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