Question

If each elements of a $2\times 2$ determinant is either $0$ or $1$, what is the probability that the value of the determinant is positive?

Answer 1

A second order determinant has $4$ entries as $\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|$

Given, each element is either $0$ or $1$

Number of such second order determinants$=\left|\begin{array}{cc}0or1& 0or1\\ 0or1& 0or1\end{array}\right|$

$=2\times 2\times 2\times 2=16$

$\therefore$ Total number of determinants$=16$

Now for the value of determinant to be positive, favourable cases are $\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|$,$\left|\begin{array}{cc}1& 1\\ 0& 1\end{array}\right|$ and $\left|\begin{array}{cc}1& 0\\ 1& 1\end{array}\right|$

Other cases $=\left|\begin{array}{cc}1& 1\\ 0& 0\end{array}\right|$,$\left|\begin{array}{cc}0& 1\\ 0& 1\end{array}\right|$,$\left|\begin{array}{cc}0& 0\\ 0& 0\end{array}\right|$,$\left|\begin{array}{cc}1& 1\\ 1& 0\end{array}\right|$,$\left|\begin{array}{cc}0& 1\\ 1& 1\end{array}\right|$, etc

In all these cases, the value of determinant is either $0$ or $-1$ and not positive.

$\therefore$ the number of favorable cases$=3$

Probability that the value of determinant is positive

$=\frac{Numberoffavorabledeterminant}{Totalnumberofdeterminants}$

$=\frac{3}{16}$