Probability Question 140
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) $ .