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