If 0 < P(A) < 1, 0 < P(B) < 1 and P(A ∪ B) = P(A) + P(B) - P(A)P(B), then

P(A ∪ B)^c = P(A^c) · P(B^c)

P(A^c ∪ B^c) = P(A^c) + P(B^c)

If 0 < P(A) < 1, 0 < P(B) < 1 and P(A ∪ B) = P(A) + P(B) -… — JEE Main 1995 Mathematics PYQ

1995

0 < P(A) < 1

0 < P(B) < 1

and

P(A \cup B) = P(A) + P(B) - P(A)P(B)

, then

A.
$P(B|A) = P(B) - P(A)$
B.
$P(A^c \cup B^c) = P(A^c) + P(B^c)$
C.
$P(A \cup B)^c = P(A^c) \cdot P(B^c)$
D.
$P(A|B) = P(A) - P(B)$

Show answer and solution

Correct option: C

Solution

The condition P(A∪B) = P(A) + P(B) - P(A)P(B) means P(A∩B) = P(A)·P(B), i.e., A and B are independent events.

For independent events, Aᶜ and Bᶜ are also independent, so P(Aᶜ∩Bᶜ) = P(Aᶜ)·P(Bᶜ). Since (A∪B)ᶜ = Aᶜ∩Bᶜ, we get P((A∪B)ᶜ) = P(Aᶜ)·P(Bᶜ), which is exactly option 3.

[1995]

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JEE Main 1995 Mathematics question on Probability