This problem involves selecting students' projects from two different sections, which can be solved using the hypergeometric distribution. The hypergeometric distribution models the probability of drawing $k$ successes (in this case, projects from a particular section) in $n$ draws from a population without replacement.

The probability mass function for the hypergeometric distribution is given by:

$P(X = k) = \frac{\binom{M}{k} \binom{N-M}{n-k}}{\binom{N}{n}}$

where:

• $N$ is the total population (total number of students),
• $M$ is the number of "successes" in the population (number of students in the section you're interested in),
• $n$ is the number of draws (number of projects graded first),
• $k$ is the number of successes (projects from the section you're interested in),
• $\binom{a}{b}$ denotes a binomial coefficient, which represents combinations.

(a) Probability that exactly 13 of the first 17 projects are from the second section:

• Total population $N = 25 + 30 = 55$ students,
• Number of students in the second section $M = 30$,
• Number of draws $n = 17$,
• Number of successes $k = 13$.

Substituting into the hypergeometric probability formula:

$P(X = 13) = \frac{\binom{30}{13} \binom{25}{4}}{\binom{55}{17}}$

(b) Probability that exactly 9 of the first 17 projects are from the first section:

• Number of students in the first section $M = 25$,
• Number of draws $n = 17$,
• Number of successes $k = 9$.

Substituting into the hypergeometric probability formula:

$P(X = 9) = \frac{\binom{25}{9} \binom{30}{8}}{\binom{55}{17}}$

(c) Probability that all 17 projects are from the same section:

We need to calculate the probability that all 17 are either from the first section or from the second section.

• All from the first section: $P(X = 17)$ where $M = 25$. $P(X = 17) = \frac{\binom{25}{17} \binom{30}{0}}{\binom{55}{17}}$

• All from the second section: $P(X = 17)$ where $M = 30$. $P(X = 17) = \frac{\binom{30}{17} \binom{25}{0}}{\binom{55}{17}}$

The total probability for this part is the sum of these two probabilities.

Let me calculate these probabilities for you.Here are the calculated probabilities:

(a) The probability that exactly 13 of the first 17 projects are from the second section is approximately 0.0222.

(b) The probability that exactly 9 of the first 17 projects are from the first section is approximately 0.1752.

(c) The probability that all 17 projects are from the same section is approximately 0.00000177 (rounded to six decimal places).

Would you like further details or have any other questions?

Here are 5 related questions you might find interesting:

1. How does the probability change if the number of projects graded first is increased to 20?
2. What would be the probability if we asked for at least 9 projects from the first section?
3. How would the results change if there were a third section with 10 students?
4. What is the expected number of projects from the first section in the first 17 graded?
5. How does this situation relate to real-world applications of the hypergeometric distribution?

Tip: When sampling without replacement, as in this problem, the hypergeometric distribution is essential for accurately calculating probabilities.