Q. 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
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) $\frac{4}{\log_{10}{4 - \log_{10}3}}$
Explanation
(a)
According to the condition $1 - \left( \frac{3}{4} \right)^{n} \geq \frac{9}{10}$
$\Rightarrow \ \left( \frac{3}{4} \right)^{n} \leq 1 - \frac{9}{10} = \frac{1}{10}$
$\Rightarrow \ \left( \frac{4}{3} \right)^{n} \geq 10$
$\Rightarrow \ n\lbrack\log_{10}{4 - \log_{10}{3\rbrack \geq \log_{10}{10 = 1}}}$
$\Rightarrow \ n \geq \frac{1}{\log_{10}{4 - \log_{10}3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.