Let $A, B$ and $C$ be three events such that the probability that exactly one of $\mathrm{A}$ and $\mathrm{B}$ occurs is $(1-\mathrm{k})$, the probability that exactly one of $\mathrm{B}$ and $\mathrm{C}$ occurs is $(1-2 \mathrm{k})$, the probability that exactly one of $C$ and $A$ occurs is $(1-k)$ and the probability of all $A, B$ and $C$ occur simultaneously is $k^{2}$, where $0<\mathrm{k}<1$. Then the probability that at least one of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ occur is :
Correct Option: , 2
$\mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B})+\mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}})=1-\mathrm{k}$
$\mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{C})+\mathrm{P}(\mathrm{A} \cap \overline{\mathrm{C}})=1-2 \mathrm{k}$
$\mathrm{P}(\overline{\mathrm{B}} \cap \mathrm{C})+\mathrm{P}(\mathrm{B} \cap \overline{\mathrm{C}})=1-\mathrm{k}$
$\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=\mathrm{k}^{2}$
$\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=1-\mathrm{k}$...(1)
$\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})-2 \mathrm{P}(\mathrm{B} \cap \mathrm{C})=1-\mathrm{k}$...(2)
$\mathrm{P}(\mathrm{C})+\mathrm{P}(\mathrm{A})-2 \mathrm{P}(\mathrm{A} \cap \mathrm{C})=1-2 \mathrm{k}$...(3)
$(1)+(2)+(3)$
$P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C)$$-P(C \cap A)=\frac{-4 k+3}{2}$
So
$P(A \cup B \cup C)=\frac{-4 k+3}{2}+k^{2}$
$P(A \cup B \cup C)=\frac{2 k^{2}-4 k+3}{2}$
$=\frac{2(\mathrm{k}-1)^{2}+1}{2}$
$\mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})>\frac{1}{2}$
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