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 :
Solutions · 1
Official solution
For x2+αx+β>0∀x∈R to hold, we should have α2−4β<0
If α=1,β can be 1, 2, 3, 4, 5, 6 i.e., 6 choices
If α=2,β can be 2, 3, 4, 5, 6 i.e., 5 choices
If α=3,β can be 3,4,5,6 i.e., 4 choices
If α=4,β can be 5 or 6 i.e., 2 choices
If α=6, No possible value for β i.e., 0 choices
Hence total favourable outcomes
\[
=6+5+4+2+0+0=17
\]
Total possible choices for α and β=6×6=36
Required probability =3617