To solve this problem, we will calculate the probability that exactly 2 of the 5 chosen students are male, given that there are 4 male and 4 female students in total.

Steps to Solve:

1. Total number of students:
   There are $4$ male students and $4$ female students, so there are $8$ students in total.

2. Total number of ways to choose 5 students from 8:
   This can be calculated using combinations. The number of ways to choose 5 students from 8 is: $\binom{8}{5} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

3. Number of ways to choose exactly 2 male students:
   We need to choose 2 male students from the 4 male students, which can be done in: $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$

4. Number of ways to choose 3 female students:
   Since we are choosing 2 males, we will need to choose $5 - 2 = 3$ female students from the 4 females, which can be done in: $\binom{4}{3} = \frac{4}{1} = 4$

5. Number of favorable outcomes:
   The number of ways to choose exactly 2 males and 3 females is the product of the two combinations: $6 \times 4 = 24$

6. Probability:
   The probability is the ratio of favorable outcomes to the total outcomes: $\text{Probability} = \frac{24}{56} = \frac{3}{7}$

Final Answer:

The probability that exactly 2 of the 5 chosen students are male is $\frac{3}{7}$.

Would you like more details, or do you have any questions?

Here are 5 follow-up questions to expand your understanding:

1. How would the probability change if there were 5 male students instead of 4?
2. What is the probability of selecting exactly 3 males and 2 females?
3. How can you generalize this problem to any number of males and females?
4. How would you calculate the probability of selecting at least one male?
5. How does the total number of combinations change if we increase the total number of students?

Tip: Remember that combinations are used when the order of selection does not matter!