Among the 16 cities that a professional society is considering for its next three annual conventions, 7 are in the western part of the US.
To avoid arguments, the selection is left to chance.
If none of the cities can be chosen more than once, what are the probabilities that

(a) none of the conventions will be held in the western U.S.
(b) that all are held in the western part of the us?

Question

Among the 16 cities that a professional society is considering for its next three annual conventions, 7 are in the western part of the US. 
To avoid arguments, the selection is left to chance.
 If none of the cities can be chosen  more than once, what are the probabilities that 

(a) none of the conventions will be held in the western U.S.
(b) that all are held in the western part of the us?

anonymous · Answer

I know I could use the hypergeometric distribution forum, but I am unsure of how to get the value of "n" since the selection is left to chance.Otherwise, so far I have: N=16, a=7, and x=0

mayankdevnani · Answer

may i help you?

mayankdevnani · Answer

@JerJason

mayankdevnani · Answer

a hyper-geometric distribution does apply here. Let u be the number of possible successful outcomes, v be the number of possible failures, and n the number of outcomes chosen from the set.

P(i successes) = [C(u, i) C(v, n - i)] / C(u + v, n)

In this case, let a western city be considered a success.
u = 7
v = 9
n = 3

P(zero western cities) = [C(7, 0) C(9, 3)] / C(16, 3)

P(three western cities) = [C(7, 3) C(9, 0)] / C(16, 3)

mayankdevnani · Answer

got it     @JerJason

anonymous · Answer

I think I understand this now. Thanks.