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Math Help - Quick conditional probability question

  1. #1
    Senior Member Danneedshelp's Avatar
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    Quick conditional probability question

    Given f(y)= 2y^{-3} , if y\geq\\ 1 and 0 otherwise, the probability

    P(2\leq\\Y\leq\\4|Y\geq\\3)=?

    I was thinking, P(2\leq\\Y\leq\\4|Y\geq\\3)=<br />
\frac{P(2\leq\\Y\leq\\4\cap\\Y\geq\\3)}{P(Y\geq\\3  )}<br />
=\frac{P(2\leq\\Y\leq\\4}{P(Y\geq\\3)}.

    I drew out the cdf and shaded the regions in question, but I am not sure how to treat the numerator.

    Thanks
    Last edited by Danneedshelp; November 4th 2009 at 04:49 PM.
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by Danneedshelp View Post
    Given f(y)= 2y^{-3} , if y\leq\\ 1 and 0 otherwise, the probability

    P(2\leq\\Y\leq\\4|Y\geq\\3)=?

    I was thinking, P(2\leq\\Y\leq\\4|Y\geq\\3)=<br />
\frac{P(2\leq\\Y\leq\\4\cap\\Y\geq\\3)}{P(Y\geq\\3  )}<br />
=\frac{P(2\leq\\Y\leq\\4}{P(Y\geq\\3)}.

    I drew out the cdf and shaded the regions in question, but I am not sure how to treat the numerator.

    Thanks
    Since the support is given as y \leq 1 the probability is clearly equal to zero. Re-check the question for a typo ....
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Since the support is given as y \leq 1 the probability is clearly equal to zero. Re-check the question for a typo ....
    My bad, I meant for the sign to be the other way.
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  4. #4
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    Well, if Y \ge 3, then certainly Y \ge 2. You need P(3 \le Y \le 4) upstairs. The rest feels good.
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