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    Member Jskid's Avatar
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    [RESOLVED]probability involving combinatorics

    A talent show judge has 6 semi finalists. The judge thinks that none of them have any talent and will randomly select the 3 winners. Suppose that in fact, there are 2 semi finalists that are talented and are equally talented. What is the probability that both will be selected in the top 3? The answer is \frac{24}{120}
    I did 6C2 because there are 6 people and 2 need to be picked from them, divided by 6C3 because that's the total possible way 3 people from the 6 can be selected. So \frac{6C2}{6C3}=\frac{3}{4}
    Last edited by Jskid; August 2nd 2010 at 07:05 PM. Reason: resolved
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    Quote Originally Posted by Jskid View Post
    A talent show judge has 6 semi finalists. The judge thinks that none of them have any talent and will randomly select the 3 winners. Suppose that in fact, there are 2 semi finalists that are talented and are equally talented. What is the probability that both will be selected in the top 3? The answer is \frac{24}{120}
    When I first read the question, I thought that cannot be the correct answer.
    I do not see how the question implies order in any way.

    However, if we take order, first, second, and third places, into account that is indeed the answer. That is a permutation \mathcal{P}(6)=6!=120. There are four ways we can include a third winner with the talented two. So we those three can finish in 3!=6 ways thus there are (4)(6)=24 possible ways.
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    Quote Originally Posted by Jskid View Post
    A talent show judge has 6 semi finalists. The judge thinks that none of them have any talent and will randomly select the 3 winners. Suppose that in fact, there are 2 semi finalists that are talented and are equally talented. What is the probability that both will be selected in the top 3? The answer is \frac{24}{120}
    I did 6C2 because there are 6 people and 2 need to be picked from them, divided by 6C3 because that's the total possible way 3 people from the 6 can be selected. So \frac{6C2}{6C3}=\frac{3}{4}
    \binom{6}{2} is the number of ways of picking different pairs of 2 from 6,

    which is not what you are looking for here.

    You want those particular two with a third person.

    So you need only select 1 from the remaining 4, which is \binom{4}{1}=4


    You were correct to calculate \binom{6}{3}

    to find the total number of ways to pick 3 from 6.

    Therefore, the probability of the two
    talented ones making the final is

    \displaystyle\huge\frac{\binom{4}{1}}{\binom{6}{3}  }=\frac{4}{\frac{6(5)4}{3!}}=4\frac{3!}{6(5)4}=\fr  ac{24}{120}
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    Member Jskid's Avatar
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    Quote Originally Posted by Archie Meade View Post
    [tex]
    So you need only select 1 from the remaining 4, which is \binom{4}{1}=4
    Could this/should this be interpreted as a permutation instead of a combination?
    i.e. 4P1
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    Quote Originally Posted by Jskid View Post
    Could this/should this be interpreted as a permutation instead of a combination?
    i.e. 4P1
    The difficulty with that is if you pick 1 from 4,
    you can't really arrange that person,
    it's the exact same as selecting 1.

    4P1=4C1

    \frac{4!}{(4-1)!}=\frac{4!}{(4-1)!1!}

    However, you can examine the order in which the finalists are chosen.

    Then you can make the calculation using permutations,
    based on who could be chosen 1st, who could be chosen next
    and who is last chosen.

    (If the two talented ones are chosen,
    then only one more finalist needs to be chosen with them
    and that requires selecting 1 from the remaining 4 contestants.)
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    Did you read my response to your OP?
    I tried to explain why that proposed answer is feasible.
    Frankly I find the second reply correct but unreasonable.

    I admit that I did editing for testing companies. So I am very much aware of the ways questions and possible responses are written. I can assure you that this question fails on both considerations: Statement and Distractions.

    Do you realize that \dfrac{24}{120}=\dfrac{1}{5}?
    So why is that proposed answer?
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    Quote Originally Posted by Jskid View Post
    A talent show judge has 6 semi finalists. The judge thinks that none of them have any talent and will randomly select the 3 winners. Suppose that in fact, there are 2 semi finalists that are talented and are equally talented. What is the probability that both will be selected in the top 3? The answer is \frac{24}{120}
    I did 6C2 because there are 6 people and 2 need to be picked from them, divided by 6C3 because that's the total possible way 3 people from the 6 can be selected. So \frac{6C2}{6C3}=\frac{3}{4}
    The logic of the analysis depends on how the judge selects the three winners. If he selects a random subset of the 6 semi-finalists the following will give the answer:

    There are 20 subsets of size 3 of a set of 6 elements ( ^6C_3 f you like but I think the notation is not particularly helpful here. It is notions not notation that is important) Of these sets 4 contain the two designated individuals (these sets contain the two designated individuals and one of the other 4 contestants, hence there are 4 such sets).

    So if the judge selects a random subset of size 3 from all the subsets of this size with each subset being equally likely, the probability that the two designated contestants are in the selected set is 4/20=1/5=24/120.

    If the judge selects a first second and third then Plato's logic applies (and gives the same answer, but it gives an answer to slightly different but equivalent question). In this question the two methods of choosing the winners are equivalent but this is not always so, different methods of selecting can give different answers and in the long run it pays to be specific.

    CB
    Last edited by CaptainBlack; August 1st 2010 at 11:26 PM. Reason: changed "distinct elements" to "elements" as we are talking sets and the elements are by definition distinct
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    There's another way, although I consider it less direct

    Probability of picking them 1st and 2nd: 2 * (1/6 * 1/5 * 4/4)

    Probability of picking them 1st and 3rd: 2 * (1/6 * 4/5 * 1/4)

    Probability of picking them 2nd and 3rd: 2 * (4/6 * 1/5 * 1/4)

    Multiplication by 2 is to account for the fact we can choose them in either of 2 orders.

    Add them up to get 1/5 as expected.
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  9. #9
    Member Jskid's Avatar
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    Quote Originally Posted by Plato View Post
    Did you read my response to your OP?
    I tried to explain why that proposed answer is feasible.
    Frankly I find the second reply correct but unreasonable.

    I admit that I did editing for testing companies. So I am very much aware of the ways questions and possible responses are written. I can assure you that this question fails on both considerations: Statement and Distractions.

    Do you realize that \dfrac{24}{120}=\dfrac{1}{5}?
    So why is that proposed answer?
    I read all replies to all of my questions. In your first post what does P(6) mean?
    Last edited by Jskid; August 2nd 2010 at 07:04 PM. Reason: added "of my" on last line
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    Quote Originally Posted by Jskid View Post
    I read all replies to all of my questions. In your first post what does P(6) mean?
    Number of permutations of {1,2,3,4,5,6} or of any set with cardinality 6. It is 6! because there are 6 ways to choose the first element, 5 ways to choose the second element, etc. You could also write as 6P6 ... if you wanted to..
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