# Probability with real number

There are two sets of real numbers: $S_1 = [0, N_1]$, $S_2 = [0, N_2]$, where $N_1$ and $N_2$ are positive integers. From $S_1$ a value $x_1$ and from $S_2$ a value $x_2$ is chosen.

What will be the probability that $x_1 + x_2 < y$, where $y$ is an integer number?

-
"From $S_1$ a value $x_1$ [..] is chosen." - Do you mean $x_1$ is drawn from $S_1$ uniformly at random? And what have you tried to solve the question? –  TMM Oct 27 '12 at 17:49
@TMM yes. I tried solving it but I can solve it if S1 is set of integer not real number. I don't know how to solve when problem contains set of real number. –  user1779660 Oct 27 '12 at 17:51
How much, if anything, do you know about continuous random variables? –  ronno Oct 27 '12 at 18:53

Compute the area $A(y)$ of the polygon $Q(y)$ defined by the equations $0\leqslant x_1\leqslant N_1$, $0\leqslant x_2\leqslant N_2$, $x_1+x_2\leqslant y$. The answer is $A(y)/(N_1N_2)$.

Note that $Q(y)$ is:

• a triangle for $0\leqslant y\leqslant\min\{N_1,N_2\}$, with area $A(y)=\frac12y^2$,
• a quadrilateral if $\min\{N_1,N_2\}\lt y\leqslant\max\{N_1,N_2\}$,
• a pentagon if $\max\{N_1,N_2\}\lt y\lt N_1+N_2$,
• the whole rectangle $[0,N_1]\times[0,N_2]$ if $y\geqslant N_1+N_2$, with area $A(y)=N_1N_2$.
-
@amitchhajer Draw a picture. These are not random polygons, for instance the pentagon is always the full rectangle minus a triangle with vertices $(N_1,N_2)$, $(N_1-t,N_2)$ and $(N_1,N_2-t)$. –  Did Oct 28 '12 at 7:42