Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.