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)

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