Question
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.
$$\frac{9}{{{{\log }_{10}}4 - {{\log }_{10}}3}}$$
C.
$$\frac{4}{{{{\log }_{10}}4 - {{\log }_{10}}3}}$$
D.
$$\frac{1}{{{{\log }_{10}}4 - {{\log }_{10}}3}}$$
Answer :
$$\frac{1}{{{{\log }_{10}}4 - {{\log }_{10}}3}}$$
Solution :
$$\eqalign{
& {\text{We have, }}P\left( {x \geqslant 1} \right) \geqslant \frac{9}{{10}} \cr
& \Rightarrow 1 - P\left( {x = 0} \right) \geqslant \frac{9}{{10}} \cr
& \Rightarrow 1 - {}^n{C_0}{\left( {\frac{1}{4}} \right)^0}{\left( {\frac{3}{4}} \right)^n} \geqslant \frac{9}{{10}} \cr
& \Rightarrow 1 - \frac{9}{{10}} \geqslant {\left( {\frac{3}{4}} \right)^n} \cr
& \Rightarrow {\left( {\frac{3}{4}} \right)^n} \leqslant \left( {\frac{1}{{10}}} \right) \cr
& {\text{Taking log to the base }}\frac{3}{4}{\text{, on both sides, we get}} \cr
& n\,{\log _{\frac{3}{4}}}\left( {\frac{3}{4}} \right) \geqslant {\log _{\frac{3}{4}}}\left( {\frac{1}{{10}}} \right) \cr
& \Rightarrow n \geqslant - {\log _{\frac{3}{4}}}10 = \frac{{ - {{\log }_{10}}10}}{{{{\log }_{10}}\left( {\frac{3}{4}} \right)}} \cr
& \Rightarrow n \geqslant \frac{1}{{{{\log }_{10}}4 - {{\log }_{10}}3}} \cr} $$