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HELP PLEASE Probability of Waste dump sites

  1. May 24, 2009 #1
    HELP PLEASE!!! Probability of Waste dump sites

    Okay, here's my problem:
    A federal agency is deciding wqhich of two waste dump projects to inviestigate. A top administrator estimates that the probability of federal law violations is 0.30 in the first site and 0.25 at the second project. He also believes the occurences of violations in these two projects are disjoint.

    #1. what is the probability of federal law violations in the first or second project
    So I'm thinking it's just p(AnB)=0.30-0.25, but I'm not sure. My prof has some examples and horrible explanaitions that don't give just probability, so I'm lost here. I also thought it could be p(AuB)= 0.3+0.25, but I am so lost as stats and math are not my strong point.

    #2. Given that there is not a federal law violation in the first project, find the probability that there is a federal law violation in the second project.
    Is this supposed to be p(B)=0.25? I know that's just the easy answer, but really, if there is non happeing in the first project, that means p(Abar) =0.70, so would it be 0.70+0.25? or 0.70-0.25? Once again, I have no idea

    #3. In reality, the administrator confused disjoint and independent, and the events are actually independent. Anser #1 and #2 with the correct information.
    Okay, so dijoint means they just don't happen at the same time, and independent means ....I don't know, how would all this change?

    Any help is appreciated. Explanaitions really work wonders, especially if you can show me with those venn diagrams. I'm just so very very lost in this class they force me to take for my degree. Kudos for those of you who rock this stuff =)
  2. jcsd
  3. May 24, 2009 #2
    Re: HELP PLEASE!!! Probability of Waste dump sites

    Let [tex]A[/tex] be the event that the first dump commits a violation.

    Let [tex]B[/tex] be the event that the second dump commits a violation.

    Disjoint means [tex] P(A \cap B) = 0 [/tex]

    It also means [tex] P(A \cap \overline{B}) = P(A) [/tex] and [tex] P( \overline{A} \cap {B}) = P(B) [/tex]

    Independence means [tex] P (A \cap B) = P(A)P(B) [/tex]

    And if [tex]A[/tex] and [tex]B[/tex] are independent, so are [tex]A[/tex] and [tex] \overline{B} [/tex], [tex] \overline{A} [/tex] and [tex]B[/tex], and [tex] \overline{A}[/tex] and [tex] \overline{B} [/tex]

    1) [tex] P(A \cup B) = P(A) + P (B) - P(A \cap B) [/tex]

    2) [tex] P(B| \overline{A}) = \frac {P(B \cap \overline{A})}{P(\overline{A})} = \frac {P(B) + P( \overline{A}) - P(B \cap \overline{A})}{P(\overline{A})} [/tex]
    Last edited: May 24, 2009
  4. May 24, 2009 #3


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    Re: HELP PLEASE!!! Probability of Waste dump sites

    "Disjoint" in this situation would mean there cannot be violations in BOTH sites- one excludes the other.
    Do you mean "independent" rather than "disjoint" here?

    First AnB is "A and B", not "A or B". Second p(AnB)= 0 if A and B are "disjoint" (mutually exclusive) or p(A)p(B), if they are independent, not P(A)- P(B).

    P(A or B)= p(A)+ p(B)- p(AnB) and so either .30+ .25= .55 if they are mutually exclusive, .30+ .25- (.30)(.25)= .55-.075= .475 if they are independent.

    If they are "disjoint" (mutually exclusive), then p(B|A) (probability of B given that A is true)= 0, by definition of "mutually exclusive". If independent, then p(B)= .25 irrelevant of A.

  5. May 24, 2009 #4
    Re: HELP PLEASE!!! Probability of Waste dump sites

    Thank you so much for putting this in perspective for me. I have a better understanding now then I did those three days in the lecture!
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