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.
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 $$
  P(E) = \int_E f \:d\lambda = \int_{-\infty}^1\int_{-\infty}^0 f(x,y) \:dx\,dy \text{.}
$$
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 &lt; 1 &amp;&amp; y &lt; 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 :) ]