Two numbers b and c are chosen at random in succession without replacement from the set {1,2,3,…,9}. Then the probability that ${\mathrm{x}}^2+\mathrm{bx}+\mathrm{c}>0,\forall \mathrm{x}\in \mathrm{ℝ}$ is

Here,${\mathrm{x}}^2+\mathrm{bx}+\mathrm{c}>0\mathrm{\forall }\mathrm{x}\in \mathrm{R}$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{D}<0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{b}}^2<4\mathrm{c}\end{array}$

value of b possible  values of c

$\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} 1<4\mathrm{c}\Rightarrow \mathrm{c}>\frac{1}{4}\Rightarrow \{1,2,3,4,5,6,7,8,9\}\\ 2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4<4\mathrm{c}\Rightarrow \mathrm{c}>1 \Rightarrow \{2,3,4,5,6,7,8,9\}\\ 3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}9<4\mathrm{c}\Rightarrow \mathrm{c}>\frac{9}{4}\Rightarrow \{3,4,5,6,7,8,9\}\\ 4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} 16<4\mathrm{c}\Rightarrow \mathrm{c}>4 \Rightarrow \{5,6,7,8,9\}\\ 5 25<4\mathrm{c}\Rightarrow \mathrm{c}>625 \Rightarrow \{7,8,9\}\\ 6 36<4\mathrm{c}\Rightarrow \mathrm{c}>9 \text{ Impossible }\end{array}$

 7       impossible

 8       impossible

 9      impossible

Number of favorable ca8e8 =9 + 8 + 7+6+3=32
Total ways=9x9=81

Required probability = 32/81