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 $xϵR$ is:

• A. $\frac{23}{81}$
• B. $\frac{7}{9}$
• C. $\frac{32}{81}$
• D. $\frac{65}{729}$

Answer 1

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

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 $\times$ 9 = 81
Now, $x^2+bx+c>0$ for all $xϵR$
$\Rightarrow D<0$
$\Rightarrow b^2-4ac<0\Rightarrow b^2-4c<0\Rightarrow b^2<4c$
Alternatively: $x^2-bx+c>0$
$\Rightarrow {(x+\frac{b}{2})}^2+\frac{4c-b^2}{4}>0$
$\Rightarrow 4c-b^2>0$
$\Rightarrow b^2<4c$
Now, the following table shows the possible values of b and c for which $b^2<4c$
$\begin{array}{ccc}c& b& Total\\ 1& 1,& 1\\ 2& 1,2& 2\\ 3& 1,2,3& 3\\ 4& 1,2,3& 3\\ 5& 1,2,3,4& 4\\ 6& 1,2,3,4& 4\\ 7& 1,2,3,4,5& 5\\ 8& 1,2,3,4,5& 5\\ 9& 1,2,3,4,5& 5\\ & & 32\end{array}$
So, favourable number of cases = 32
Hence required probability = $\frac{32}{81}$