Question

Let, 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

• 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 1

The correct option is D $\frac{1}{{\log }_{10}4-{\log }_{10}3}$

Probability of at least one success $=1-Nosuccess=1{-}^n{C}_n{q}^n$

where $q=1-p=\frac{3}{4}$

we want $1-{\left(\frac{3}{4}\right)}^n\ge \frac{9}{10}$

$⟹\frac{1}{10}\ge {\left(\frac{3}{4}\right)}^n⟹{\left(\frac{3}{4}\right)}^n\le \frac{1}{10}$

Taking logarithm on base 10 we have

$n{\log }_{10}\left(\frac{3}{4}\right)\le {\log }_{10}{\left(10\right)}^{-1}⟹n({\log }_{10}3-{\log }_{10}4)\le -1⟹n({\log }_{10}4-{\log }_{10}3)\ge 1⟹n\ge \frac{1}{{\log }_{10}4-{\log }_{10}3}$