Explanation

(c)
Given, mean $np = 2$ ...(i)
And variance $npq = 1$ ...(ii)
From Eqs.(i) and (ii), we get
$q = \frac{1}{2}$
$\therefore\ p = 1 - q = \frac{1}{2}$
From Eq.(i), $n \times \frac{1}{2} = 2$
$\therefore\ \ n = 4$
The binomial distribution is $\left( \frac{1}{2} + \frac{1}{2} \right)^{4}$
Now, $P(X > 1) = P(X = 2) + P(X = 3) + P(X = 4)$
$= {\ ^{4}C}_{2}\left( \frac{1}{2} \right)^{2}\left( \frac{1}{2} \right)^{2} +^{4}C_{3}\left( \frac{1}{2} \right)^{3}\left( \frac{1}{2} \right)^{1} +^{4}C_{4}\left( \frac{1}{2} \right)^{4}$
$= \frac{6 + 4 + 1}{16}$
$= \frac{11}{16}$