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Math Help - Probability help??

  1. #1
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    Probability help??

    I wondered if some one can help me and show working out the calks to this question...

    A contractor has to excavate for foundations in an area for which there are no plans of existing services. It is anticipated that he may cut across one or more of the service lines whilst excavating. The following probabilities have been calculated for each of the following services:

    Water supply - 0.08
    Gas - 0.05
    Telephone Cable – 0.04
    Drainage – 0.12
    Electricity – 0.05

    Calculate the probability of cutting across

    a) at least one of the services
    b) both water and drainage
    c) electricity or telephone cable

    I have the answers but I don’t know how to get them no putting them on here so I can check your answers too.

    Hope someone can help

    Regards

    J

    The ansers I have are

    a = 0.2986
    b = 0.0096
    c = 0.088

    .....?
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  2. #2
    GAMMA Mathematics
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    Water supply - 0.08
    Gas - 0.05
    Telephone Cable – 0.04
    Drainage – 0.12
    Electricity – 0.05

    Calculate the probability of cutting across

    a) at least one of the services
    b) both water and drainage
    c) electricity or telephone cable
    A) P(at least one) = 1 - P(at most one)
    = 1 - P(0) - P(1)
    =1 - .92(.95).96(.88)(.95) - .08(.95)(.96)(.88)(.95) - .05(.92)(.96)(.88)(.95)  - .04(.92)(.95)(.88)(.95) - .12(.92).95(.96).95 - .05(.92).95(.96)(.88)

    B) P(both water and drainage) = P(W)P(D)
    = .12(.08)

    I will leave the last one up to you.
    Last edited by colby2152; December 4th 2007 at 07:59 AM.
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  3. #3
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    Unhappy Hmm

    I get the answer B easily its just the product of P(W) and P(D) but Im still confused as to whats going on. I see you have taken the eventually of each of the servies in turn. but I think the brackets have confused me. as to what you do to these numbers Grr. Probability was never my strong point *sob sob*
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  4. #4
    GAMMA Mathematics
    colby2152's Avatar
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    Quote Originally Posted by juangambino View Post
    I get the answer B easily its just the product of P(W) and P(D) but Im still confused as to whats going on. I see you have taken the eventually of each of the servies in turn. but I think the brackets have confused me. as to what you do to these numbers Grr. Probability was never my strong point *sob sob*
    P(A \cap B) = P(A)P(B) if independence

    P(A or B) = P(A \cup B)
    = P(A) + P(B) - P(A \cap B)
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  5. #5
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    I get the answer B easily its just the product of P(W) and P(D) but Im still confused as to whats going on. I see you have taken the eventually of each of the servies in turn. but I think the brackets have confused me. as to what you do to these numbers Grr. Probability was never my strong point *sob sob*
    If I understand your problem right, you are being confused by the invisible multiplication signs. These sit between any numbers not separated by another operation (an operation is +,- etc). You will see (or not see them :P) these more and more as you progress in maths.

    However, Colby is mistaken anyway.

    a) the probability of at least one happening = 1 - the probability none happen.

    To get the probability of none happening, just multiply together the probability of each one not happening. So

    1-[(1-0.08)(1-0.05)(1-0.04)(1-0.12)(1-.05)] should get you your answer.

    B) P(both water and drainage) = P(W)P(D)
    =
    colby got part b) right and pointed you in the right direction for part c, but just in case you haven't \cap before I will explain in words.


    the probability of either of two things happening is the sum of their probabilities minus the probability of them both happening. We subtract the probability of them both happening because we have counted this part twice; it is included in the probability of both events.
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