If $X$ follows Binomial distribution with parameters $n=5, p$ and $P(X=2)=9 P(X=3)$, then $p$ is equal to ............. .
$$\begin{array}{lrl} \because & P(X=2)=9 \cdot P(X=3) & \text { (where, } n=5 \text { and } q=1-p) \\ \Rightarrow & { }^5 C_2 p^2(1-p)^3=9 \cdot{ }^5 C_3 p^3(1-p)^2 & \end{array}$$
$$\begin{aligned} & \Rightarrow \quad \frac{5!}{2!3!} p^2(1-p)^3=9 \cdot \frac{5!}{3!2!} p^3(1-p)^2 \\ & \Rightarrow \quad \frac{p^2(1-p)^3}{p^3(1-p)^2}=9 \\ & \Rightarrow \quad \frac{(1-p)}{p}=9 \Rightarrow 9 p+p=1 \\ & \therefore \quad p=\frac{1}{10} \end{aligned}$$
If X be a random variable taking values $x_1, x_2, x_3, \ldots, x_{\mathrm{n}}$ with probabilities $\mathrm{P}_1, \mathrm{P}_2, \mathrm{P}_3, \ldots, \mathrm{P}_{\mathrm{n}}$, respectively. Then, $\operatorname{Var}(x)$ is equal to ......... .
$$\begin{aligned} \operatorname{Var}(X) & =E(X)^2-[E(X)]^2 \\ & =\sum_{i=1}^n X^2 P(X)-\left[\sum_{i=1}^n X P(X)\right]^2 \\ & =\Sigma P_i x_i^2-\left(\Sigma P_i x_j\right)^2 \end{aligned}$$
Let $A$ and $B$ be two events. If $P(A / B)=P(A)$, then $A$ is ........... of $B$.
$$\begin{array}{lr} \because & P(A / B)=\frac{P(A \cap B)}{P(B)} \\ \Rightarrow & P(A)=\frac{P(A \cap B)}{P(B)} \\ \Rightarrow & P(A) \cdot P(B)=P(A \cap B) \end{array} $$ So, $A$ is independent of $B$.