The coefficients of a quadratic equation ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0(\mathrm{a}\ne \mathrm{b}\ne \mathrm{c})$ are chosen from first three prime numbers, the probability that roots of
the equation are real is

Solution:

 First three prime numbers are 2, 3 and 5
Total ways of choosing a, b, c = 3 x 2 x 1= 6   $[\because \mathrm{a}\ne \mathrm{b}\ne \mathrm{c}]$
For roots to be real ,${\mathrm{b}}^2-4\mathrm{ac}\ge 0$

Value of b              Possible values of a and c

2                             $\begin{array}{l}4-4\mathrm{ac}\ge 0\Rightarrow \mathrm{ac}\le 1\\ \mathrm{a}=\mathrm{c}=1\end{array}$

                                (No values of a and c)

3                             $\begin{array}{l}9-4\mathrm{ac}\ge 0\Rightarrow \mathrm{ac}\le \frac{9}{4}\\ \mathrm{ac}=1,\mathrm{ac}=2\\ \mathrm{a}=\mathrm{c}=1,\mathrm{a}=2,\mathrm{c}=1\\ \mathrm{a}=1,\mathrm{c}=2\end{array}$

                                (No values of a and c)

5                            $\begin{array}{l}25-4\mathrm{ac}\ge 0\Rightarrow \mathrm{ac}\le \frac{25}{4}\\ \therefore \mathrm{ac}=1,2,3,4,5,6\\ \mathrm{ac}\ne 1,2,3,4,5\\ \text{ for }\mathrm{ac}=6\\ (2,3)(3,2)\end{array}$

Favorable ways = 2

Required probability $=\frac{\text{ Favourable ways }}{\text{ Total ways }}$

                                 $=\frac{2}{6}=\frac{1}{3}$