Question

In a binomial distribution $B\left(n,p=\frac{1}{4}\right)$, if the probability of at least one success is greater than equal to $\frac{9}{10}$, then $n$ is greater than

• 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. $\frac{4}{{\log }_{10}{}^4-{\log }_{10}{}^3}$

Answer 1

Answer: (A)

$1-q^n\ge \frac{9}{10}$
$\Rightarrow {\left(\frac{3}{4}\right)}^n\le \frac{1}{10}$
$\Rightarrow n\ge {\log }_{\frac{3}{4}}10$
$\Rightarrow n\ge \frac{1}{{\log }_{10}{}^4-{\log }_{10}{}^3}$