Question

Let $X$ be a binomially distributed random variable with mean $4$ and variance $\frac{4}{3}$. Then, $54P(X\le 2)$ is equal to

• A. $\frac{146}{27}$
• B. $\frac{126}{81}$
• C. $\frac{146}{81}$
• D. $\frac{73}{27}$

Answer 1

The correct option is A $\frac{146}{27}$

Mean $\mu =4=np$
Variance $={\sigma }^2=np(1-P)=\frac{4}{3}$
$4(1-P)=\frac{4}{3}$
$P=\frac{2}{3}$
$n\times \frac{2}{3}=4$
$n=6$

$P(X=k)={}^n{C}_k{P}^k(1-P{)}^{n-k}$

$P(X\le 2)=P(X=0)+P(X=1)+P(X=2)$

$={}^6{C}_0{P}^0(1-P{)}^6+{}^6{C}_1{P}^1(1-P{)}^5+{}^6{C}_2{P}^2(1-P{)}^4$

$={}^6{C}_0{\left(\frac{1}{3}\right)}^6+{}^6{C}_1\left(\frac{2}{3}\right){\left(\frac{1}{3}\right)}^5+{}^6{C}_2{\left(\frac{2}{3}\right)}^2{\left(\frac{1}{3}\right)}^4$

$={\left(\frac{1}{3}\right)}^6[1+12+60]=\frac{73}{3^6}$

$54P(X\le 2)=\frac{73}{3^6}\times 54=\frac{146}{27}$