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Let $A$ and $B$ be any two arbitrary events, then, which one of the following is TRUE?

1. $P (A \cap B) = P(A)P(B)$
2. $P (A \cup B) = P(A)+P(B)$
3. $P (A \mid B) = P(A \cap B)P(B)$
4. $P (A \cup B) \leq P(A) + P(B)$
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Why should it be? The operator is multiplication not division.

$(a)$ is true only if events are independent.

$(b)$ is true only if events are mutually exclusive i.e. $P (A \cap B) = 0$

$(c)$ is false everywhere

$(d)$ is always true as $P (A \cup B) = P(A)+P(B)-P (A \cap B)$

Since, $P (A \cap B) >0$, $P (A \cup B) \leq P(A)+P(B)$
edited
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in the case of  option 1 what is P(A ∩ B) if a and b are not independent?

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In general, P(A $\cap$ B) = P(A|B) * P(B).

If A and B are independent, then knowing B doesn't affect probability of A, hence P(A|B) = P(A)

So P(A $\cap$ B) = P(A) * P(B) when A and B are independent.
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Arbitrary means anything you can take it may be mutually exclusive or mutually independent.
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Why is C false??
Option D is Ans

If A and B be two arbitrary events, then

P (A ∪ B) = P (A) +P (B) - P(A $\cap$ B)

Let say 10=7+8-5

So now  10 <= 7+8

So option D ....P (A ∪ B) <= P (A) +P (B) is correct Ans.
If A and B are mutually exclusive events then:

A$\cap$B = $\Phi$ and A$\cup$B = A+B

If A and B are non mutually exclusive events then:

A$\cap$B != $\Phi$ and A$\cup$B = A+B-A$\cap$B (principle of inclusion-exclusion)

therefore A$\cup$B <= A + B

edited
What are arbitrary events?If they are independent then Option A and if disjoint then Option C.
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Option A is correct
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Option D should be correct.
(d)P(A∪B)≤P(A)+P(B)

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