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Math Help - Probability question...

  1. #1
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    Probability question...

    Takeisha and Antoine are part of a group of 8 women and 8 men who will participate in a chess tournament. They will be paired up randomly into teams of two. (Takeisha is a female and Antoine is a male.)

    a) What is the probability that Takeisha and Antoine will be paired together as partners?
    b) What is the probability that Takeisha will be paired with a woman?
    c) What is the probability that each of them will be paired with a person of the same sex, i.e. Takeisha with a woman and Antoine with a man?
    d) What is the probability that each of them will be paired with a woman?

    Can anyone help me with this, I'm stuck!! =/....thank you!!!!
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  2. #2
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    Quote Originally Posted by nikkik View Post
    Takeisha and Antoine are part of a group of 8 women and 8 men who will participate in a chess tournament. They will be paired up randomly into teams of two. (Takeisha is a female and Antoine is a male.)

    a) What is the probability that Takeisha and Antoine will be paired together as partners?
    b) What is the probability that Takeisha will be paired with a woman?
    c) What is the probability that each of them will be paired with a person of the same sex, i.e. Takeisha with a woman and Antoine with a man?
    d) What is the probability that each of them will be paired with a woman?

    Can anyone help me with this, I'm stuck!! =/....thank you!!!!
    In these sorts of problems the probability is:

    (number of favourable cases)/(total number of cases)

    This is because we set things up so each case is equally likely.

    a) T can be pired with any of the 17 others, this is the total number of cases. In only one of these is T paired with A, this is the number of
    favourable cases. So the required probability is:

    p = 1 /17 ~= 0.0588.

    b) T can be pired with any of the 17 others, this is the total number of cases. As there are six other women, in 6 of these cases T is paired with
    another woman, this is the number of favourable cases. So the required probability is:

    p = 6 /17 ~= 0.3529.

    RonL
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  3. #3
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    Thank you very much!!! =)

    I also had another question I was having problems solving

    Suppose there is a ski club with 4 members. Each member normally hangs his (or her) ski poles on a hook, with each member using a separate hook. If the cleaning person takes down all the poles to wash the wall, and puts them back up randomly with two poles to each of the 4 hooks, what is the probability that the poles will be paired correctly, i.e. that each hook will hold two poles belonging to the same member?
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  4. #4
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    Hello, nikkik!

    Here's one approach to this problem.
    . . (There are many others.)


    Suppose there is a ski club with 4 members. Each member normally hangs
    his (or her) ski poles on a hook, with each member using a separate hook.
    If the cleaning person takes down all the poles to wash the wall,
    and puts them back up randomly with two poles to each of the 4 hooks,
    what is the probability that the poles will be paired correctly,
    i.e. that each hook will hold two poles belonging to the same member?

    Consider the four hooks: 1, 2, 3, 4
    . . and the eight ski poles: A, A, B, B, C, C, D, D.


    Two poles will be placed on Hook-1.
    The first can be any pole: . \frac{8}{8}\:=\:1
    The second must be its mate: . \frac{1}{7}

    Two poles will be placed on Hook 2.
    The first can be any of the remaining 6 poles: . \frac{6}{6}\:=\:1
    The second must be its mate: . \frac{1}{5}

    Two poles will be placed on Hook 3.
    The first can be any of the remaining 4 poles: . \frac{4}{4}\:=\:1
    The second must be its mate: . \frac{1}{3}

    Two poles will be placed on Hook 4.
    The remaining two poles are a matching pair: \frac{2}{2}\:=\:1


    The probability is: . 1\cdot\frac{1}{7}\cdot1\cdot\frac{1}{5}\cdot1\cdot  \frac{1}{3}\cdot1\;=\;\boxed{\frac{1}{105}}

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  5. #5
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    Thanks so much, that was very helpful!! I'm still having trouble with the last two parts of the first question I posted though, I got the first two but can't seem to figure these out->

    Takeisha and Antoine are part of a group of 8 women and 8 men who will participate in a chess tournament. They will be paired up randomly into teams of two. (Takeisha is a female and Antoine is a male.)

    What is the probability that each of them will be paired with a person of the same sex, i.e. Takeisha with a woman and Antoine with a man?

    What is the probability that each of them will be paired with a woman?
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  6. #6
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    There are D = \frac{{16!}}{{\left( {8!} \right)\left( {2^8 } \right)}} ways to form unordered pairs of contestants. There are 7 ways that T can be paired with another female; likewise 7 ways that A can be paired with another male. This leaves 12 unassigned people. There are \frac{{12!}}{{\left( {6!} \right)\left( {2^6 } \right)}} ways to form unordered pairs of these.

    What is the answer to your question?

    Post Script
    Note that there are \frac{{8!}}{{\left( {4!} \right)\left( {2^4 } \right)}} = {105} ways to form unordered pairs of the ski poles.
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