Question

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

$\frac{3}{16}$

Explanation for the correct option:

Finding the probability that the determinant:

The maximum number of second-order determinants that can be created using $0,1$

$$\begin{gathered} =2\times 2\times 2\times 2 \\ =16 \end{gathered}$$

Because each of the four elemental positions can be filled in one of two ways: by $0$ or $1$

$\therefore \mathrm{n}\left(\mathrm{S}\right)=16$

Let $\mathrm{E}$ denote the occurrences of obtaining a non-negative determinant.
Then $\mathrm{E}$ denotes the occurrence of a negative determinant.
Clearly positive second-order determinants that may be created with $0,1$ are $\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right],\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right],\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]$

$\begin{array}{rcl}\therefore \mathrm{P}\left(\mathrm{E}\right)& =& \frac{\mathrm{n}\left(\mathrm{E}\right)}{\mathrm{n}\left(\mathrm{S}\right)}\\ & =& \frac{3}{16}\end{array}$

Hence, the correct option is C.