Two events A and B will be independent, if

Two events A and B are said to be independent, if P(AB) = P(A) × P(B)

Consider the result given in alternative B.

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

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

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

$\Rightarrow P(A \cup B)=P(A)+P(B)-P(A) \cdot P(B)$

$\Rightarrow P(A)+P(B)-P(A B)=P(A)+P(B)-P(A) \cdot P(B)$

$\Rightarrow P(A B)=P(A) \cdot P(B)$

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

Distracter Rationale

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

A and B are mutually exclusive.

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

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

However, $\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})=m n \neq 0$

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

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

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

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

$\Rightarrow P(A) \cdot P(B) \neq P(A B)$

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 A and B are independent.

Thus, the correct answer is B.