a. For 0 ≤ y ≤ 3, find the conditional density $f(x | y)$ of $X$, given $Y = y$.
Using area of $x$ and $y$ region $\cdot$ density $= 1$
I got area of triangle $\cdot$ density $= 1$ AKA $\cfrac{1}{2}\cdot3 \cdot 3 \cdot density = 1$ AKA $4.5 \cdot density = 1 $
so density $= f(x,y) = \cfrac{2}{9}$
Then $f(y) = \int_0^{3-y}\cfrac{2}{9}dx = \cfrac{2}{9}(3-y)$ for $0\leq y \leq 3$
so $f(x|y) = \cfrac{\cfrac{2}{9}}{\cfrac{2}{9}(3-y)} = \cfrac{1}{(3-y)}$ for $0\leq x, 0\leq y, x+y\leq3 $
b. Find the conditional probability that $X ≤ 1$, given $Y = 1$.
$P(X\leq1|Y=1) = \int_0^1\cfrac{1}{3-1}dx=\cfrac{1}{2}$
c. Find the conditional probability that $X ≤ 1$, given $Y ≤ 1$.
$P(X\leq1|Y\leq1) = ???$
I would appreciate it if someone could tell me if I have the bounds and calculations right for parts a) and b) and if someone could give me a hint on how to solve c) I don't know what to do if the given part $Y\leq1$ is an inequality and not $=$ to something, so any tips or help would be appreciated.
Thank You!
It's all correct. By a simple drawing yo get that
$$\mathbb{P}(X\leq 1|Y\leq 1)=\frac{\mathbb{P}(X\leq 1;Y\leq 1)}{\mathbb{P}(Y\leq 1)}=\frac{1}{2+\frac{1}{2}}=\frac{2}{5}$$
Purple area divided by the area of the red trapezoid
Observation:
The conditional density found in (a) is
$$f(x|y)=\frac{1}{3-y}$$
where $y \in[0;3)$ and $x \in [0;3-y]$
This means that $X$ conditioned to a value of a given $y$ is uniform in $[0;3-y]$ thus any probability can be immediately calcualted without integrals