I have the joint density function $f(y_1,y_2) = 3y_1, 0 \leq y_2 \leq y_1 \leq 1.$ And $0$ elsewhere.
The conditional density for $f(y_1|y_2)$ is defined over $y_2 \leq y_1 \leq 1$, provided that $y_2 \leq 0 $ .
The question is about Y1, the proportion of items stocked and Y2, proportions of items sold.
I have to find the probability that more than 1/2 is sold given that 3/4 is stocked.
My atempt is to define $f(y_1|y_2)$:
$f(y_1|y_2)= \frac{f(y_1,y_2}{f_2(y_2)} = \frac{1/2}{3/4 y_2} = \frac{2}{3y_2}$
Then calculate $P(y_1 > 1/2 \; | \; 3/4) $ With the integral $\int_{\frac{1}{2}}^{1} \frac{2}{3y_2} dy = \frac{1}{6}$.
But the result should be 1/3 according to my textbook, so I'm doing something wrong somewhere. Hope someone can help me with figuring out where I'm going wrong.