Homework Question. Joint Probability Distribution. Here is the question.
The joint PDF of X and Y is given by $f_{XY}(x,y) = {\frac 14} e^{-|x|-|y|}$. Find $P(X \le 1 ,and,  Y \le 0)$
Solving the problem I first found the marginal probabilities of X and Y. Can you please explain what I should do next.
 A: To find the probability of some event $E$, you have to compute $$
  P(E) = \int_E \:dF
$$
where $F$ is your probability distribution. In your case, since you probability distribution has a density, you can also express that as $$
  P(E) = \int_E f \:d\lambda
$$
where $\lambda$ is the lebesgue measure on your probability space (which then is a subset of some $\mathbb{R}^d)$. The lebesgue measure is the ususal measure or length/area/volume/$\ldots$, so this is just a plain old integral. Additionally, in your case $E = (-\infty,1]\times(-\infty,0]$ is rectangular, which makes the integration especially simple. You get $$
  P(E) = \int_E f \:d\lambda = \int_{-\infty}^1\int_{-\infty}^0 f(x,y) \:dx\,dy \text{.}
$$
A: In general you need to use the joint PDF of $(X,Y)$ to calculate such probabilities. To that end, let $A=(-\infty,1]\times(-\infty,0]\subseteq\mathbb{R}^2$, then
$$
P(X\leq 1,\, Y\leq 0)=P((X,Y)\in A)=\iint_A f_{(X,Y)}(x,y)\,\mathrm dx\,\mathrm dy\tag{1}
$$
which you can easily calculate. 
Note that since you have found the marginal PDFs of $X$ and $Y$ you have probably found out that $X$ and $Y$ are independent since $f_{(X,Y)}(x,y)=f_X(x)\cdot f_Y(y)$ for all $x,y$. Hence
$$
P(X\leq 1,Y\leq 0)=P(X\leq 1)P(Y\leq 0)=\left(\int_{-\infty}^1f_X(x)\,\mathrm dx\right)\left(\int_{-\infty}^0f_Y(y)\,\mathrm dy\right)
$$
which is exactly what you get if you calculate $(1)$.
A: Solution using the mathStatica / Mathematica combo:
The joint pdf $f(x,y)$ is given as:
f = Exp[-Abs[x] - Abs[y]]/4;     domain[f] = {{x, -∞, ∞}, {y, -∞, ∞}};

You seek:  
Prob[x < 1 && y < 0, f]

returns: $\frac 12 - \frac{1}{4 e}$
[ Even if this is homework, I guess your lecturer will want to see some workings, but this will give you something to aim at :) ]
