Probability Question

Question

 Consider a group of nine individuals selected from the AARP (American Association of Retired People) of 50-year-olds and above in the United States. The number of persons in this sample who suffer from diabetes is a binomial random variable with parameters n = 9 and p = 0.186. 

1. What is the probability that exactly five of the individuals in the sample suffer from diabetes? 
2. What is the probability that two of them have diabetes?
3. Without regard to order, in how many ways can you select three individuals from this group of nine? 

My main issue is that I'm not sure which formula to use/how to calculate this.

Answer 1

For 3, it is number of combinations of 3 out of 9: 9!

-------

3!6!

For 1 and 2 probability of number of successes x out of n (9) when chance of success is p (0.186):

9!

----- (0.186)^x (1-0.186)^(9-x)

x! (9-x)!

For 1 plug in 5 for x and for 2 plug in 2 for x.