The random variable [X] follows binomial distribution [B(n, p)], for which the difference …

Question

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 :

Accepted Answer

$$
\begin{aligned}
& n p-n p q=1 \\
\Rightarrow & n p(1-q)=1 \\
\Rightarrow & n p^2=1
\end{aligned}
$$

$$
\begin{aligned}
& 2 P(X=2)=3 P(X=1) \\
& 2 \cdot{ }^n C_2 p^2 q{ }^{n-2}=3 \cdot{ }^n C_1 p \cdot q^{n-1} \\
\Rightarrow & 2 \cdot \frac{n \cdot(n-1)}{2} \cdot p=3 \cdot n \cdot q \\
\Rightarrow & (n-1) p=3(1-p) \\
\Rightarrow & \left(\frac{1}{p^2}-1\right) p=3(1-p) \\
\Rightarrow & \frac{(1-p)(1+p)}{p}=3(1-p) \\
\Rightarrow & 1+p=3 p \\
\Rightarrow & p=\frac{1}{2} \\
\therefore & n=4
\end{aligned}
$$

$$
\begin{aligned}
n^2 P(x>1) & =n^2(1-P(x=1)-P(x=0)) \\
& =16\left(1-{ }^4 C_1 \cdot\left(\frac{1}{2}\right)^4-\left(\frac{1}{2}\right)^4\right)=11
\end{aligned}
$$

Solutions · 1