For three events A,B,C P(Exactly one of A or B occurs)=P(Exactly one of B or C occurs) =P(Exactly one of C or A occurs)=1/4 and P (All the three events occur simultaneously) = 1/16. Then the probability that at least one of the events occurs, is

$$\begin{gathered} P\text{ (Exactly one of }A\text{ or }B\text{ occurs })=P(A\cup B)-P(A\cap B)=\frac{1}{4} \\ \Rightarrow P(A)+P(B)-2P(A\cap B)=\frac{1}{4} \end{gathered}$$

$$\begin{gathered} P\left(\text{ Exactly one of }B\text{ or }C\text{ occurs }\right)=P(B\cup C)-P(B\cap C)=\frac{1}{4} \\ \Rightarrow P\left(B\right)+P\left(C\right)-2P(B\cap C)=\frac{1}{4} \end{gathered}$$

$$\begin{gathered} P\left(\text{ Exactly one of }A\text{ or }C\text{ occurs }\right)=P(A\cup C)-P(A\cap C)=\frac{1}{4} \\ \Rightarrow P\left(C\right)+P\left(A\right)-2P(A\cap C)=\frac{1}{4} \end{gathered}$$

Adding above 3 equations

$2P\left(A\right)+2P\left(B\right)+2P\left(C\right)-2P(A\cap B)-2P(B\cap C)-2P(A\cap C)=\frac{3}{4}$

$\Rightarrow P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(A\cap C)=\frac{3}{8}$

$\begin{array}{r}P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)\\ \end{array}$

$\text{ So, }P(A\cup B\cup C)=\frac{3}{8}+\frac{1}{16}=\frac{7}{16}$