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Math Help - review question,stumped

  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by dan123 View Post
    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
    Last edited by noob mathematician; May 24th 2009 at 07:20 PM.
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  3. #3
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    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}

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