Probability that the sum of two discrete random variables is at most 2 Given 2 Random variables and the following sample spaces: $$X = \{0,1,2,3\}$$ and $$Y=\{0,1,2,3\}$$

How could I find the probability that: $$P(X+Y \le 2)$$

My attempt is:$$P(X+Y \le2)=P(X=0)+P(Y=1)*P(X=0)+P(Y=2)*P(X=1)+P(Y=0)*P(X=1)+P(Y=1)*P(X=2)+P(Y=0)$$

EDIT: Wow I came back and had all these responses.... I apologize. But $X$ and $Y$ are the random variables that represent the amount of times a head was flipped from 3 coin tosses. Coin A has 50/50 chance of getting heads or tails. Coin B has 1/4 chance of getting heads.
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  "Assuming $\ds{X}$ and $\ds{Y}$ are uniformly distributed over their ranges" as $\texttt{@Fimpellizieri}$ said:


The sum is over $\ds{m,n\ \in \braces{0,1,2,3}}$ with the constrain
$\ds{m + n = s}$ for a given $\ds{s \in \braces{0,1,2}}$. The final result is found by adding contributions over $\ds{s \in \braces{0,1,2}}$:
\begin{align}
&amp;\mrm{P}\pars{X + Y \leq 2} = \sum_{s = 0}^{2}\braces{%
{1 \over 4}\sum_{m = 0}^{3}
{1 \over 4}\sum_{n = 0}^{3}\bracks{m + n = s}} =
{1 \over 16}\sum_{s = 0}^{2}\sum_{m = 0}^{3}\sum_{n = 0}^{3}\bracks{n = s - m}
\\[5mm] = &amp;\
{1 \over 16}\sum_{s = 0}^{2}\sum_{m = 0}^{3}\bracks{0 \leq s - m \leq 3} =
{1 \over 16}\sum_{s = 0}^{2}\sum_{m = 0}^{s}1 =
{1 \over 16}\sum_{s = 0}^{2}\pars{s + 1} = {6 \over 16} =
\bbox[5px,border:0.1em groove navy]{3 \over 8}
\end{align}
$P(X+Y\leq2)=$

$1-P(X+Y\geq3)=$

$1-\frac{4}{7}=$

$\frac{3}{7}$.

there are four winning cases  : $3,4,5,6$

between the following $7$ possible sums:

$0,1,2,3,4,5,6$.