# Probability that the product of two random numbers chosen in (0,1) is less than some value

Specifically I want to find the probability that the product of two numbers chosen randomly on (0,1) interval is less than $\frac{2}{9}$. The numbers are independent to each other.

That's what I done:

I define two random variables $X_1$ and $X_2$ which have Uniform distribution on $(0,1)$ interval.

Then the probability mass functions are

$f_{X_1}(x) = P(X_1 = x) = \left \{ \begin{matrix} 1 & \mbox{if x$\epsilon$(0,1)} \\ 0 & \mbox{$otherwise$}\end{matrix}\right.$

$f_{X_2}(x) = P(X_2 = x) = \left \{ \begin{matrix} 1 & \mbox{if x$\epsilon$(0,1)} \\ 0 & \mbox{$otherwise$}\end{matrix}\right.$

Solving the inequality equation

$X_1X_2 \lt \frac{2}{9} \Rightarrow X_1 \lt \frac{2}{9X_2}$

Then integrating the mass probability functions

$P(X_1X_2 < \frac{2}{9}) = \displaystyle\int_{0}^{1} \displaystyle\int_{0}^{\frac{2}{9x_2}}\ f_{X_1}(x_1) f_{X_2}(x_2)dx_1dx_2 = \displaystyle\int_{0}^{1} \displaystyle\int_{0}^{\frac{2}{9x_2}}\ dx_1dx_2 = \frac{2}{9} \displaystyle\int_{0}^{1} \frac{dx_2}{x_2}$

Is easy to see that appears a Ln function. We can't solve Ln(0) so... What's wrong and how do I solve this? Thank you

When you got $X_1 \lt \frac 2{9X_2}$ you lost the fact that you also need $X_1 \lt 1$. You can't just use $\frac 2{9x_2}$ as the upper limit of the integral. In the range $0 \le x_2 \le \frac 29$ the upper limit is $1$. That removes the divergence.
• Thanks! Good answer. Can you tell me if it is good? I need to do the sum of two double integrals. The first with $X_2$ in $(0, \frac{2}{9})$ and $X_1$ in $(0,1)$. The second with $X_2$ in $(\frac{2}{9}, 1)$ and $X_1$ in $(0, \frac{2}{9x_2})$ ? – DN_Euler Nov 27 '17 at 21:46