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

A) $\frac{1}{3}$
B) $\frac{1}{2}$
C) $\frac{2}{3}$
D) $\frac{3}{4}$
Explanation

(a)
Here, $n = 6$
According to the question
$\ ^{6}C_{2}p^{2}q^{4} = 4.{\ ^{6}C}_{4}p^{4}q^{2}$
$\Rightarrow \ q^{2} = 4p^{2}$
$\Rightarrow \ (1 - p)^{2} = 4p^{2}$
$\Rightarrow \ 3p^{2} + 2P - 1 = 0$
$\Rightarrow \ (p + 1)(3p - 1) = 0$
$\Rightarrow p = \frac{1}{3}(\because p\ cannot\ be\ negative)$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.