Example 1:
Consider choosing a 5 member committee consisting of grade 8 and grade 9 students.There are 7 grade 8's, and 6 grade 9's to choose from. Let the random variable X denote the number of grade 8's on the committe. What is the probability of selecting exactly 2 grade 8 students to be on the committe ( P(X=2) )?
Solution:
The number of ways to choose 2 grade 8 students is 7C2=21, since there are 7 grade 8 students to choose from. The 3 other members on the committe must be grade 9's, and there are 6 that we can choose from. So there are 6C3=20 ways to choose the remaining memebers. If there were no restrictions, the number of ways to select a committe of 5 out of 7+6=13 students is 13C5=1287.
So P(X=2)=(7C2) (6C3) / (13C5)=0.3263
A: What is the probability of selecting a 5 member committe if there has to be only 1 grade 8 on the committe?
What is the probability that there is at least 4 grade 8's on the committe?
P(X>=4)= P(4) +P(5)
= (7C4)(6C1)/(13C5) + (7C5)/(13C5)
=(210+21) / 1287=0.1795
B: What is the probability that there is at most 2 grade 8's on the committe?
Solution
where x = max(0, n- (N-r)), ..., min(n,r) 
