Answer in Statistics and Probability for Jac #248496

Given the joint density function

f(x,y) = {6-x-y / 8 ; 0<x<2, 2<y<4,

{0, elsewhere,

find P(1<Y<3 | X=1).

Solution.

Such as a range of Y is (2,4), then

“P(1<Y<3|X=1)=P(1<Y<2|X=1) +P(2<Y<3|X=1)n= 0+ P(2<Y<3|X=1) newlinentext{}”

It is known that the conditional probability can be calculated by the formula

“f(y|x)=frac{f(x,y)}{g(x)}.”

Find “g(x)=int_2^4frac{6-x-y}{8}dy=frac{3-x}{4},” where 0<x<2.

Find f(y|x), where x=1:“P(2<Y<3|X=1)=int_2^3(frac{6-x-y}{8}:frac{3-x}{4})dy|_{x=1}=(frac{5}{4}y-frac{y^2}{8})|_2^3=frac{15}{4}-frac{9}{8}-frac{10}{4}+frac{4}{8}=frac{5}{4}-frac{5}{8}=frac{5}{8}.” Answer. “frac{5}{8}.”