How Do You Calculate Probabilities in Waste Dump Site Investigations?

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The discussion focuses on calculating probabilities related to federal law violations at two waste dump sites, with initial confusion over whether the events are disjoint or independent. The correct approach for calculating the probability of violations at either site involves using the formula P(A or B) = P(A) + P(B) - P(A and B), which varies based on the relationship between the events. If the events are disjoint, the probability of violations occurring at both sites is zero, while if they are independent, the probabilities can be multiplied. The clarification that the administrator confused disjoint with independent significantly impacts the calculations, leading to a better understanding of how to approach the problem. Overall, the conversation highlights the importance of correctly identifying the relationship between events in probability calculations.
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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 =)
 
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Let A be the event that the first dump commits a violation.

Let B be the event that the second dump commits a violation.


Disjoint means P(A \cap B) = 0

It also means P(A \cap \overline{B}) = P(A) and P( \overline{A} \cap {B}) = P(B)


Independence means P (A \cap B) = P(A)P(B)

And if A and B are independent, so are A and \overline{B}, \overline{A} and B, and \overline{A} and \overline{B}


1) P(A \cup B) = P(A) + P (B) - P(A \cap B)

2) 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})}
 
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Fear_of_Math said:
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.
"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?

#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.
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.

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.
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.

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?
I did not notice until after my previous answers the whole problem is about a confusion of "disjoint" and "independent" so I have alread answered this!

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 =)
 


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!
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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