The correct option is C 3281
Since b and c each can assume 9 values from 1 to 9.
So, total number of ways of choosing b and c is 9 × 9 = 81
Now, x2+bx+c>0 for all xϵR
⇒D<0
⇒b2−4ac<0⇒b2−4c<0⇒b2<4c
Alternatively: x2−bx+c>0
⇒(x+b2)2+4c−b24>0
⇒4c−b2>0
⇒b2<4c
Now, the following table shows the possible values of b and c for which b2<4c
cbTotal11,121,2231,2,3341,2,3351,2,3,4461,2,3,4471,2,3,4,5581,2,3,4,5591,2,3,4,5532
So, favourable number of cases = 32
Hence required probability = 3281