Question

Let 'head' means one and `tail' means two, and the coefficient of the equation $a{x}^2+bx+c=0$ are chosen by tossing a coin. The probability that the roots of the equation are non-real, is equal to

• A. $\frac{5}{8}$
• B. $\frac{7}{8}$
• C. $\frac{3}{8}$
• D. $\frac{1}{8}$

Answer 1

The correct option is B $\frac{7}{8}$

Since there are 3 coefficients, there will be total 3 tosses and the sample space will contain $2\times 2\times 2=8$ elements.
Out of these, the outcome $(H,T,H)$ corresponding to $(a,b,c)=(1,2,1)$ is unfavourable.
Since, $a=1$, $b=2$ and $c=1$
$\Rightarrow b^2-4ac$$=0$
For imaginary roots $b^2-4ac<0$.
The other cases are favourable cases. There are 7 such cases.
Therefore, the required probability is $\frac{7}{8}$.