The random variable $\mathrm{X}$ follows binomial distribution $\mathrm{B}(\mathrm{n}, \mathrm{p})$, for which the difference of the mean and the variance is 1 . If $2 \mathrm{P}(\mathrm{X}=2)=3 \mathrm{P}(\mathrm{X}=1)$, then $n^{2} \mathrm{P}(\mathrm{X}>1)$ is equal to :

$$ \begin{aligned} & n p-n p q=1 \\\\ \Rightarrow & n p(1-q)=1 \\\\ \Rightarrow & n p^2=1 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & 2 P(X=2)=3 P(X=1) \\\\ & 2 \cdot{ }^n C_2 p^2 q{ }^{n-2}=3 \cdot{ }^n C_1 p \cdot q^{n-1} \\\\ \Rightarrow & 2 \cdot \frac{n \cdot(n-1)}{2} \cdot p=3 \cdot n \cdot q \\\\ \Rightarrow & (n-1) p=3(1-p) \\\\ \Rightarrow & \left(\frac{1}{p^2}-1\right) p=3(1-p) \\\\ \Rightarrow & \frac{(1-p)(1+p)}{p}=3(1-p) \\\\ \Rightarrow & 1+p=3 p \\\\ \Rightarrow & p=\frac{1}{2} \\\\ \therefore & n=4 \end{aligned} $$ <br/><br/>$$ \begin{aligned} n^2 P(x>1) & =n^2(1-P(x=1)-P(x=0)) \\\\ & =16\left(1-{ }^4 C_1 \cdot\left(\frac{1}{2}\right)^4-\left(\frac{1}{2}\right)^4\right)=11 \end{aligned} $$