Choose randomly two points $a,b$ over $[0,1]$ find the probability such that $a,b,\frac{1}{2}$ are traingle sides.

Choose randomly two points $a,b$ over $[0,1]$ find the probability such that $a,b,\frac{1}{2}$ are traingle sides.

My solution:

$$\frac{1}{2}\leq a+b\implies \frac{1}{2}-a\leq b$$

$$\int_0^\frac{1}{2} da \int_{\frac{1}{2}-a}^1 db=\int_0^\frac{1}{2} \frac{1}{2}+a da=\frac{3}{8}$$

I would be grateful for feedback.

Thanks !

Three inequalities must be met:

$$1/2\leq a+b\\ a\leq 1/2+b\\ b\leq 1/2+a.$$

The area of the shaded region is your answer.

If $a>\frac{1}{2}$, then $a+b>\frac{1}{2}$ is met. The condition is $\frac{1}{2}+b>a$. Symmetrically for the case $b>\frac{1}{2}$. If $a,b\leqslant \frac{1}{2}$, then we must have $\frac{1}{2}<a+b$.

As an integral, the probability you are looking for is $$\int _0^{1/2} \int _{-a+\frac{1}{2}} ^{a+\frac{1}{2}}\mathrm{d}b\mathrm{d}a + \int _{1/2}^1 \int _{a-\frac{1}{2}}^1 \mathrm{d}b\mathrm{d}a $$

The integral is really unnecessary, though. The three conditions cut out triangles of area $1/8$. Hence the probability you are looking for is $5/8$.

The integral you are computing neglects the region $[0.5,1]\times [0,1]$ (of $ab$ plane) entirely.