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Thread: Probabilities

  1. #1
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    Probabilities

    Okay I have an answer for this one, but it just seems to easy , Can somone tell me if I did this right. I feel like theres more to solving the problem then what I did. Heres the problem:


    The table below gives the probabilities of combinations of religion and political parties in a major U.S city.



    Political parites Protestant (A) Catholic (B) Jewish (C) Other(D)

    Democrat (E) .35 .10 .03 .02

    Republican ( F) .27 .09 .02 .01

    Independent (G) .05 .03 .03 .01


    What is the probability that a randomly selected person would be a democrat who was not jewish?


    The probability I got was .47, all I did was p(.35)+p(.10)+p(.02)= .47
    but it just seems to easy.
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  2. #2
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    Quote Originally Posted by jax760 View Post
    Okay I have an answer for this one, but it just seems to easy , Can somone tell me if I did this right. I feel like theres more to solving the problem then what I did. Heres the problem:


    The table below gives the probabilities of combinations of religion and political parties in a major U.S city.



    Political parites Protestant (A) Catholic (B) Jewish (C) Other(D)

    Democrat (E) .35 .10 .03 .02

    Republican ( F) .27 .09 .02 .01

    Independent (G) .05 .03 .03 .01


    What is the probability that a randomly selected person would be a democrat who was not jewish?


    The probability I got was .47, all I did was p(.35)+p(.10)+p(.02)= .47
    but it just seems to easy.
    You need the concept of conditional probability: $\displaystyle \Pr(E \, | \, C') = \frac{\Pr(E \cap C')}{\Pr(C')} = \frac{\Pr(E \cap C')}{1 - \Pr(C)}$.

    Now substitute the data (you have already unwittingly calculated the numerator: 0.47).


    Edit: NO you don't. Your answer is correct. Conditional probability is not relevant in this question.
    Last edited by mr fantastic; Jan 17th 2009 at 06:10 PM.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    You need the concept of conditional probability: $\displaystyle \Pr(E \, | \, C') = \frac{\Pr(E \cap C')}{\Pr(C')} = \frac{\Pr(E \cap C')}{1 - \Pr(C)}$.

    Now substitute the data (you have already unwittingly calculated the numerator: 0.47).
    I guess it's matter of interpretation, but it seems to me the question asks for $\displaystyle \Pr(E \cap C')$, in which case the answer given in the OP is correct.
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    Quote Originally Posted by awkward View Post
    I guess it's matter of interpretation, but it seems to me the question asks for $\displaystyle \Pr(E \cap C')$, in which case the answer given in the OP is correct.
    I agree completely with that.
    What would your answer be if the question were “What is the probability that a randomly selected person is a Jewish Democrat?”
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    Quote Originally Posted by Plato View Post
    I agree completely with that.
    What would your answer be if the question were “What is the probability that a randomly selected person is a Jewish Democrat?”
    $\displaystyle \Pr(E\cap C) = 0.03$.
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  6. #6
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    Quote Originally Posted by awkward View Post
    $\displaystyle \Pr(E\cap C) = 0.03$.
    Again I agree with you. I agreed with you the first time that the OP is correct.
    I don't see anything Bayesian about this.
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  7. #7
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    Quote Originally Posted by Plato View Post
    Again I agree with you. I agreed with you the first time that the OP is correct.
    I don't see anything Bayesian about this.
    Both you guys are right. My mistake.
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