• # question_answer In a binomial distribution $B\left( n,p=\frac{1}{4} \right),$if  the probability of at least one success is greater than or equal to$\frac{9}{10}$, then n is greater than     AIEEE  Solved  Paper-2009 A)                                                     $\frac{1}{\log _{10}^{4}-\log _{10}^{3}}$           B) $\frac{1}{\log _{10}^{4}+\log _{10}^{3}}$C)                        $\frac{9}{\log _{10}^{4}-\log _{10}^{3}}$           D) $\frac{4}{\log _{10}^{4}-\log _{10}^{3}}$

$P(x\ge 1)\ge \frac{9}{10}$ $\Rightarrow$ $1-P(x=0)\le \frac{9}{10}$ $\Rightarrow$ $\frac{1}{10}\ge {{\left( \frac{3}{4} \right)}^{n}}$ $\Rightarrow$ ${{\left( \frac{3}{4} \right)}^{n}}\le \frac{1}{10}$ $\Rightarrow$ $n[\log _{10}^{3}-\log _{10}^{4}]\le -1$ $\Rightarrow$ $n\ge \frac{1}{\log _{10}^{4}-\log _{10}^{3}}$