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Math Help - 3 of 10 samples are contaminated

  1. #1
    Junior Member Bartimaeus's Avatar
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    Arrow 3 of 10 samples are contaminated

    3 of 10 samples are contaminated. 2 are randomly chosen. What is the probability of both of the chosen samples are contaminated?

    Hope you can help.
    Thanks.

    EDITlease include all working.
    Last edited by Bartimaeus; August 14th 2006 at 11:44 PM. Reason: to add more info
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Bartimaeus
    3 of 10 samples are contaminated. 2 are randomly chosen. What is the probability of both of the chosen samples are contaminated?

    Hope you can help.
    Thanks.
    Probability that first sample is contaminated P1=3/10.

    Probability that second sample is contaminate given the first
    is contaminated is P2=2/9.

    Probability both contaminated=P1*P2

    RonL
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  3. #3
    Junior Member Bartimaeus's Avatar
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    Thanks RonL
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  4. #4
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    Hello, Bartimaeus!

    RonL's solution is the best one.
    Here's an alternate approach (if you're familiar with Combinations).


    Three of ten samples are contaminated. .Two are randomly chosen.
    What is the probability of both of the chosen samples are contaminated?

    There are \binom{10}{2} = 45 possible outcomes.

    To get two contaminated samples, there are: \binom{3}{2} = 3 ways.

    Therefore: . P(\text{both contaminated}) \:=\:\frac{3}{45} \:=\:\frac{1}{15}

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  5. #5
    Junior Member Bartimaeus's Avatar
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    Smile

    I'm not really familiar with Combinations.
    Could you explain them further to me, please?
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  6. #6
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    Quote Originally Posted by Bartimaeus
    I'm not really familiar with Combinations.
    Could you explain them further to me, please?
    Say there are 5 bullets:
    .36 .357 .308 .45 .50
    And you are told to select two of them, the possibilities are,
    Code:
    .36 and .357
    .36 and .308
    .36 and .45
    .36 and .50
    .357 and .308
    .357 and .45
    .357 and .50
    .308 and .45
    .308 and .50
    .45 and .50
    In total there are 10 possibilities.
    You chose 2 from 5. We write,
    {5 \choose 2}=10 another notation,
    _5C_2=10.

    The mathematical formula is,
    {n\choose m}=\frac{n!}{m!(n-m)!}

    Note: Sometimes the combinations formula is refferred to as the binomial coefficients. Because the coefficients in the binomial expansion follow the combinations formula.
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