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Math Help - probability help!

  1. #1
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    Post probability help!

    I need help in some probability based problems...

    Q1) A 2007 study by the National center on Addiction and substance abuse at the Columbia University reported that for college students, the estimated probability of being a binge drinker was .50 for males and .34 for females. Using notation, express each of the these as a conditional probabilities. -------> I know we have to conditional probability P(A/B) = P(A or B)/P(B)
    So, I think its going to be P(F/M) = P(F) = .34
    Similarly, P(M/F) = P(M) = .50, if anyone could please make me understand.


    Q2) A top administrator estimated that the probability of Federal Law violation is .30 at the first project and .25 at the second project. Also, he believes the occurrences of violations in these two projects are disjoint.

    1) What is the federal law violation in the first or second project.
    -------> I think its just P(A or B) = P(A) + P(B) = .55

    2) Given that there is not a federal law violation in the first project, find the probability that there is federal law violatoin in the second project.
    --------->I am not sure how to deal with this problem.

    3) What would be the change if we call both of the above options as independent....
    ---------> is it just going to be P(A intersect B) = P(A) . P(B)...since both of their occurrences depend upon each other..

    Thanks for the help!
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    Quote Originally Posted by Vedicmaths View Post
    I need help in some probability based problems...

    Q1) A 2007 study by the National center on Addiction and substance abuse at the Columbia University reported that for college students, the estimated probability of being a binge drinker was .50 for males and .34 for females. Using notation, express each of the these as a conditional probabilities. -------> I know we have to conditional probability P(A/B) = P(A or B)/P(B)
    So, I think its going to be P(F/M) = P(F) = .34
    Similarly, P(M/F) = P(M) = .50, if anyone could please make me understand.

    Mr F says: Pr(B | M) = 0.5, Pr(B | F) = 0.34. Context should make the notation obvious.


    Q2) A top administrator estimated that the probability of Federal Law violation is .30 at the first project and .25 at the second project. Also, he believes the occurrences of violations in these two projects are disjoint.

    1) What is the federal law violation in the first or second project.
    -------> I think its just P(A or B) = P(A) + P(B) = .55

    Mr F says: Correct, because {\color{red} \Pr(A \cap B) = 0} by definition of disjoint.

    2) Given that there is not a federal law violation in the first project, find the probability that there is federal law violatoin in the second project.
    --------->I am not sure how to deal with this problem.

    Mr F says: If events A and B are disjoint then they're mutually exclusive .... if one happens then the other can't happen. Let A be the event 'violation is first project' and B the event 'violation in second project'. Then Pr(B | A') = 0.25/0.7 = 5/14. A Venn diagram makes this very clear. Alternatively, {\color{red} \Pr(B | A') = \frac{\Pr(B \cap A')}{\Pr(A')} = \frac{0.25}{0.25 + 0.45} = \, ....}

    3) What would be the change if we call both of the above options as independent....
    ---------> is it just going to be P(A intersect B) = P(A) . P(B)...since both of their occurrences depend upon each other..

    Mr F says: If events A and B are independent then:

    Pr(B | A) = Pr(B | A') = Pr(B) = 0.25.

    {\color{red}\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B) = \Pr(A) + \Pr(B) - \Pr(A) \cdot \Pr(B) = \, ..... }

    Thanks for the help!
    ..
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    Quote Originally Posted by Vedicmaths View Post
    I need help in some probability based problems...

    Q1) A 2007 study by the National center on Addiction and substance abuse at the Columbia University reported that for college students, the estimated probability of being a binge drinker was .50 for males and .34 for females. Using notation, express each of the these as a conditional probabilities. -------> I know we have to conditional probability P(A/B) = P(A or B)/P(B)

    [snip]
    That should be P(A and B) / P(B).
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    Sir, thank you very much for the help. But could you please tell me how did you get .25/.7--- where this .7 came from ? I understand the concept but its just hard to me to get those number.

    Thanks!
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    Quote Originally Posted by Vedicmaths View Post
    Sir, thank you very much for the help. But could you please tell me how did you get .25/.7--- where this .7 came from ? I understand the concept but its just hard to me to get those number.

    Thanks!
    There are many ways of seeing it. I suggested using a Venn diagram as one possible way. Another way: Pr(A') = 1 - Pr(A) = ......
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