Statistics And Probability Question 576

Question: If A and B are arbitrary events, then

[DCE 2002]

Options:

A) $ P(A\cap B)\ge P(A)+P(B) $

B) $ P(A\cup B)\le P(A)+P(B) $

C) $ P(A\cap B)=P(A)+P(B) $

D) None of these

Show Answer

Answer:

Correct Answer: B

Solution:

  •                  $ P(A\cup B)=P(A)+P(B)-P(A\cap B)\le P(A)+P(B) $ ,                                                                                                  $ (\because P(A\cap \,B)\ge \,0) $ .