1. **Stating the problem:** We have three insurance options A, B, and C with probabilities $\frac{1}{4}$, $\frac{1}{3}$, and $\frac{5}{12}$ respectively. Each employee can choose exactly two of these options or none at all. We need to find the probability that a randomly selected employee does not choose any additional insurance. 2. **Formula and rules:** The total probability of all possible choices must sum to 1. Since employees choose either two options or none, the sum of probabilities of choosing exactly two options plus the probability of choosing none equals 1. 3. **Calculate the probability of choosing exactly two options:** The pairs are (A and B), (A and C), and (B and C). Assuming independence and that the probabilities given are marginal probabilities, the probability of choosing two options is the sum of the probabilities of these pairs. 4. **Calculate each pair's probability:** $$P(A \cap B) = P(A) \times P(B) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12}$$ $$P(A \cap C) = P(A) \times P(C) = \frac{1}{4} \times \frac{5}{12} = \frac{5}{48}$$ $$P(B \cap C) = P(B) \times P(C) = \frac{1}{3} \times \frac{5}{12} = \frac{5}{36}$$ 5. **Sum the probabilities of choosing two options:** $$\frac{1}{12} + \frac{5}{48} + \frac{5}{36} = \frac{4}{48} + \frac{5}{48} + \frac{20}{144}$$ Convert all to a common denominator 144: $$\frac{12}{144} + \frac{15}{144} + \frac{20}{144} = \frac{47}{144}$$ 6. **Calculate the probability of choosing no additional insurance:** $$P(\text{none}) = 1 - P(\text{two options}) = 1 - \frac{47}{144} = \frac{144}{144} - \frac{47}{144} = \frac{97}{144}$$ 7. **Final answer:** The probability that a randomly selected employee does not choose any additional insurance is $$\boxed{\frac{97}{144}}$$. Note: The options given (4/14 and 1/2) do not match this result, so the problem might have additional context or assumptions not stated here.