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Math Help - Picking numbers from a lot

  1. #1
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    Picking numbers from a lot

    Say I guess 4 numbers from 1 to 30, then 4 numbers are draw randomly from the list, without replacement, what are the chances that I will get at least 3 correct?

    My idea so far:

    So to get all correct, P(all 4 correct) =  4! \frac {1}{30} \frac {1}{29} \frac {1}{28} \frac {1}{27}

    and to have 3 correct, it is P(3 correct) =  \frac {4!}{3!} \frac {1}{30} \frac {1}{29} \frac {1}{28} \frac {26}{27}

    So do I just add them up together?
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Say I guess 4 numbers from 1 to 30, then 4 numbers are draw randomly from the list, without replacement, what are the chances that I will get at least 3 correct?

    My idea so far:

    So to get all correct, P(all 4 correct) =  4! \frac {1}{30} \frac {1}{29} \frac {1}{28} \frac {1}{27}

    and to have 3 correct, it is P(3 correct) =  \frac {4!}{3!} \frac {1}{30} \frac {1}{29} \frac {1}{28} \frac {26}{27}

    So do I just add them up together?

    Your logic sounds fine to me.
    But P(3) case is not tallying with your calculation.
    P(3 correct) =  \frac {4C_3.26C_1}{30C_4} (as per me)
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