}^{n}}}{{{6}^{n}}} \right)+\left( \frac{{{5}^{n-1}}}{{{6}^{n-1}}} \right)\left( \frac{1}{6} \right)$ C) $\left( \frac{{{5}^{n}}}{{{6}^{n}}} \right)+n\left( \frac{{{5}^{n-1}}}{{{6}^{n-1}}} \right)\left( \frac{1}{6} \right)$ D) $\left( \frac{{{5}^{n}}}{{{6}^{n}}} \right)+n\left( \frac{{{5}^{n-1}}}{{{6}^{n-1}}} \right)\left( \frac{1}{{{6}^{2}}} \right)$

 ${{B}_{1}}{{B}_{2}}....{{B}_{n}}$ Case I Di never show 1 Probability $={{\left( \frac{5}{6} \right)}^{n}}$ Case II ${{D}_{2}}$shows 1 (one time) then ${{D}_{1}}$ Probability $=\left\{ {}^{n}{{C}_{1}}{{\left( \frac{5}{6} \right)}^{n-1}}\left( \frac{1}{6} \right) \right\}\left( \frac{1}{6} \right)$ Total probability $={{\left( \frac{5}{6} \right)}^{n}}+n\left( \frac{{{5}^{n-1}}}{{{6}^{n-1}}} \right)\left( \frac{1}{{{6}^{2}}} \right)$