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Math Help - Selecting pair of items

  1. #1
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    Selecting pair of items

    Suppose there are 10 different pairs of socks in a drawer. In other words, there are 20 singleton socks in total in the drawer. I randomly pick 2 socks out from the drawer. What is the probability that the 2 socks that I picked out is a matching pair?


    My answer to this question is 1/2. But I am not sure if what I have done is correct because 1/2 seems quite unbelievable.

    My interpretation is this:

    To get one particular pair of sock, I need this: \frac{1}{20}\cdot\frac{1}{19}

    Since there are 20 socks in the drawer and I am picking 2 out of it, I take \frac{1}{20}\cdot\frac{1}{19}\cdot\binom{20}{2} = \frac{1}{2} .

    But is this correct?
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  2. #2
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    Re: Selecting pair of items

     P = \frac{10}{\binom{20}{2}}=\frac{10}{190}=\frac{1}{1  9}
    Thanks from xEnOn
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  3. #3
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    Re: Selecting pair of items

    Quote Originally Posted by xEnOn View Post
    Suppose there are 10 different pairs of socks in a drawer. In other words, there are 20 singleton socks in total in the drawer. I randomly pick 2 socks out from the drawer. What is the probability that the 2 socks that I picked out is a matching pair?
    There are \binom{20}{2}=190 ways to pick pairs,
    Ten of those match,

    Ponder this \frac{10}{190}=\frac{1}{19} which is \frac{20}{20}\cdot\frac{1}{19}
    Thanks from xEnOn
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  4. #4
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    Re: Selecting pair of items

    ohh yeah... Now that when you mention the 10ways to pick versus the total 190 ways, it makes sense.

    Somehow, I am having this tendency to think of this problem being able to solve with the binomial distribution kind of technique, where I formulate the probability for one set of success and then multiply by the number of possible permutated ways. This was what I thought to get my earlier answer.
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  5. #5
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    Re: Selecting pair of items

    Quote Originally Posted by xEnOn View Post
    I am having this tendency to think of this problem being able to solve with the binomial distribution kind of technique, where I formulate the probability for one set of success and then multiply by the number of possible permutated ways. This was what I thought to get my earlier answer.
    It cannot be binomial distribution because the operations are not independent.
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  6. #6
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    Re: Selecting pair of items

    Hello, xEnOn!

    Suppose there are 10 different pairs of socks in a drawer.
    In other words, there are 20 singleton socks in total in the drawer.
    I randomly pick 2 socks out from the drawer.
    What is the probability that the 2 socks that I picked out is a matching pair?

    In your first draw, you can get ANY sock (it doesn't matter).

    Of the remaining 19 socks, how many match the first sock?

    Then what is the probability of a matching pair?
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  7. #7
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    Re: Selecting pair of items

    yea, from this perspective, it really is 1/19 because I needed just one more sock from the drawer of 19 socks left. Thanks a lot for all the enlightenments.
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