# 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

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The expression is right. – André Nicolas Apr 17 '14 at 5:01
@AndréNicolas Good! Thank you for confirmation! – Display Name Apr 17 '14 at 5:02

## 1 Answer

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.

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