If the numbers appeared on the two throws of a fair six faced die are $\alpha$ and $\beta$, then the probability that $x^{2}+\alpha x+\beta>0$, for all $x \in \mathbf{R}$, is :
Solution
For $x^{2}+\alpha x+\beta>0 \forall x \in R$ to hold, we should have $\alpha^{2}-4 \beta<0$
<br/><br/>
If $\alpha=1, \beta$ can be 1, 2, 3, 4, 5, 6 i.e., 6 choices
<br/><br/>
If $\alpha=2, \beta$ can be 2, 3, 4, 5, 6 i.e., 5 choices
<br/><br/>
If $\alpha=3, \beta$ can be $3,4,5,6$ i.e., 4 choices
<br/><br/>
If $\alpha=4, \beta$ can be 5 or 6 i.e., 2 choices
<br/><br/>
If $\alpha=6$, No possible value for $\beta$ i.e., 0 choices<br/><br/>
Hence total favourable outcomes
<br/><br/>
$$
\begin{aligned}
&=6+5+4+2+0+0 \\\\
&=17
\end{aligned}
$$
<br/><br/>
Total possible choices for $\alpha$ and $\beta=6 \times 6=36$
<br/><br/>
Required probability $=\frac{17}{36}$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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