Let X and Y be continuous random variables with joint density function
$f(x,y)=24xy$ on the domain $x,y>0, 0<x+y<1$ and is $0$ otherwise
What is the conditional probability?
$$P(X>1/2 | X+Y<3/4)$$
Any help would be much appreciated. Thank you.
I can find $P(X>1/2)$ and $P(X+Y<3/4)$ but I don’t know how find the conditional probability.
Answer is given as $\frac{1}{9}$. I have no idea how to get that.
Best Answer
Hint:
The conditional probability would be calculated as follows (from the definition):
$$P(X>1/2 | X+Y < 3/4) = \frac{P(X>1/2 \cap X + Y <3/4)}{P(X+Y < 3/4)}$$
You’ve already solved for the denominator, so you need to formulate the correct double integral for the numerator :
$$\int_{1/2}^{3/4} \int_0^{3/4-x} 24xy\;dydx $$
Can you take from here?