Question

If $p$ and $q$ are chosen randomly from the set $\{1,2,3,4,5,6,7,8,9,10\}$, with replacement, determine the probability that the roots of the equation $x^2+px+q=0$ are real

• A. $0.62$
• B. $0.50$
• C. $0.31$
• D. $0.73$

Answer 1

The correct option is A $0.62$

The required probability $=1-$(probability of the venet that the roots of $x^2+px+q=0$ are non real)
The roots of $x^2+px+q=0$ will be non-real if and only if $p^2-4q<0\Rightarrow p^2<4q$
The possible values of $p$ and $q$ can be possible according to the following table

value of $q$	value of $p$	no. of pairs of $p,q$
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
10	1,2,3,4,5,6	6

Therefore, the number of possible pairs$=$$38$
Also, the total number of possible pairs is $10\times 10=100$
Therefore the required probability $=1-\frac{38}{100}=0.62$