MathematicsMedium75×since 2002Q4962Let E₁ and E₂ be two events such that the conditional probabilities P(E1∣E2)=12P({E_1}|{E_2}) = {1 \over 2}P(E1∣E2)=21, P(E2∣E1)=34P({E_2}|{E_1}) = {3 \over 4}P(E2∣E1)=43 and P(E1∩E2)=18P({E_1} \cap {E_2}) = {1 \over 8}P(E1∩E2)=81. Then :AP(E1∩E2)=P(E1) . P(E2)P({E_1} \cap {E_2}) = P({E_1})\,.\,P({E_2})P(E1∩E2)=P(E1).P(E2)BP(E1′∩E2′)=P(E1′) . P(E2)P(E{'_1} \cap E{'_2}) = P(E{'_1})\,.\,P(E{_2})P(E1′∩E2′)=P(E1′).P(E2)CP(E1∩E2′)=P(E1) . P(E2)P({E_1} \cap E{'_2}) = P({E_1})\,.\,P({E_2})P(E1∩E2′)=P(E1).P(E2)DP(E1′∩E2)=P(E1) . P(E2)P(E{'_1} \cap {E_2}) = P({E_1})\,.\,P({E_2})P(E1′∩E2)=P(E1).P(E2)Check answerSkip