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Thread: Quick conditional probability question

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

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

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

    I was thinking, $\displaystyle P(2\leq\\Y\leq\\4|Y\geq\\3)=
    \frac{P(2\leq\\Y\leq\\4\cap\\Y\geq\\3)}{P(Y\geq\\3 )}
    =\frac{P(2\leq\\Y\leq\\4}{P(Y\geq\\3)}$.

    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; Nov 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 $\displaystyle f(y)= 2y^{-3}$ , if $\displaystyle y\leq\\ 1$ and 0 otherwise, the probability

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

    I was thinking, $\displaystyle P(2\leq\\Y\leq\\4|Y\geq\\3)=
    \frac{P(2\leq\\Y\leq\\4\cap\\Y\geq\\3)}{P(Y\geq\\3 )}
    =\frac{P(2\leq\\Y\leq\\4}{P(Y\geq\\3)}$.

    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 $\displaystyle y \leq 1$ 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 $\displaystyle y \leq 1$ 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 $\displaystyle Y \ge 3$, then certainly $\displaystyle Y \ge 2$. You need $\displaystyle P(3 \le Y \le 4)$ upstairs. The rest feels good.
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