Joint distribution probabilities I have a question that is similar to the following(made up here):
The construction of a tower of cards is done is two stages, procrastination and the actual building. The time in minutes needed to complete each stage are independent discrete random variables, X and Y, with probability functions;
$f_X(x) = \frac{7}{10}$ if $ x = 2, \frac{3}{10}$ if $x = 3$, and $0$ otherwise.
$f_Y(x) = \frac{2}{5}$ if $x = 3, \frac{2}{5}$ if $x = 4, \frac{1}{5}$ if $x = 5$ $0$ otherwise
What is the probability the task took more than six minutes to complete?
Now I haven't dealt with joint distribution problems before. But I can see 3 scenarios that yield more than 6 minutes of time elapsed. $f_X(2) $ then $f_Y(5)$  or $f_X(3)$ then $f_Y(4)$ or $f_X(3)$ then $f_Y(5)$
Can I simply then take $(\frac{7}{10}*\frac{1}{5} + \frac{3}{10}*\frac{2}{5} + \frac{3}{10}*\frac{1}{5})$? This seems right at $.32$. Furthermore if I do the other three scenarios I get a total probability of one, which increases my confidence with it once again. Any hints or confirmation? Thank you for your time

This explicit computation may help:
\begin{align}
   P[X+Y > 6]
&= P[X +Y > 6, X=2] + P[X +Y > 6, X=3] \\
&= P[Y = 5, X = 2] + P[ Y \geq 4, X=3 ] \\
&= P[Y = 5]P[ X = 2] + P[ Y \geq 4]P[ X=3 ] \\
&= \left(\frac{1}{5} \right)\left(\frac{7}{10} \right) + 
   \left(\frac{3}{5} \right)\left(\frac{3}{10} \right) \\
&= \frac{8}{25}.
\end{align}
So, you did in fact make the right computations.
