Question

If $X$ follows a binomial distribution with parameters $n=6$ and $p$ and $4P\left(X=4\right)=P\left(X=2\right)$, then $p=$

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{6}$
• D. $\frac{1}{3}$

Answer 1

The correct option is D

$\frac{1}{3}$

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 $4P\left(X=4\right)=P\left(X=2\right)$ can be written as:

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

Step 2. Find the value of $p$.

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

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

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 D.