Probability Ques 18
If $p$ and $q$ are chosen randomly from the set $1,2,3,4,5$, $6,7,8,9$ and $10$ with replacement, determine the probability that the roots of the equation $x^{2}+p x+q=0$ are real.
(1997, 5M)
Show Answer
Answer:
Correct Answer: 18.$(0.62)$
Solution:
Formula:
- The required probability $=1-$ (probability of the event that the roots of $x^{2}+p x+q=0$ are non-real).
The roots of $x^{2}+p x+q=0$ will be non-real if and only if $p^{2}-4 q<0$, i.e. if $p^{2}<4 q$
The possible values of $p$ and $q$ can be possible according to the following table.
| Value of $q$ | Value of $p$ | Number of pairs of $p, q$ |
|---|---|---|
| 1 | 1 | 1 |
| 2 | 1,2 | 3 |
| 3 | $1,2,3$ | 3 |
| 5 | $1,2,3$ | 4 |
| 6 | $1,2,3,4$ | 4 |
| 7 | $1,2,3,4$ | 5 |
| 9 | $1,2,3,4,5$ | 5 |
| 10 | $1,2,3,4,5$ | 5 |
Therefore, the number of possible pairs $=38$
Also, the total number of possible pairs is $10 \times 10=100$
$\therefore$ The required probability $=1-\frac{38}{100}=1-0.38=0.62$
BETA
