Probability Question 306
Question: In a binomial distribution $ B( n,p=\frac{1}{4} ) $ , if the probability of at least one success is greater than or equal to $ \frac{9}{10} $ , then n is greater than:
Options:
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} $
Show Answer
Answer:
Correct Answer: D
Solution:
We have $ P(x\ge 1)\ge \frac{9}{10} $
$ \Rightarrow 1-p(x=0)\ge \frac{9}{10} $
$ \Rightarrow 1{{-}^{n}}C_0{{( \frac{1}{4} )}^{0}}{{( \frac{3}{4} )}^{n}}\ge \frac{9}{10} $
$ \Rightarrow 1-\frac{9}{10}\ge {{( \frac{3}{4} )}^{n}}\Rightarrow {{( \frac{3}{4} )}^{n}}\le ( \frac{1}{10} ) $ Taking log to the base ¾, on both sides, we get $ n{\log_{3/4}}( \frac{3}{4} )\ge {\log_{3/4}}( \frac{1}{10} ) $
$ \Rightarrow n\ge -{\log_{3/4}}10=\frac{-{\log_{10}}10}{{\log_{10}}( \frac{3}{4} )} $
$ \Rightarrow n\ge \frac{1}{{\log_{10}}4-{\log_{10}}3} $
BETA
