Question

A determinant of second order is made with the elements $0$ and $1$.

What is the probability that the determinant is positive?

• A. $\frac{7}{12}$
• B. $\frac{11}{12}$
• C. $\frac{3}{16}$
• D. $\frac{15}{16}$

Answer 1

The correct option is C

$\frac{3}{16}$

An explanation for the correct option:

Step 1: Find the number of possible ways a determinant of second-order can be made with the elements $0$ and $1$.

We have been given that, a determinant of second order is made with the elements $0$ and $1$.

We need to find the probability that the determinant is positive.

the number of possible ways a determinant of second-order can be made with the elements $0$ and $1$ :

The $4$ places in the matrix can be filled by ${}^{2}C_1$ ways each.

therefore the sample space will be $n\left(S\right)=2^4$

$=16$

Step 2: Find the number of possible ways that the determinant is positive.

Let event $A:$the determinant is positive.

If a matrix is $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ then $ad-cb$ is the determinant and for it to be positive $ad=1$ and $cb=0$

Therefore, The determinant is positive for $\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|,\left|\begin{array}{cc}1& 0\\ 1& 1\end{array}\right|,\left|\begin{array}{cc}1& 1\\ 0& 1\end{array}\right|$

So, the possible number of ways that the determinant is positive.

$n\left(A\right)=3$

Step 3: Find the required probability.

the probability that the determinant is positive:

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}$

$\mathbf{\therefore }\mathit{P}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{=}\frac{\mathbf{3}}{\mathbf{16}}$

Therefore, option (C) is the correct answer.