# Thread: Quick conditional probability question

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

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

Thanks

2. Originally Posted by Danneedshelp
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 ....

3. Originally Posted by mr fantastic
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.

4. Well, if $Y \ge 3$, then certainly $Y \ge 2$. You need $P(3 \le Y \le 4)$ upstairs. The rest feels good.