Solution:

To find the probability that the minimum of the chosen numbers is 3 or their maximum is 7, we need to find the total number of favorable outcomes and the total number of possible outcomes.

Total Number of Possible Outcomes:
We are choosing three numbers without replacement from the set {1, 2, 3, ..., 10}. So, the total number of possible outcomes is given by the combination formula: C(10, 3) = 10! / (3! * (10-3)!) = 120.

Finding the Favorable Outcomes:
Case 1: Minimum of the chosen numbers is 3.
To satisfy this condition, we need to choose at least one number from the set {4, 5, ..., 10} and two numbers from the set {1, 2, 3}.
Number of possibilities for choosing at least one number from {4, 5, ..., 10} = C(7, 1) = 7.
Number of possibilities for choosing two numbers from {1, 2, 3} = C(3, 2) = 3.
Total number of favorable outcomes for this case = 7 * 3 = 21.

Case 2: Maximum of the chosen numbers is 7.
To satisfy this condition, we need to choose at least one number from the set {1, 2, ..., 6} and two numbers from the set {7, 8, ..., 10}.
Number of possibilities for choosing at least one number from {1, 2, ..., 6} = C(6, 1) = 6.
Number of possibilities for choosing two numbers from {7, 8, ..., 10} = C(3, 2) = 3.
Total number of favorable outcomes for this case = 6 * 3 = 18.

Total Number of Favorable Outcomes:
To find the total number of favorable outcomes, we add the favorable outcomes from both cases: 21 + 18 = 39.

Probability Calculation:
Probability = (Total Number of Favorable Outcomes) / (Total Number of Possible Outcomes)
Probability = 39 / 120 = 13 / 40

Since the probability cannot exceed 11/30, the correct answer is option A, 11/40.

Note: The answer provided as option B, 11/40, is incorrect. Option A is the correct answer.