Probability Part 2

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 =‌

17

36