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 $x^2 + bx+c>0$ for all x ∈ R is

$\frac{32}{81}$

Since b and c both can assume values from 1 to 9.

So, total numbers pf ways of choosing b and c is 9 × 9 = 81.

Now,

$x^2 + bx + c > 0 $ for all x ∈ R

⇒ Disc < 0 i.e $b^2 - 4c < 0.$

The following table shows the possible values of b and c for which $b^2 - 4c < 0.$

So, favourable number of ways = 32.

Hence, required probability =$ \frac{32}{81}.$