What is the probability that the sum of two randomly chosen…
What is the probability that the sum of two randomly chosen positive proper fractions is greater than $1$ and their product is greater than $\frac29$?
Explanation of Method
The two fractions $x,y$ are independent and uniform on $(0,1)$, so $(x,y)$ is a uniform point in the unit square and the probability equals an area. The conditions describe the region above the line $x+y=1$ and above the hyperbola $xy=\frac29$. The line and the hyperbola cross at $x=\frac13$ and $x=\frac23$; between these points the line is the higher boundary, outside them the hyperbola is. We integrate the height of the region in three pieces.
Step 1: The model
A "randomly chosen positive proper fraction" is a number uniformly distributed in $(0,1)$. For two independent fractions $x$ and $y$ the point $(x,y)$ is uniform in the unit square (area $1$), so
\[ P = \text{area of } \Big\{(x,y): x+y\gt1,\ xy\gt\tfrac29\Big\}. \]
Step 2: Where the boundaries meet
On the line $y=1-x$ and the hyperbola $y=\frac{2}{9x}$ at the same point: $x(1-x)=\frac29$, i.e. $9x^2-9x+2=0$, so
\[ x = \frac13 \quad\text{or}\quad x=\frac23. \]
For $\frac13\lt x\lt\frac23$ the line is higher (at $x=\frac12$: $1-x=0.5\gt\frac49$); otherwise the hyperbola is higher. The hyperbola reaches $y=1$ at $x=\frac29$, so for $x\le\frac29$ no point satisfies $xy\gt\frac29$.
- both conditions: above the line and above the hyperbola
Step 3: The area in three pieces
For each $x$ the favourable $y$ run from the higher boundary up to $1$:
\[ \begin{aligned} P &= \int_{2/9}^{1/3}\Big(1-\frac{2}{9x}\Big)dx + \int_{1/3}^{2/3}\big(1-(1-x)\big)\,dx + \int_{2/3}^{1}\Big(1-\frac{2}{9x}\Big)dx. \end{aligned} \]
- First: $\Big[x-\tfrac29\ln x\Big]_{2/9}^{1/3} = \tfrac19 - \tfrac29\ln\tfrac32$.
- Second: $\Big[\tfrac{x^2}{2}\Big]_{1/3}^{2/3} = \tfrac12\left(\tfrac49-\tfrac19\right) = \tfrac16$.
- Third: $\Big[x-\tfrac29\ln x\Big]_{2/3}^{1} = \tfrac13 - \tfrac29\ln\tfrac32$.
Step 4: The answer
\[ \boxed{P = \frac19+\frac16+\frac13 - \frac49\ln\frac32 = \frac{11}{18}-\frac49\ln\frac32 \approx 0.431.} \]
A simulation with two million random pairs gave $0.4310$. ✓
Step 5: The textbook version "not greater"
In many problem collections this exercise asks that the sum be not greater than $1$ and the product not greater than $\frac29$. Then the region is the triangle $x+y\le1$ (area $\frac12$) minus the part above the hyperbola, $\int_{1/3}^{2/3}\big(1-x-\frac{2}{9x}\big)dx = \frac16-\frac29\ln2$:
\[ P = \frac12-\frac16+\frac29\ln2 = \frac13+\frac29\ln2\approx0.487. \]
Result
| Version | Probability |
|---|---|
| sum $\gt1$ and product $\gt\tfrac29$ (as stated) | $\frac{11}{18}-\frac49\ln\frac32\approx0.431$ |
| sum $\le1$ and product $\le\tfrac29$ (textbook version) | $\frac13+\frac29\ln2\approx0.487$ |