Find $P(Y\ge 1+\frac{4}{3}X)$ when pdf is given

Question

Let X,Y be two continuous random variable and their joint pdf is given by: $$f(x,y)=\frac{2}{625}(20-x-2y); 0 \le x \le 5;0 \le y \le 5$$ $$f(x,y)=0; \quad otherwise$$

Find $P(Y\ge 1+\frac{4}{3}X)$.

I cannot figure out how to tackle this problem. I know that I have to do double integration but not sure how to get the bounds as I thought $1+\frac{4}{3}X$ needs to be incorporated into the bounds as well. Any insight is much appreciated.

Answer 1

0 < y < 5
0 < x < 5

now you need to find P(Y > 1 + 4x/3)
draw the region [0,5] * [0*5] and the line y = 1 + 4x/3
you have to take the upper region bounded by line y = 5 and y = 1 +4x/3 and x= 0

hence limits are
y varies from 1 + 4x/3 to 5
and x varies from 0 to 3

https://www.symbolab.com/solver/multiple-integrals-calculator/%5Cint_%7B0%7D%5E%7B3%7D%5Cint_%7B1%2B%5Cfrac%7B4x%7D%7B3%7D%7D%5E%7B5%7D%20%20%5Cfrac%7B2%7D%7B625%7D%5Cleft(20-x-2y%20%5Cright)dy%20dx

Integrating we get required probability as 28/125 = 0.224