Let $$A$$, $$B$$, $$C$$ be three events such that the probability that exactly one of $$A$$ and $$B$$ occurs is $$(1-k)$$, the probability that exactly one of $$B$$ and $$C$$ occurs is $$(1-2k)$$, 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 $$0 < k < 1$$. Then the probability that at least one of $$A$$, $$B$$ and $$C$$ occurs is:

Let

$$x_A,x_B,x_C$$ be the probabilities of outcomes belonging only to $$A,B,C$$ respectively.

Let

$$y_{AB},y_{BC},y_{CA}$$ be the probabilities of outcomes belonging to exactly two events, and let $$z=P(A\cap B\cap C)=k^2$$

Then,

$$x_A+x_B+y_{BC}+y_{CA}=1-k\quad\cdots(1)$$

$$x_B+x_C+y_{AB}+y_{CA}=1-2k\quad\cdots(2)$$

$$x_C+x_A+y_{AB}+y_{BC}=1-k\quad\cdots(3)$$

Adding

$$(1),(2),(3),$$

$$2(x_A+x_B+x_C+y_{AB}+y_{BC}+y_{CA})=3-4k$$

So,

$$x_A+x_B+x_C+y_{AB}+y_{BC}+y_{CA}=\frac32-2k$$

Now,

$$P(A\cup B\cup C)=x_A+x_B+x_C+y_{AB}+y_{BC}+y_{CA}+z$$

Substituting

$$z=k^2,$$

$$P(A\cup B\cup C)=\frac32-2k+k^2$$

Rewriting,

$$P(A\cup B\cup C)=k^2-2k+\frac32$$

$$=(k-1)^2+\frac12$$

Since

$$0<k<1,$$

we have

$$0<(k-1)^2<1$$

Therefore,

$$P(A\cup B\cup C)>\frac12$$

Hence, the required probability is

$$\boxed{\frac32-2k+k^2}$$ and it is always greater than $$\boxed{\frac12}$$