1. We need the quadratic $f(x)=x^2+\alpha x+\beta$ to satisfy $f(x)>0 \quad \text{for all } x\in \mathbb{R}.$

2. For a quadratic $ax^2+bx+c$ to be positive for all real $x$, we need:

• $a>0$, and
• discriminant $<0$.

Here, $a=1>0$, so we only need $\alpha^2-4\beta<0.$ That is, $\alpha^2<4\beta.$

3. Since $\alpha$ and $\beta$ are outcomes of two throws of a fair die, $\alpha,\beta\in\{1,2,3,4,5,6\}.$ Total number of ordered pairs is $6\times 6=36.$

4. Count pairs $(\alpha,\beta)$ satisfying $\alpha^2<4\beta.$ We check each possible value of $\alpha$:

• If $\alpha=1$: $1<4\beta$ true for all $\beta=1,2,3,4,5,6$. Count $=6$.

• If $\alpha=2$: $4<4\beta \iff \beta>1$ so $\beta=2,3,4,5,6$. Count $=5$.

• If $\alpha=3$: $9<4\beta \iff \beta>\frac{9}{4}$ so $\beta=3,4,5,6$. Count $=4$.

• If $\alpha=4$: $16<4\beta \iff \beta>4$ so $\beta=5,6$. Count $=2$.

• If $\alpha=5$: $25<4\beta \iff \beta>\frac{25}{4}=6.25$ impossible. Count $=0$.

• If $\alpha=6$: $36<4\beta \iff \beta>9$ impossible. Count $=0$.

5. Total favorable cases: $6+5+4+2+0+0=17.$

6. Therefore, the required probability is $\frac{17}{36}.$

7. Comparing with the stored correct answer: stored answer is A, i.e. $\frac{17}{36}$, which matches.