Two events $A$ and $B$ are said to be independent, if $P(A B)=P(A) \times P(B)$

Consider the result given in alternative $B$.

$\mathrm{P}\left(\mathrm{A} \mathrm{B}^{\prime}\right)=[1-\mathrm{P}(\mathrm{A})][1-\mathrm{P}(\mathrm{B})]$

$\Rightarrow \mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=1-\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{A}) . \mathrm{P}(\mathrm{B})$

$\Rightarrow 1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})=1-\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$

$\Rightarrow \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})$

$\Rightarrow \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{AB})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$

$\Rightarrow \mathrm{P}(\mathrm{AB})=\mathrm{P}(\mathrm{A}). \mathrm{P}(\mathrm{B})$

This implies that $\mathrm{A}$ and $\mathrm{B}$ are independent, if $\mathrm{P}\left(\mathrm{AB}^{\prime}\right)=[1-\mathrm{P}(\mathrm{A})][1-\mathrm{P}(\mathrm{B})]$

Distracter Rationale

$A.$  Let $\mathrm{P}(\mathrm{A})=\mathrm{m}, \,\mathrm{P}(\mathrm{B})=\mathrm{n}, \,0<\mathrm{m}, \,\mathrm{n}<1$

$A$ and $B$ are mutually exclusive.

$\therefore \mathrm{A} \cap \mathrm{B}=\phi$

$\Rightarrow \mathrm{P}(\mathrm{AB})=0$

However, $\mathrm{P}(\mathrm{A}) . \mathrm{P}(\mathrm{B})=\mathrm{mn} \neq 0$

$\therefore \mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \neq \mathrm{P}(\mathrm{AB})$

$C.$ Let $A:$ Event of getting an odd number on throw of a die $=\{1,3,5\}$

$\Rightarrow P(A)=\frac{3}{6}=\frac{1}{2}$

$B:$  Event of getting an even number on throw of a die $=\{2,4,6\}$

$P(B)=\frac{3}{6}=\frac{1}{2}$

Here, $A \cap B=\phi$

$\mathrm{P}(\mathrm{AB})=0$

$P(A) \cdot P(B)=\frac{1}{4} \neq 0$

$\mathrm{P}(\mathrm{A}) .\mathrm{P}(\mathrm{B}) \neq \mathrm{P}(\mathrm{AB})$

$D.$ From the above example, it can be seen that, $P(A)+P(B)=\frac{1}{2}+\frac{1}{2}=1$

However, it cannot be inferred that $\mathrm{A}$ and $\mathrm{B}$ are independent.

Thus, the correct answer is $B$.