Find: $P(X + Y > 1/2)$
I know the answer is $63/64$. But how do you find that answer? I'm really struggling with what the bounds of integration would be for a question like this. Can someone show me? I tried $\int_{1/2}^1\int_0^{1-x}180x^2y^2dydx$ but this gives me 11/32 which isn't the correct answer?
What do I need to do to derive $63/64$?
Thank you for any help
Watch lectures on calculating area, volume using integrals there are tons of them on youtube. Bounds are easy to figure out if you draw graphs. I will show you how it is so easy to solve these types of questions using graphs.
$P(X+Y\ge\frac{1}{2})=1-P(X+Y<\frac{1}{2})=\int\int f(x,y)dy dx$
The region inside the red outlined triangle is your required area where you have to figure out bounds.
Draw a vertical strip like shown below and go bottom to up(You can also start drawing a horizontal strip and go left to right).
From the bottom, we see $y$ touches the $x$-axis so $y=0$ and strip goes up and touches $y=\frac{1}{2}-x$ and drag thiis vertical strip left to right we see $x=0$ initially and goes till $x=\frac{1}{2}$
so we have just found our bounds.
$1-\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}-x} f(x,y)dy dx$ is required probability.