Each coefficient in the equation $ax^2 + bx + c = 0 $ is determined by throwing an ordinary six faced die. The probability that the equation will have real roots, is
Answer & explanation
Correct answer: option 3
Since each of the coefficient a, b and c can take the values from 1 to 6.
∴ Total numbers of equations = 6 × 6 × 6 = 216.
The roots of the equation $ax^2 + bx + c = 0 $ will be real if $b^2 - 4 ac ≥ 0 ⇒ b^2 ≥ 4 ac.$
The favourable number of elementary events can be enumerated as follows:
|
ac |
a |
c |
4ac |
b(so that $ b^2 ≥ 4ac $ |
No. of ways |
|
1 |
1 |
1 |
4 |
2, 3, 4, 5, 6 |
1 × 5 = 5 |
|
2 |
$\left\{\begin{matrix}1\\2\end{matrix}\right.$ |
2 1 |
8 |
3, 4, 5, 6 |
2 × 4 = 8 |
|
3 |
$\left\{\begin{matrix}1\\3\end{matrix}\right.$ |
3 1 |
12 |
4, 5, 6 |
2 × 3 = 6 |
|
4 |
$\left\{\begin{matrix}1\\4\\2\end{matrix}\right.$ |
4 1 2 |
16 |
4, 5, 6 |
3×3=9 |
|
5 |
$\left\{\begin{matrix}1\\5\end{matrix}\right.$ |
5 1 |
20 |
5, 6 |
2 × 2 = 4 |
|
6 |
$\left\{\begin{matrix}1\\6\\2\\3\end{matrix}\right.$ |
6 1 3 2 |
24 |
5, 6 |
4 ×2 = 8 |
|
7 |
ac is not possible |
|
|
|
0 |
|
8 |
$\left\{\begin{matrix}2\\4\end{matrix}\right.$ |
4 2 |
32 |
6 |
2 ×1= 2 |
|
9 |
3 |
3 |
36 |
6 |
1 |
Total = 43
Since $b^2≥ 4ac$ and since the maximum value of $b^2$ is 36, therefore ac = 10, 11, 12 ...etc. is not possible.
∴ Total number of favourable elementary events = 43
Hence, required probability = $\frac{43}{216}$