Quick conditional probability question

Danneedshelp · #1 $Old$ November 4th, 2009, 04:33 PM

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

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

Thanks

mr fantastic · #2 $Old$ November 4th, 2009, 06:41 PM

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

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 $y \leq 1$ the probability is clearly equal to zero. Re-check the question for a typo ....