Question

If $X$ is a Binomial variate with the range $\left\{0,1,2,3,4,5,6\right\}$ and $P\left(X=2\right)=4P\left(X=4\right)$, then the parameter $p$ of $X$ is

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{3}{4}$

Answer 1

The correct option is A

$\frac{1}{3}$

Explanation for the correct option.

Step 1. Form the equation.

For binomial distribution for parameters $n$ and $p$, $P\left(X=r\right)$ is expanded as: $P\left(X=r\right)={}^{n}C_r{p}^r{q}^{n-r}$ where $q=1-p$.

Now, here $n=6$. So the equation $P\left(X=2\right)=4P\left(X=4\right)$ can be written as:

$$\begin{gathered} {}^{6}C_2{p}^2{q}^{6-2}=4\times {}^{6}C_4{p}^4{q}^{6-4} \\ \Rightarrow {}^{6}C_{6-2}{p}^2{q}^4=4\times {}^{6}C_4{p}^4{q}^2\left[{}^{n}C_r={}^{n}C_{n-r}\right] \\ \Rightarrow q^2=4p^2 \end{gathered}$$

Step 2. Find the value of $p$.

In the equation $q^2=4p^2$, substitute $1-p$ for $q$ and solve the quadratic equation for $p$.

$$\begin{gathered} {\left(1-p\right)}^2=4p^2 \\ \Rightarrow 4p^2-p^2-1+2p=0 \\ \Rightarrow 3p^2+2p-1=0 \\ \Rightarrow 3p^2+3p-p-1=0 \\ \Rightarrow 3p(p+1)-(p+1)=0 \\ \Rightarrow (p+1)(3p-1)=0 \end{gathered}$$

So either $p=-1$ or $p=\frac{1}{3}$.

But the value of $p$ cannot be negative, so $p=-1$ is rejected.

Thus the value of $p$ is $\frac{1}{3}$.

Hence, the correct option is A.