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Math Help - very basic probability question

  1. #1
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    very basic probability question

    I was just confused if I am thinking this one right.

    so the question is:

    "In a set of numbers from 1 to 10. If two numbers are to be selected from these 10 numbers with replacement, then what is the probability that at least one of them is even?"

    So I thought. there are four possible outcome,

    Even*Odd
    Odd*Even
    Odd*Odd
    Even*Even

    and since 3 out of them contains even number So I thought the probability
    was 3/4. Is this the right way of thinking this problem??

    please help. thank you.
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  2. #2
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    Quote Originally Posted by pandakrap View Post
    I was just confused if I am thinking this one right.

    so the question is:

    "In a set of numbers from 1 to 10. If two numbers are to be selected from these 10 numbers with replacement, then what is the probability that at least one of them is even?"

    So I thought. there are four possible outcome,

    Even*Odd
    Odd*Even
    Odd*Odd
    Even*Even

    and since 3 out of them contains even number So I thought the probability
    was 3/4. Is this the right way of thinking this problem??

    please help. thank you.
    Calculate 1 - Pr(none are even) = 1 - \frac{{5 \choose 2} \cdot {5 \choose 0}}{{10 \choose 2}}.
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  3. #3
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    thank you for your response.

    and I'm sorry, but could you explain the formula you have put up? I don't think I clearly understand the logics of that formula.

    thank you
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  4. #4
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    Quote Originally Posted by pandakrap View Post
    thank you for your response.

    and I'm sorry, but could you explain the formula you have put up? I don't think I clearly understand the logics of that formula.

    thank you
    Have you learned about combinatorics eg. the number of ways of choosing r objects from n objects etc.?
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  5. #5
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    yes I have I know why you did 5 choose 2, because there are 5 even and 5 odd, and i know why you did 10 choose 2, because there are total of 10 numbers and you need to choose 2,

    but What I don't get is why you did 1 minus, that whole rationalized combinatorics.

    So I guess, I kind of can understand why you did each part of the equation. but I can't seem to understand why you have decided to put those numbers together.
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  6. #6
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    Quote Originally Posted by pandakrap View Post
    yes I have I know why you did 5 choose 2, because there are 5 even and 5 odd, and i know why you did 10 choose 2, because there are total of 10 numbers and you need to choose 2,

    but What I don't get is why you did 1 minus, that whole rationalized combinatorics.

    So I guess, I kind of can understand why you did each part of the equation. but I can't seem to understand why you have decided to put those numbers together.
    Read my first reply again .... The part where I said "Calculate 1 - Pr(none are even)" .....
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  7. #7
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    oh......... okokok hahaha, now I feel stupid.. thanks, I just got that..

    thank you very much
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  8. #8
    MHF Contributor matheagle's Avatar
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    actually the choose is incorrect, that's the hypergeometric situation which is correct when you are choosing WITHOUT replacement

    This is binominal, since it is with replacement.

    The 3/4 is correct, which can be achieved via 1-(1/2)(1/2) as well.
    Last edited by matheagle; March 3rd 2010 at 07:09 PM. Reason: typo
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  9. #9
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    With replacement,

    P(both\  odd)=\frac{\binom{5}{1}}{\binom{10}{1}}\frac{\bino  m{5}{1}}{\binom{10}{1}}=\frac{5}{10}\ \frac{5}{10}=\frac{1}{2}\ \frac{1}{2}=\frac{1}{4}

    Therefore,

    P(at\ least\ 1\ even)=1-P(both\ odd)=1-\frac{1}{4}
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  10. #10
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    Quote Originally Posted by matheagle View Post
    actually the choose is incorrect, that's the hypergeometric situation which is correct when you are choosing WITHOUT replacement

    This is binominal, since it is with replacement.

    The 3/4 is correct, which can be achieved via 1-(1/2)(1/2) as well.
    I could have sworn it said without .... So I'm going blind as well ....
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  11. #11
    MHF Contributor matheagle's Avatar
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    and who calls me oldeagle?
    am eagled eyed as well
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