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. 0.31
• B. 0.62
• C. 0.5
• D. 0.84

Answer 1

The correct option is B 0.62

The required probability $=$ $1-$ (probability of the event 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 required probability is $1-\frac{38}{100}=1-0.38=0.62$