Question

a rectangle is drawn with its adjacent sides chosen from independent and identically distributed uniform distributions over the interval (0,1) . What is the probability that the area of the rectangle is more than 0.5

Answer 1

Let X and Y be random variables representing the side lengths of the rectangle. Then the area of the rectangle is given by the random variable XY.

We have

$\displaystyle P(XY > 0.5) = \int_{0.5}^1 \int_{0.5/y}^1 dx \, dy \\\\ = \int_{0.5}^1 \left(1-\frac1{2y}\right) \, dy \\\\ = \left(1 - \frac12 \ln(1)\right) - \left(\frac12 - \frac12 \ln\left(\frac12\right)\right) \\\\ = \boxed{\frac{1-\ln(2)}2}$

or approximately 0.1534.