Two numbers are independently and uniformly chosen from the interval (0,1). What is the probability that the sum of the numbers is less than 1 and the product of the numbers is less than 2/9? (Note that both conditions hold simultaneously.)
Given than $n_1\sim\operatorname{uniform}(0,1)$ and $n_2\sim\operatorname{uniform}(0,1)$
$$Z = n_1\cdot n_2 \sim \operatorname{uniform}(0,1)$$
If $X\sim \operatorname U(0,1)$ then
$$P(X < x) = \frac{x-a}{b-a}$$ $$P\left(z < \frac{2}{9}\right) = \frac{\frac{2}{9}-0}{1-0} = \frac{2}{9}$$
Is this correct or am i making an error somewhere?
In other words, what is the area of the region in the square $0\le x\le 1,\,0\le y\le 1$ satisfying $x+y<1,\,xy<\tfrac29$? It's the double integral$$\begin{align}\int_0^1dx\int_0^{\min\left\{1-x,\,\tfrac{2}{9x}\right\}}dy&=\int_0^1dx\min\left\{1-x,\,\tfrac{2}{9x}\right\}\\&=\int_0^{1/3}(1-x)dx+\int_{1/3}^{2/3}\tfrac{2}{9x}dx+\int_{2/3}^1(1-x)dx\\&=\tfrac13+\tfrac29\ln2\\&\approx0.4874.\end{align}$$You'll want to double-check my arithmetic, but Monte Carlo agrees.