What is the probability for $x$ to be positive only?

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 $dfrac{frac{13}{3}}{1+frac{13}{3}}=boxed{frac{13}{16}}$.

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".