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# In a binomial distribution $B \bigg( n, P = \large\frac{1}{4} \bigg)$ If the probability of at least one success is greater than or equal to $\large\frac{9}{10}$ then $n$ is greater than

$\begin {array} {1 1} (A)\;\large\frac{1}{log_{10}4-log_{10}3} & \quad (B)\;\large\frac{1}{log_{10}4+log_{10}3} \\ (C)\;\large\frac{9}{log_{10}4-log_{10}3} & \quad (D)\;\large\frac{4}{log_{10}4-log_{10}3} \end {array}$

According to the condition $1- \bigg( \large\frac{3}{4} \bigg)^n \geq \large\frac{9}{10}$
$\Rightarrow \bigg( \large\frac{3}{4} \bigg)^n \leq 1- \large\frac{9}{10} = \large\frac{1}{10}$
$\Rightarrow \bigg( \large\frac{4}{3} \bigg)^n \geq 10$
$\Rightarrow n [ \log\: 4 - \log \: 3 ] \geq \log_{10}10=1$
$n \geq \large\frac{1}{\log_{10}4-\log_{10}3}$
Ans : (A)