If the numbers appeared on the two throws of a fair six faced die are α and β , then the probability that x 2 + α x + β 0 , for all x ∈ R , is

Question

If the numbers appeared on the two throws of a fair six faced die are α and β , then the probability that x 2 + α x + β > 0 , for all x ∈ R , is

TestHub · Accepted Answer

Total sample space of two dice will be equal to $6 \times 6 = 36$
Now for $x^{2} + α x + β > 0$ its discriminant $D$ must be less than zero,
So $D = b^{2} - 4 a c < 0$ or $α^{2} - 4 β < 0$ $\Rightarrow α^{2} < 4 β$
Now, if $α = 1, β$ can take values $1, 2, 3, 4, 5, 6$
If $α = 2, β$ can take values $2, 3, 4, 5, 6$
If $α = 3, β$ can take values $3, 4, 5, 6$
If $α = 4, β$ can take values $5, 6$
If $α = 5$ and $6$ , no value of $β$ possible
So number of favourable ways $= 17$
Required probability $= \frac{favourable ways}{total sample space}$ $= \frac{17}{36}$

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