Random Variables X and Y follow a joint distribution:

f(x,y)=(6-x-y)/8 if 0<x<2, 2<y<4
and 0, else
Find P(1<Y<3|X=1)

I first got the marginal distribution for x and got (3-x)/4 (Integrated from 2 to 4)and then got the conditional dist and got (5-y)/4 and then integrated that from 1 to 3 and got 1.5, which means it's wrong. Can someone tell me what. I'm am I doing wrong?

To find P(1<Y<3|X=1), you need to use the conditional probability formula:

P(Y=y|X=1) = f(X=1, Y=y) / f(X=1)

To calculate the value of f(X=1, Y=y), we need to evaluate the joint distribution function at X=1 and Y=y:

f(X=1, Y=y) = (6-1-y)/8 = (5-y)/8

Now, to find the marginal distribution of X, you correctly calculated it as (3-X)/4 by integrating the joint distribution over the range of Y. However, when evaluating the marginal distribution at X=1 (not integrating it), you get:

f(X=1) = (3-1)/4 = 1/2

Finally, substitute these values into the conditional probability formula:

P(Y=1<Y<3|X=1) = f(X=1, Y=1<Y<3) / f(X=1)

= ∫[(5-y)/8]dy from 1 to 3 / (1/2)

= [(5/8)(y) - (1/8)(y^2)] from 1 to 3 / (1/2)

= [(5/8)(3) - (1/8)(9)] / (1/2)

= (15/8 - 9/8) / (1/2)

= 6/8 / (1/2)

= 6/8 * 2/1

= 0.75

Therefore, P(1<Y<3|X=1) = 0.75.