Question

The mean and standard deviation of a binomial variate $X$ are $4$ and $\sqrt{3}$ respectively. Then $P\left(X\ge 1\right)=$

• A. $1-{\left(\frac{1}{4}\right)}^{16}$
• B. $1-{\left(\frac{3}{4}\right)}^{16}$
• C. $1-{\left(\frac{2}{3}\right)}^{16}$
• D. `$1-{\left(\frac{1}{3}\right)}^{16}$

Answer 1

Answer: (A)

$1-{\left(\frac{1}{4}\right)}^{16}$

For Binomial distribution, Mean$\left(\mu \right)=np$ and Standard Deviation$\left(\sigma \right)=\sqrt{np(1-p)}$

According to question, Mean = 4 and Variance = $\sqrt{3}$

Thus, $np=\mathrm{4.....}\left(1\right)\sqrt{np(1-p)}=\sqrt{3}\Rightarrow np(1-p)=\mathrm{3.......}\left(2\right)$

Putting the value of np from first equation in the second equation.

$4\times (1-p)=3\Rightarrow 1-p=\frac{3}{4}\Rightarrow p=\frac{1}{4}$

Now, putting the value of p in equation(1), $n\times \frac{1}{4}=4\Rightarrow n=16$

$\because \Sigma P\left(X_i\right)=1$

$\therefore P(X\ge 1)=1-P(X<1)=1-P(X=0)=1-{}^{16}{C}_0{p}^0(1-p{)}^{16-0}$

$=1-\frac{16!}{0!(16-0)!}\times {\left(\frac{3}{4}\right)}^0\times {\left(\frac{1}{4}\right)}^{16}=1-{\left(\frac{1}{4}\right)}^{16}$