Two events A and B will be independent, then

Two events A and B are independent when one event happening doesn't affect the other event's chances.

For independent events A and B:

$P(A \cap B) = P(A) \times P(B)$

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For Option 1: $P(\bar{A} \cap \bar{B}) = [1-P(A)][1-P(B)]$

$\bar{A}$ means "A does NOT happen" and $\bar{B}$ means "B does NOT happen"

$P(\bar{A}) = 1 - P(A)$

$P(\bar{B}) = 1 - P(B)$

If A and B are independent, then $\bar{A}$ and $\bar{B}$ are also independent.

Therefore:

$P(\bar{A} \cap \bar{B}) = P(\bar{A}) \times P(\bar{B})$

$P(\bar{A} \cap \bar{B}) = [1-P(A)][1-P(B)]$

This is correct.

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For Option 2: $P(A) + P(B) = 1$

Independence doesn't mean probabilities add up to 1.

Counter-example: Tossing a coin where A = heads and B = getting 6 on a die.

$P(A) = 0.5$ and $P(B) = \frac{1}{6}$

They're independent, but $0.5 + \frac{1}{6} \neq 1$

This is incorrect.

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For Option 3: $P(A) = P(B)$

Independent events don't need equal probabilities.

Counter-example: $P(A) = 0.3$ and $P(B) = 0.7$

They can still be independent even though $0.3 \neq 0.7$

This is incorrect.

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For Option 4: $P(A) + P(B) = 0$

This only happens if $P(A) = 0$ and $P(B) = 0$ (impossible events).

This is not what independence means.

This is incorrect.

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Therefore, the answer is Option 1.