1. **Understanding the problem:** We are given the probabilities of two players, Encik Wilson and Encik Alvin, making a "hole in one" to hole S. The probability that Encik Wilson makes the hole in one is $\frac{5}{12}$. The probability that Encik Alvin makes the hole in one is $\frac{7}{12}$. We want to find the probability that either Encik Wilson or Encik Alvin makes the hole in one (i.e., the probability that at least one of them succeeds). 2. **Defining the events:** Let $W$ be the event that Encik Wilson makes a hole in one. Let $A$ be the event that Encik Alvin makes a hole in one. Given: $$P(W)=\frac{5}{12}, \quad P(A)=\frac{7}{12}$$ 3. **Probability of either event occurring:** The probability that either $W$ or $A$ happens is given by: $$P(W \cup A) = P(W) + P(A) - P(W \cap A)$$ Assuming these events are independent (the performance of one does not affect the other), we have: $$P(W \cap A) = P(W) \times P(A) = \frac{5}{12} \times \frac{7}{12} = \frac{35}{144}$$ 4. **Calculating $P(W \cup A)$:** $$P(W \cup A) = \frac{5}{12} + \frac{7}{12} - \frac{35}{144} = \frac{12}{12} - \frac{35}{144} = 1 - \frac{35}{144} = \frac{144}{144} - \frac{35}{144} = \frac{109}{144}$$ 5. **Final answer:** The probability that either Encik Wilson or Encik Alvin makes a hole in one to hole S is: $$\boxed{\frac{109}{144}}$$