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
Correct Answer: B
Solution :
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\le P(A)+P(B)\], \[(\because \,\,P(A\cap \,B)\ge \,0)\].
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