If A and B are independent, then P (exactly one of $\mathrm{A}, \mathrm{B}$ occurs)
$$=P(A) P\left(B^{\prime}\right)+P(B) P\left(A^{\prime}\right) \text {. }$$
If $A$ and $B$ are two events such that $P(A)>0$ and $P(A)+P(B)>1$, then $P(B / A) \geq 1-\frac{P\left(B^{\prime}\right)}{P(A)}$
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 ............ .
$$\begin{aligned} &\text { Here, }\\ &\begin{aligned} & P\left(A^{\prime} \cup B^{\prime}\right)=\frac{2}{3} \text { and } P(A \cup B)=\frac{5}{9} \\ & P\left(A^{\prime} \cup B^{\prime}\right)=1-P(A \cap B) \end{aligned}\\ &\Rightarrow \quad \frac{2}{3}=1-P(A \cap B)\\ &\Rightarrow \quad P(A \cap B)=1-\frac{2}{3}=\frac{1}{3} \end{aligned}$$
$$\begin{aligned} \because \quad P\left(A^{\prime}\right)+P\left(B^{\prime}\right) & =1-P(A)+1-P(B) \\ & =2-[P(A)+P(B)] \\ & =2-[P(A \cup B)+P(A \cap B)] \\ & =2-\left(\frac{5}{9}+\frac{1}{3}\right)=2-\left(\frac{5+3}{9}\right) \\ & =\frac{18-8}{9}=\frac{10}{9} \end{aligned}$$
