Seven college students have applied for internships at a local firm. Three of them will be selected for interviews. Part 1 of 2 In how many ways can three students be selected for interviews? There are 35 ways that three students can be selected for interviews. Part: $1 / 2$ Part 2 of 2 Four of the seven students are from Middle Georgia State College. What is the probability that all three of the interviewed students are from Middle Georgia State College? Express your answer as a reduced fraction or decimal rounded to at least four decimal places. The probability that all three are from Middle Georgia State College is

Step 1 ：The first part of the question is asking for the number of ways to select 3 students out of 7 for interviews. This is a combination problem, because the order in which the students are selected does not matter. The formula for combinations is \(C(n, k) = \frac{n!}{k!(n-k)!}\), where n is the total number of items, k is the number of items to choose, and ! denotes factorial. So, \(C(7, 3) = \frac{7!}{3!(7-3)!} = 35\). Therefore, there are 35 ways that three students can be selected for interviews.

Step 2 ：The second part of the question is asking for the probability that all three students selected are from Middle Georgia State College. There are 4 students from this college, so we need to find the number of ways to select 3 students out of these 4. This is also a combination problem, and we can use the same formula. So, \(C(4, 3) = \frac{4!}{3!(4-3)!} = 4\). The probability is then the number of ways to select 3 students from Middle Georgia State College divided by the total number of ways to select 3 students out of 7. Therefore, the probability is \(\frac{4}{35} = 0.11428571428571428\).

Step 3 ：Final Answer: The probability that all three students selected for interviews are from Middle Georgia State College is approximately \(\boxed{0.1143}\) or \(\boxed{\frac{4}{35}}\) in reduced fraction form.