Question

If $p$ and $q$ are selected randomly with replacement from first $10$ natural number. Find the probability that their roots of equation $x^2+px+q=0$ are real ?

Answer 1

$x^2+px+q=0$

For real roots

$p\ge 0$

$p^2-4q\ge 0$

$p^2\ge 49$

For r, $q=1$

$p=2,3,4,5,6,7,8,9$ or $10$

Probability of event $=\frac{1}{10}\times \frac{9}{10}$

For $q=2$

$p=3,4,5,6,7,8,9$ or $10$

Probability of event$=\frac{1}{10}\times \frac{8}{10}$

For $q=3$, $p=4,5,6,7,8,9$ or $10$

Probability of event$=\frac{1}{10}\times \frac{7}{10}$

For $q=4$, $p=5,6,7,8,9$ or $10$

Probability of event$=\frac{1}{10}\times \frac{6}{10}$

For $q=5,p=5,6,7,8,9$ or $10$

Probability of event$=\frac{1}{10}\times \frac{5}{10}$

For $q=4$, $p=4,5,6,7,8,9$ or $10$

Probability of event$=\frac{1}{10}\times \frac{4}{10}$

Adding probability of all these events:-

$\frac{1}{100}[9+8+7+7+6+6+5+5+5+4]=\frac{62}{100}=0.62$.