Mixed Distribution Question Let
$$f\left(x,y\right)= \begin{cases} \frac{1}{3}, & x = 1, \, 0 \le y \le 1 \\ \frac{1}{6}, & x = 2, \, 0 \le y \le 2 \\ \frac{1}{9}, & x = 3, \, 0 \le y \le 3 \end{cases}$$
I want to find
$\mathbb{P}(X >1 ,Y < 1),\mathbb{P}(X > 2,Y < 2),\mathbb{P}(X > 2,Y < 3)$
Answer:-
X is having discrete distribution and Y is having Continuous distribution.So how to answer this question? Your guidance is needed.If anyone answer this question correctly, it is very good.
 A: The problem describes a random variable $(X,Y)$ that does not have a density function.  However, conditional on the value of $X$, $Y$ will have a density function (provided $X$ is one of $1,2,3$) as specified in the question.
We may visualize this as a density that is comprised of three parts: one part for $X=1$, another for $X=2$, and a third for $X=3$.  As always, the total area under the graph of the density function must be $1$.  Here is a picture showing all three parts with the areas beneath them filled in:

The nonzero area of each region is a rectangle.  The rectangles have areas $1/3\times 1=1/3$, $1/6\times 2=1/3$, and $1/9 \times 3=1/3$ for a total of $1$, as expected.
We may depict any event by showing which combinations of $X$ and $Y$ belong to it.  For example, the event abbreviated by "$X\gt 1, Y \lt 1$" consists of all points on any of the three y-axes for which $x\gt 1$ and $y \lt 1$.  Evidently that rules out anything where $x=1$ as well as any portion where $y\ge 1$.  To show this, the next figure blots out the entire plot for $x=1$ and it shades (in red) the portions of the other two plots for which $y \lt 1$.

The desired probability is the sum of all red areas that have not been shaded out.  There are two such red areas: one for $x=2$, a rectangle of base $1$ and height $1/6$; and one for $x=3$, a rectangle of base $1$ and height $1/9$.  The sum of their areas
$$1 \times 1/6 + 1 \times 1/9 = 5/18$$
is the probability of this event.
The other questions may be solved in the same way.

The correctness of this visualization is assured by the axioms of probability.  Any event $E\subset\mathbb{R}^2$ can be partitioned into four non-overlapping events $E_1 = \{(x,y)\,|\, x=1\}$, $E_2 = \{(x,y)\,|\, x=2\}$, $E_3 = \{(x,y)\,|\, x=3\}$, and $E_0=\mathbb{R}^2 \setminus (E_1 \cup E_2 \cup E_3)$ (which consists of everything else).  We worked out that $\Pr(E_1)=\Pr(E_2)=\Pr(E_3)=1/3$, implying (by the law of total probability) that $\Pr(E_0) = 1-(1/3+1/3+1/3)=0$.  That permits us to ignore $E_0$ altogether.  Consequently, the figures represent the conclusion that for all events $E$,
$$\Pr(E) = \Pr(E_1) + \Pr(E_2) + \Pr(E_3).$$
Moreover, for $i=1,2,3$, the definition of conditional probability tells us
$$\Pr(E_i) = \Pr(E_i\,|\, X=i)\,\Pr(X=i).$$
The function $f$ is none other than a specification of the densities for each of those conditional probabilities, each multiplied by $\Pr(X_i)$.  Consequently,
$$\Pr(E) = \sum_{i=1}^3 \int_{E\cap \{x=i\}} f(i,y) dy.$$
The figures depict those three integrals.
