Ooooone more math question?

Question

Ok seriously I am having a huge math brain block.

A committee is formed to investigate what activities teenagers have available in small communities. The committee is to have 7 members, chosen randomly from interested community members. There are 10 Parents, 5 teens, and 4 adults with no children. All have expressed interest.

What is the probability that the adults no children all end up on the committee. Express to the nearest 10th of a percent.

So far I have
PA=10
PB=5
PC=4
I have no idea where to go

Answer 1

Total number of combinations is
19 Choose 7

Total number of those combinations that include the four Parents_No_Children:
15 Choose 3
(basically - all of these combinations automatically include the 4 PNC - there are only 3 more openings, with 15 to choose from)

Assuming every combination of 7 has equal probability:

(15 Choose 3)
-------------------
(19 Choose 7)

This is the probability. ( If you don't know what Choose means, you'll have to look it up. A good explanation is here:
http://en.wikipedia.org/wiki/Combinatorics)

Though if you haven't studied the 'choose' function in your class, there is probably some other solution.

Answer 2

There are 19 total candidates
The odds that an adult w/o children will be picked 1st is 4/19.
The odds that another will be picked next is 3/18
The odds a third one will be picked next is 2/17
The odds that the fourth on is picked next is 1/16
The toatal odss are 4/19*3/18*2/17*1/16 = 1/7752 = 0.0129 %

Answer 3

I think the probability that no children are on the council would match the probably that all four children are on the council. That's 4P19.

It might be 4C19... (I get permutations & combinations confused.)

Answer 4

Its been a while since I did probability problems, but I think you add up all the adults. 14 and divide it by the total of people.
14
---- = .77777 x 100 = 77.8%
18