Question

A determinant is chosen at random from the set of all determinants of order $2$ with elements $0$ or $1$ only.

The probability that the value of the determinant chosen is positive is

• A. $\frac{3}{16}$
• B. $\frac{3}{8}$
• C. $\frac{1}{4}$
• D. None of these

Answer 1

The correct option is A

$\frac{3}{16}$

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 is chosen at random from the set of all determinants of order $2$ with elements $0$ or $1$ only.

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

the number of possible ways a determinant of second-order can be made with the elements $0$ and $1$that the $4$ places in the matrix can be filled by ${}^{2}C_1$ ways each.

Let $S$ be the sample space. Therefore the sample space will ve

$$\begin{gathered} \Rightarrow n\left(S\right)=2^4 \\ \Rightarrow n\left(S\right)=16 \end{gathered}$$

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

Let $A$be the event that the determinant chose is positive.

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

So the possible choices are: $\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|,\left|\begin{array}{cc}1& 0\\ 1& 1\end{array}\right|,\left|\begin{array}{cc}1& 1\\ 0& 1\end{array}\right|$

then the possible number of ways that the determinant is positive.

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

Step 3: Find the required probability:

To find the probability that the determinant chosen is positive,

We know that $\mathit{P}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{A}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}$

$\Rightarrow P\left(A\right)=\frac{3}{16}$

Therefore, option (A) is the correct answer.