A) \[P\,(A)+P\,(B)=0\]
B) \[P\,(A)+P\,(B)=P\,(A)\,P\,\left( \frac{B}{A} \right)\]
C) \[P\,(A)+P\,(B)=2\,P\,(A)\,P\,\left( \frac{B}{A} \right)\]
D) None of these
Correct Answer: C
Solution :
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\] \[\Rightarrow P(A\cap B)=P(A)+P(B)-P(A\cap B)\] \[\{\because \,\,\,P(A\cap B)=P(A\cup B)\}\] \[\Rightarrow 2P(A\cap B)=P(A)+P(B)\] \[\Rightarrow 2P(A)\,.\,\frac{P(A\cap B)}{P(A)}=P(A)+P(B)\] \[\Rightarrow 2P(A).P\left( \frac{B}{A} \right)=P(A)+P(B)\].You need to login to perform this action.
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