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

• 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. $\frac{4}{{\log }_{10}4-{\log }_{10}3}$

Answer 1

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

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

$P(x\ge 1)\ge \frac{9}{10}$
$\Rightarrow 1-P(x=0)\ge \frac{9}{10}$

$P(x=0){=}^n{C}_0{p}^0{q}^n={\left(\frac{3}{4}\right)}^n$
$\Rightarrow \frac{1}{10}\ge (\frac{3}{4}{)}^n$
$\Rightarrow (\frac{3}{4}{)}^n\le \frac{1}{10}$
$\Rightarrow n[{\log }_{10}3-{\log }_{10}4]\le -1$
$\Rightarrow n\ge \frac{1}{{\log }_{10}4-{\log }_{10}3}$