Joint PMF of two independent random variable Let $X$ and $Y$ be independent random variables. Each of them has a
  geometric distribution with $E[X] = 2$ and $E[Y] = 3$.
  
  (a) Find the joint p.m.f. of $X$ and $Y$.
  
  (b) Compute the probability that $X + Y \le 4$.
  
  (c) Deﬁne two new random variables by $W = \min{ \left( X, Y \right ) }$ and $Z = \max{ \left( X, Y \right ) }$.
  Find the joint p.m.f. of $W$ and $Z$.


Ans:

$X$ and $Y$ are independent random variables. Each of them has a geometric distribution with $E[X] = 2$ and $E[Y] = 3$. 

So joint p.m.f. of $X$ and $Y$,

$E[XY] = E[X] E[Y] = 2 * 3 = 6$

Is this correct?

And for $(b)$ and $(c)$ can anyone suggest how to proceed??

for (c), s min(X; Y ) is a geometric random variable with parameter 1 - (1 - p)^2 = 2p - p^2
Hint
The pmf for geometric distribution (the probability that it will take $k$ trials to get the first success) is given by
$$\Pr(X = k) = (1-p)^{k-1}\,p\,$$
where $p$  is the success probability in each trial.
It is also known that the mean of $X$ is $\mathrm{E}(X) = \frac{1}{p}$.
Can you get it from here?
$P=1/E[X]=0.5$

$P=1/E[y]=0.333$

$PX(x)=(1-0.5)^{k-1}0.5$

$PY(y)=(1-0.33)^{k-1}0.333$

The Joint Probability:

$E[XY]=E[X]*E[y]=2\cdot3=6$

$p=1/E[XY]=0.167$
$PX,Y(x,y)=(1-0.167)^{k-1}0.167 ~~~  K=1,2,3.......$