Question

In a Binomial distribution $B\left(n,p=\frac{1}{4}\right)$, if the probability of atleast one success is greater than or equal to $\frac{9}{10}$, then $n$ is greater than

• A. $\frac{1}{{\log }_{10}\left(4\right)-{\log }_{10}\left(3\right)}$
• B. $\frac{1}{{\log }_{10}\left(4\right)+{\log }_{10}\left(3\right)}$
• C. $\frac{9}{{\log }_{10}\left(4\right)-{\log }_{10}\left(3\right)}$
• D. $\frac{4}{{\log }_{10}\left(4\right)-{\log }_{10}\left(3\right)}$

Answer 1

The correct option is A

$\frac{1}{{\log }_{10}\left(4\right)-{\log }_{10}\left(3\right)}$

Explanation for the correct option.

Find the value of $n$.

Now it is given that the probability of success is $p=\frac{1}{4}$. So the probability of failure is:

$\begin{array}{rcl}q& =& 1-p\\ & =& 1-\frac{1}{4}\\ & =& \frac{3}{4}\end{array}$

Now the probability of $x$ success in $n$ trials is given as:

$\begin{array}{rcl}\mathit{P}\left(X=x\right)& \mathbf{=}& {}^{\mathbf{n}}\mathit{C}_{\mathbf{x}}{\mathit{q}}^{\mathbf{n}\mathbf{-}\mathbf{x}}{\mathit{p}}^{\mathbf{x}}\\ & =& {}^{n}C_x{\left(\frac{3}{4}\right)}^{n-x}{\left(\frac{1}{4}\right)}^x\end{array}$

Now, probability of atleast one success is given as:

$\begin{array}{rcl}P\left(X\ge 1\right)& =& 1-P\left(X=0\right)\\ & =& 1-{}^{n}C_0{\left(\frac{3}{4}\right)}^{n-0}{\left(\frac{1}{4}\right)}^0\\ & =& 1-1\times {\left(\frac{3}{4}\right)}^n\times 1\\ & =& 1-{\left(\frac{3}{4}\right)}^n\end{array}$

It is given that the probability of atleast one success is greater than or equal to $\frac{9}{10}$. So the inequality is:

$$\begin{gathered} P\left(X\ge 1\right)\ge \frac{9}{10} \\ \Rightarrow 1-{\left(\frac{3}{4}\right)}^n\ge \frac{9}{10} \\ \Rightarrow -{\left(\frac{3}{4}\right)}^n\ge \frac{9}{10}-1 \\ \Rightarrow -{\left(\frac{3}{4}\right)}^n\ge -\frac{1}{10} \\ \Rightarrow {\left(\frac{3}{4}\right)}^n\le \frac{1}{10} \end{gathered}$$

Now take the logarithm of base $10$ both sides and solve for $n$.

$$\begin{gathered} {\log }_{10}{\left(\frac{3}{4}\right)}^n\le {\log }_{10}\left(\frac{1}{10}\right) \\ \Rightarrow n\left[{\log }_{10}\left(3\right)-{\log }_{10}\left(4\right)\right]\le -1\left[log{a}^m=mloga;log\left(\frac{a}{b}\right)=loga-logb;lo{g}_a\left(\frac{1}{a}\right)=-1\right] \\ \Rightarrow -n\left[{\log }_{10}\left(4\right)-{\log }_{10}\left(3\right)\right]\le -1 \\ \Rightarrow n\left[{\log }_{10}\left(4\right)-{\log }_{10}\left(3\right)\right]\ge 1 \\ \Rightarrow n\ge \frac{1}{{\log }_{10}\left(4\right)-{\log }_{10}\left(3\right)} \end{gathered}$$

So, $n$ is greater than $\frac{1}{{\log }_{10}\left(4\right)-{\log }_{10}\left(3\right)}$.

Hence, the correct option is A.