Question

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 $x\in R$ is

• A. $\frac{17}{123}$
• B. $\frac{32}{81}$
• C. $\frac{82}{125}$
• D. $\frac{45}{143}$

Answer 1

The correct option is B $\frac{32}{81}$

Here, $x^2+bx+c>0\forall x\in R$
$\therefore D<0$
$\Rightarrow b^2<4c$

$\begin{array}{cccc}Valueofb& Possiblevaluesofc& & \\ 1& 1<4c& \Rightarrow c>\frac{1}{4}& \Rightarrow \{1,2,3,4,5,6,7,8,9\}\\ 2& 4<4c& \Rightarrow c>1& \Rightarrow \{2,3,4,5,6,7,8,9\}\\ 3& 9<4c& \Rightarrow c>\frac{9}{4}& \Rightarrow \{3,4,5,6,7,8,9\}\\ 4& 16<4c& \Rightarrow c>4& \Rightarrow \{5,6,7,8,9\}\\ 5& 25<4c& \Rightarrow c>6.25& \Rightarrow \{5,6,7,8,9\}\\ 6& 36<4c& \Rightarrow c>9& Impossible\\ 7& Impossible& & \\ 8& Impossible& & \\ 9& Impossible& & \end{array}$
$\therefore$ Number of favourable cases = 9 + 8 + 7 + 5 + 3
=32
Total ways $=9\times 9=81$
$\therefore$ Required probability = $\frac{32}{81}$