<s>[INST] What is the probability for $x$ to be positive only? If $x,y,z\in \mathbb R$ and $x+y+z=5,\, xy+yz+zx=3$, what is the probability that $x>0$ ? $$(a)\quad\frac3{16}\qquad (b)\quad\frac5{16}\qquad (c)\quad\frac{13}{16}\qquad (d)\quad \frac{15}{16}$$
I've tried forming a cubic equation and then trying to analyze its roots.
How to approach these type of questions?
 [/INST] The solutions of your two equations form a circle in $xyz$ space, the intersection of the sphere $x^2 + y^2 + z^2 = 19$ with the plane $x+y+z=5$, which can be parametrized as
$$ \eqalign{x &= \frac{5}{3} - \frac{8}{3} \cos(t)\cr
            y &= \frac{5}{3} + \frac{4}{\sqrt{3}} \sin(t) + \frac{4}{3} \cos(t)\cr
            z &= \frac{5}{3} - \frac{4}{\sqrt{3}} \sin(t) + \frac{4}{3} \cos(t)\cr} $$
It's not at all clear how to interpret "probability" in this context,  but there is no $t$ for which $x > 0$ while $y \le 0$ and $z \le 0$.  So if $x > 0$ only means $x > 0$ while $y \le 0$ and $z \le 0$, the answer must be $0$. On the other hand, if you asked for the probability that $x > 0$, one possible interpretation would be the probability that $x>0$ if you took a random $t$ uniformly from $[-\pi,\pi]$.
Then you'd get $x > 0$ iff $\cos(t) < 5/8$, i.e. with probability $1 - \frac{\arccos(5/8)}{\pi} \approx 0.7149010415$.  This is an irrational number. So again "none of the above".
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