# Thread: Quick conditional probability question

1. ## 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

2. Originally Posted by Danneedshelp
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 ....

3. Originally Posted by mr fantastic
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.

4. 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.