More probability

[frusic]

The committee has 7 members. 10 parents, 5 teens, 4 adults without children.

What is the probability that the adults without children are all on the same committee?

I've got that adults without children make up 21% of the eligible committee members, I'm not sure if that's relevant or where to go from there if it is

 

[soroban]

Hello, frusic!

You left out a key phrase . . .

The committee has 7 members selected from 10 parents, 5 teens, 4 adults without children.

What is the probability that the adults without children are all on the same committee?

There are 19 people and 7 will be selected.
. . There are \(\displaystyle C(19,7)\:=\:\frac{19!}{7!\,12!}\;=\;50,388\) choices.

If all 4 Adults are on the committe, the other 3 member are chosen from the 15 Others.
. . There are \(\displaystyle C(15,3)\:=\:\frac{15!}{3!\,12!}\:=\:455\) choices.

Therefore, \(\displaystyle P(\text{4 adults})\;\L=\;\frac{455}{50,388}\:=\:\frac{35}{3876}\)

 

[galactus]

I would think you would use 4 choices from the 4 adults without kids multiplied by 3 choices from the remaining 15 people, all divided by 7 choices from the total of 19 people.

\(\displaystyle \frac{C(4,4)C(15,3)}{C(19,7)}=.9%\) chance that all 4 non-kid adults will be on the committee.