Question

If $X$ is a binomial variable with $E\left(X\right)=2$ and $V\left(X\right)=\frac{4}{3}$,
the probability of $x$ successes is

• A. ${}^6{C}_x{\left(\frac{1}{3}\right)}^x{\left(\frac{2}{3}\right)}^{6-x}$
• B. ${}^6{C}_x{\left(\frac{2}{3}\right)}^x{\left(\frac{1}{3}\right)}^{6-x}$
• C. ${}^9{C}_x{\left(\frac{1}{3}\right)}^x{\left(\frac{2}{3}\right)}^{6-x}$
• D. ${}^9{C}_x{\left(\frac{1}{3}\right)}^{6-x}{\left(\frac{2}{3}\right)}^x$

Answer 1

Answer: (A)

${}^6{C}_x{\left(\frac{1}{3}\right)}^x{\left(\frac{2}{3}\right)}^{6-x}$

Given E(X)=2 and V(X)= $\frac{4}{3}$
E(X) = Mean = np = 2
V(X) = Variance = np(1-p) = $\frac{4}{3}$
$\Rightarrow 2(1-p)=\frac{4}{3}$
$\Rightarrow 1-p=\frac{2}{3}$
$\Rightarrow p=\frac{1}{3}$
$\Rightarrow np=2$
$\Rightarrow n=\frac{2}{p}=2\times 3=6$
The probability of x successes is $\Rightarrow P(X=x){=}^n{C}_x{p}^x(1-p{)}^{n-x}$
$={}^6{C}_x\left(\frac{1}{3}{)}^x\right(1-\frac{1}{3}{)}^{6-x}$

$={}^6{C}_x\left(\frac{1}{3}{)}^x\right(\frac{2}{3}{)}^{6-x}$