If the numbers appearing on two throws of a fair six-faced die are and , then the probabil…

Question

If the numbers appearing on two throws of a fair six-faced die are $\alpha$ and $\beta$, then the probability that $x^2+\alpha x+\beta>0$ for all $x\in\mathbb R$, is:

Accepted Answer

We need $\alpha^2-4\beta<0$, i.e. $\beta>\dfrac{\alpha^2}4$.

- $\alpha=1$: $\beta>0.25$, all $\beta=1,\dots,6$ work --- 6 choices.
- $\alpha=2$: $\beta>1$, $\beta=2,\dots,6$ --- 5 choices.
- $\alpha=3$: $\beta>2.25$, $\beta=3,\dots,6$ --- 4 choices.
- $\alpha=4$: $\beta>4$, $\beta=5,6$ --- 2 choices.
- $\alpha=5$: $\beta>6.25$ --- 0 choices.
- $\alpha=6$: $\beta>9$ --- 0 choices.

Total favourable $=6+5+4+2=17$; total outcomes $=36$.

$$P=\frac{17}{36}.$$

Solutions · 1