Question

The mean and variance of a binomial distribution are $\frac{4}{3}$ and $\frac{8}{9}$ respectively. Find $P(x\ge 1)$.

Answer 1

Given the mean and variance of a binomial distgribution are $\frac{4}{3}$ and $\frac{8}{9}$ respectively.

Let the parameters be $n,P$.

So, $np=\frac{4}{3}$ and $npq=\frac{8}{9}$ where $q=1-p$

$q=\frac{\frac{8}{9}}{\frac{4}{3}}=\frac{2}{3}$

So, $p=1-q=1-\frac{2}{3}=\frac{1}{3}$

$np=\frac{4}{3}\Rightarrow n=4$

$\therefore$ The binomial distribution is ,

$P(x=r){=}^4{C}_r\left(\frac{1}{3}{)}^r\right(\frac{2}{3}{)}^{4-r}$

Where $r=0,1,2,3,4.$\

$P(x\ge 1)={}^4{C}_1\left(\frac{1}{3}\right){\left(\frac{2}{3}\right)}^3+{}^4{C}_2{\left(\frac{1}{3}\right)}^2{\left(\frac{2}{3}\right)}^2+{}^4{C}_3{\left(\frac{1}{3}\right)}^3\left(\frac{2}{3}\right)+{}^4{C}_4{\left(\frac{1}{3}\right)}^4\left(\frac{2}{3}\right).$