Problem source

The random variable $U$ takes the values $+1$, $0$ and $-1\,$, each with probability $\frac13\,$. The random variable $V$ takes the values $+1$ and $-1$ as follows:
\begin{center}
\begin{tabular}{ll}
if $U=1\,$,&then $\P(V=1)= \frac13$ and $\P(V=-1)=\frac23\,$;\\[2mm]
if $U=0\,$,&then $\P(V=1)= \frac12$ and $\P(V=-1)=\frac12\,$;\\[2mm]
if $U=-1\,$,&then $\P(V=1)= \frac23$ and $\P(V=-1)=\frac13\,$.
\end{tabular}
\end{center}
\begin{questionparts}
\item Show that  the probability that both roots of the equation $x^2+Ux+V=0$ are real is $\frac12\;$.
\item Find the expected value of the larger root of the equation  $x^2+Ux+V=0\,$, given that both roots  are real.
\item Find the probability that the roots of the equation
$$x^3+(U-2V)x^2+(1-2UV)x + U=0$$ are all positive.
\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$ \begin{align*}
&& \mathbb{P}(\text{both roots real}) &= \mathbb{P}(\Delta \geq 0) \\
&&&= \mathbb{P}(U^2 \geq 4V) \\
&&&= \mathbb{P}(V = -1) \\
&&&= \tfrac13 ( \tfrac23 + \tfrac12 + \tfrac13) \\
&&&= \tfrac13 \cdot \tfrac 32  = \frac12
\end{align*}
\item Our equations will be:

$x^2+x-1 = 0$ with larger root $\frac{-1 + \sqrt{5}}{2}$
$x^2-1 = 0$ with larger root $1$
$x^2-x-1 = 0$ with larger root $\frac{1 + \sqrt5}{2}$

and the expected value is \begin{align*}
&& \E[\text{larger root}|\text{both real}] &= \frac23 \left ( \frac23 
\cdot \frac{-1+\sqrt5}{2}  + \frac12 \cdot 1 + \frac13 \cdot \frac{1+\sqrt5}{2} \right) \\
&&&= \frac23 \left ( \frac{2+3\sqrt5}{6} \right) \\
&&&= \frac{2+3\sqrt5}{9}
\end{align*}

\item Suppose we have $x^3+(U-2V)x^2+(1-2UV)x + U = 0$, then for all roots to be positive, we need $U < 0 \Rightarrow U = -1$ (otherwise there is a root at or below zero).

Therefore our two possible cubics are:

$x^3 -3x^2+3x-1 = (x-1)^3$ (all roots positive)
$x^3+x^2-x-1 = (x-1)(x+1)^2$ (not all roots positive!)

Therefore the probability is $\frac13 \cdot \frac23 = \frac29$
\end{questionparts}