A) \[\frac{1}{{{\log }_{10}}4+{{\log }_{10}}3}\]
B) \[\frac{9}{{{\log }_{10}}4-{{\log }_{10}}3}\]
C) \[\frac{4}{{{\log }_{10}}4-{{\log }_{10}}3}\]
D) \[\frac{1}{{{\log }_{10}}4-{{\log }_{10}}3}\]
Correct Answer: D
Solution :
[d] We have \[P(x\ge 1)\ge \frac{9}{10}\] \[\Rightarrow 1-p(x=0)\ge \frac{9}{10}\] \[\Rightarrow 1{{-}^{n}}{{C}_{0}}{{\left( \frac{1}{4} \right)}^{0}}{{\left( \frac{3}{4} \right)}^{n}}\ge \frac{9}{10}\] \[\Rightarrow 1-\frac{9}{10}\ge {{\left( \frac{3}{4} \right)}^{n}}\Rightarrow {{\left( \frac{3}{4} \right)}^{n}}\le \left( \frac{1}{10} \right)\] Taking log to the base ¾, on both sides, we get \[n{{\log }_{3/4}}\left( \frac{3}{4} \right)\ge {{\log }_{3/4}}\left( \frac{1}{10} \right)\] \[\Rightarrow n\ge -{{\log }_{3/4}}10=\frac{-{{\log }_{10}}10}{{{\log }_{10}}\left( \frac{3}{4} \right)}\] \[\Rightarrow n\ge \frac{1}{{{\log }_{10}}4-{{\log }_{10}}3}\]
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