Let us denote the probabilities of the 8 disjoint regions in the Venn diagram as follows:

• $x_1 = P(A\text{ only})$
• $x_2 = P(B\text{ only})$
• $x_3 = P(C\text{ only})$
• $x_4 = P(A\cap B\text{ only})$
• $x_5 = P(B\cap C\text{ only})$
• $x_6 = P(C\cap A\text{ only})$
• $x_7 = P(A\cap B\cap C)$
• $x_0 = P(\text{none})$

We are given:

1. Probability that exactly one of $A$ and $B$ occurs is $1-k$.

   “Exactly one of $A$ and $B$” means $A\oplus B$, i.e. $P(A\oplus B)=P(A\cap B^c)+P(A^c\cap B).$ This includes the regions: $x_1+x_6+x_2+x_5=1-k \quad \cdots (1)$

2. Probability that exactly one of $B$ and $C$ occurs is $1-2k$: $x_2+x_4+x_3+x_6=1-2k \quad \cdots (2)$

3. Probability that exactly one of $C$ and $A$ occurs is $1-k$: $x_3+x_5+x_1+x_4=1-k \quad \cdots (3)$

4. Probability that all three occur simultaneously is: $x_7=k^2.$

We need: $P(A\cup B\cup C)=x_1+x_2+x_3+x_4+x_5+x_6+x_7.$

Step 1: Add the three given equations

Adding (1), (2), and (3):

So,

$2(x_1+x_2+x_3+x_4+x_5+x_6)=3-4k.$

Hence,

$x_1+x_2+x_3+x_4+x_5+x_6=\frac{3-4k}{2}.$

Therefore,

$P(A\cup B\cup C)=\frac{3-4k}{2}+k^2.$

So let $f(k)=k^2-2k+\frac{3}{2}=(k-1)^2+\frac{1}{2}.$

Thus,

$P(A\cup B\cup C)=(k-1)^2+\frac{1}{2}.$

Step 2: Use the condition $0<k<1$

Since $0<k<1$, we have $(k-1)^2>0.$

Therefore,

$P(A\cup B\cup C)=\frac{1}{2}+(k-1)^2>\frac{1}{2}.$

So the probability that at least one of $A,B,C$ occurs is greater than $\frac12$.

Step 3: Match with the options

• A: greater than $\frac18$ but less than $\frac14$ — false
• B: greater than $\frac12$ — true
• C: greater than $\frac14$ but less than $\frac12$ — false
• D: exactly equal to $\frac12$ — false

Hence, the correct option is:

$\boxed{\text{B}}$

Verification with stored answer

Stored correct answer: B

Our derived answer: B

They agree.