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 =‌