<s>[INST] Prove that the probability that $x+y\leq 1,$ given that $x^2+y^2\geq \frac{1}{4}$ is $\frac{8-\pi}{16-\pi}$. Let $2$ positive real numbers $x,y$ satisfy $x\leq 1,y\leq 1$ are chosen at random.Prove that the probability that $x+y\leq 1,$ given that $x^2+y^2\geq \frac{1}{4}$ is $\frac{8-\pi}{16-\pi}$.
Since the $x+y\leq 1$ and $x^2+y^2\geq \frac{1}{4}$,so the required probability is area outside circle of radius $\frac{1}{2}$,centered at $(0,0)$ and inside the line segment whose $x-$intercept and $y-$intercept are 1 each in the first quadrant.so probability$=\frac{1}{2}-\frac{\pi}{16}=\frac{8-\pi}{16}$
But my answer does not match the given answer.Somewhere i must have faltered,Can you please guide me to the right method?
 [/INST] For the conditional probability you need to divide by the "given" area.
in this case $ Area(x^2+y^2 \ge \frac 14)=1-\frac{\pi}{16}$
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