If P (A ∪ B) = P (A ∩ B) for any two events A and B, then A. P (A

1.	If P (A ∪ B) = P (A ∩ B) for any two events A and B, then A. P (A) = P B. (B) P (A) > P (B) C. P (A) < P (B) D. none of these
Answer» We have, P(A ⋃ B) = P(A ⋂ B) By General Addition Rule, P(A) + P (B) – P(A ⋂ B) = P(A ⋃ B) ⇒ P(A) + P (B) – P(A ⋂ B) = P(A ⋂ B) [given] ⇒ [P(A) – P(A ⋂ B)] + [P(B) – P(A ⋂ B)] = 0 But P(A) – P(A ⋂ B) ≥ 0 and P(B) – P(A ⋂ B) ≥ 0 [∵ P(A ⋂ B) ≤ P(A) or P(B)] ⇒ P(A) – P(A ⋂ B) = 0 and P(B) – P(A ⋂ B) = 0 ⇒ P(A) = P(A ⋂ B) …(i) and P(B) = P(A ⋂ B) …(ii) From (i) and (ii), we get ∴ P(A) = P(B) Hence, the correct option is (A).

If P (A ∪ B) = P (A ∩ B) for any two events A and B, then A. P (A) = P B. (B) P (A) > P (B) C. P (A) < P (B) D. none of these

Answer»

We have, P(A ⋃ B) = P(A ⋂ B)

By General Addition Rule,

P(A) + P (B) – P(A ⋂ B) = P(A ⋃ B)

⇒ P(A) + P (B) – P(A ⋂ B) = P(A ⋂ B) [given]

⇒ [P(A) – P(A ⋂ B)] + [P(B) – P(A ⋂ B)] = 0

But P(A) – P(A ⋂ B) ≥ 0

and P(B) – P(A ⋂ B) ≥ 0

[∵ P(A ⋂ B) ≤ P(A) or P(B)]

⇒ P(A) – P(A ⋂ B) = 0

and P(B) – P(A ⋂ B) = 0

⇒ P(A) = P(A ⋂ B) …(i)

and P(B) = P(A ⋂ B) …(ii)

From (i) and (ii), we get

∴ P(A) = P(B)

Hence, the correct option is (A).

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If P (A ∪ B) = P (A ∩ B) for any two events A and B, then A. P (A) = P B. (B) P (A) &gt; P (B) C. P (A) &lt; P (B) D. none of these

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If P (A ∪ B) = P (A ∩ B) for any two events A and B, then A. P (A) = P B. (B) P (A) > P (B) C. P (A) < P (B) D. none of these