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?
On
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".
Given that $x+y+z=5$ and $xy+yz+zx=3$, we find that $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=19$.
Now we find the range of $x$. For this, we write $y+z=5-x$ and $y^2+z^2=19-x^2$.
Now by a basic application of A.M-G.M Inequality, or Titu's Lemma, we can write $y^2+z^2 \geq \dfrac{(y+z)^2}{2} \implies 19-x^2 \geq \frac{(5-x)^2}{2} \iff (x+1)\left(x-\frac{13}{3}\right)\leq 0$.
Therefore $x \in \left[-1,\frac{13}{3}\right]$, thus the probability that $x>0$ is $\dfrac{\frac{13}{3}}{1+\frac{13}{3}}=\boxed{\frac{13}{16}}$.