Two numbers b and c are chosen at random (with replacement from the numbers 1,2,3,4,5,6,7,8 and 9). The probability that x2+bx+c>0for all x∈R is

Here,x2+bx+c>0∀x∈R∴    D<0⇒    b2<4cvalue of b possible  values of c1         1<4c ⇒c>14 ⇒{1,2,3,4,5,6,7,8,9}2     4<4c ⇒c>1   ⇒{2,3,4,5,6,7,8,9}3     9<4c ⇒c>94 ⇒{3,4,5,6,7,8,9}4        16<4c ⇒c>4   ⇒{5,6,7,8,9}5   25<4c ⇒c>625  ⇒{7,8,9}6   36<4c ⇒c>9    Impossible  7       impossible 8       impossible 9      impossibleNumber of favorable ca8e8 =9 + 8 + 7+6+3=32Total ways=9x9=81Required probability = 32/81