1. ## review question,stumped

A committee is to be formed to investigate what activities teenagers have in small communities. The committee is to have 7 members, chosen randomly. There are 10 parents,5 teenagers,and 4 adults without children.

What is the probability that the adults without children are chosen for the committee.

I have it at 29C4,but then there is still 3 spots left in the committee,so do you have to calculate everyone elses chances of slotting in,and THEN factor that in to the 4 adults without children?

2. Originally Posted by dan123
A committee is to be formed to investigate what activities teenagers have in small communities. The committee is to have 7 members, chosen randomly. There are 10 parents,5 teenagers,and 4 adults without children.

What is the probability that the adults without children are chosen for the committee.

I have it at 29C4,but then there is still 3 spots left in the committee,so do you have to calculate everyone elses chances of slotting in,and THEN factor that in to the 4 adults without children?
Hi,

Are there 29 people or 19 people to choose from?
Suppose there are 19: 10 parents, 5 teenagers, and 4 adults
We find the probability that no adults without children are chosen
$P(\text{ no adults without children})= \frac{{15 \choose7}}{{19 \choose 7}}$ (Assuming that we are finding at least one adult without children)

Then find the complement:
$P(\text{at least one adults without children})=1-\frac{{15 \choose7}}{{19 \choose 7}}=\frac{1127}{1292}$
I may be wrong

3. Hello, Dan!

A committee is to have 7 members, chosen randomly,
from 10 parents, 5 teenagers, and 4 adults without children.

What is the probability that the adults without children are chosen for the committee?

I assume that this means that all four AWOCs are on the committee.
There are 19 people and 7 are selected.
. . There are: . $_{19}C_7$ possible committees.

There are 4 AWOCs and 15 Others.
. . There is 1 way to have the 4 AWOCs on the committee.
. . There are: $_{15}C_3$ choices for the other 3 members.
Hence, there are: . $_{15}C_3$ committees with the 4 AWOCs.

Therefore, the probability is: . $\frac{_{15}C_3}{_{19}C_7} \;=\;\frac{15!}{3!\,12!}\cdot\frac{7!\,12!}{19!} \;=\;\frac{35}{3876}$