Answer:
a) P=0.1721
b) P=0.3528
c) P=0.3981
Step-by-step explanation:
This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.
a) In this case we have a sample size of n=15.
The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.
The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.
This probability can be calculated as:

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.
We apply the same formula but as a sum:

Then we have:

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).
The first (k less or equal to 5) is already calculated.
We have to calculate for k equal to 10 or more.

Then we have

The sum of the probabilities is
