If $A, B$ and $C$ are three independent events such that
$$P(A)=P(B)=P(C)=p,$$
then $P$ (atleast two of $A, B$ and $C$ occur) $=3 p^2-2 p^3$.
If $A$ and $B$ are two events such that $P(A / B)=p, P(A)=p, P(B)=\frac{1}{3}$ and $P(A \cup B)=\frac{5}{9}$, then $p$ is equal to ............... .
Here, $$P(A)=p_1 P(B)=\frac{1}{3} \text { and } P(A \cup B)=\frac{5}{9}$$
$\because P(A / B)=\frac{P(A \cap B)}{P(B)}=p \Rightarrow P(A \cap B)=\frac{p}{3}$
$$\begin{aligned} \text{and }\quad P(A \cup B) & =P(A)+P(B)-P(A \cap B) \\ \Rightarrow\quad \frac{5}{9} & =p+\frac{1}{3}-\frac{p}{3} \Rightarrow \frac{5}{9}-\frac{1}{3}=\frac{2 p}{3} \end{aligned}$$
$$\Rightarrow \quad \frac{5-3}{9}=\frac{2 p}{3} \Rightarrow p=\frac{2}{9} \times \frac{3}{2}=\frac{1}{3}$$
If $A$ and $B$ are such that $P\left(A^{\prime} \cup B^{\prime}\right)=\frac{2}{3}$ and $P(A \cup B)=\frac{5}{9}$, then $P\left(A^{\prime}\right)+P\left(B^{\prime}\right)$ is equal to ............ .
$$\begin{aligned} &\text { Here, }\\ &\begin{aligned} & P\left(A^{\prime} \cup B^{\prime}\right)=\frac{2}{3} \text { and } P(A \cup B)=\frac{5}{9} \\ & P\left(A^{\prime} \cup B^{\prime}\right)=1-P(A \cap B) \end{aligned}\\ &\Rightarrow \quad \frac{2}{3}=1-P(A \cap B)\\ &\Rightarrow \quad P(A \cap B)=1-\frac{2}{3}=\frac{1}{3} \end{aligned}$$
$$\begin{aligned} \because \quad P\left(A^{\prime}\right)+P\left(B^{\prime}\right) & =1-P(A)+1-P(B) \\ & =2-[P(A)+P(B)] \\ & =2-[P(A \cup B)+P(A \cap B)] \\ & =2-\left(\frac{5}{9}+\frac{1}{3}\right)=2-\left(\frac{5+3}{9}\right) \\ & =\frac{18-8}{9}=\frac{10}{9} \end{aligned}$$
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 ......... .
$$\begin{aligned} \operatorname{Var}(X) & =E(X)^2-[E(X)]^2 \\ & =\sum_{i=1}^n X^2 P(X)-\left[\sum_{i=1}^n X P(X)\right]^2 \\ & =\Sigma P_i x_i^2-\left(\Sigma P_i x_j\right)^2 \end{aligned}$$