Question

A determinant of the second order is made with the element $0$ and $1$. What is the probability that the determinant made is non-negative?

• A. Required probability is $P\left(E\right)=\frac{3}{16}$
• B. Required probability is $P\left(E\right)=\frac{10}{16}$
• C. Required probability is $P\left(E\right)=\frac{13}{16}$
• D. Required probability is $P\left(E\right)=\frac{15}{16}$

Answer 1

The correct option is C Required probability is $P\left(E\right)=\frac{13}{16}$

The number of determinant of the second order that can be formed with $0,1=2\times 2\times 2\times 2=16$
Since each of the four places of elements can be filled in $2$ ways: by $0$ or $1$
$\therefore n\left(S\right)=16$
Let $E=$ the events of getting non-negative determinant.
Then $E^{\prime }=$ the event of getting negative determinant.
Clearly negative determinant of second order that can be made with $0,1$ are $\left|\begin{array}{cc}0& 1\\ 1& 0\end{array}\right|,\left|\begin{array}{cc}1& 1\\ 1& 1\end{array}\right|,\left|\begin{array}{cc}1& 1\\ 1& 0\end{array}\right|$
$\therefore P\left(E^{\prime }\right)=\frac{n\left(E^{\prime }\right)}{n\left(S\right)}=\frac{3}{16}$
$\therefore P\left(E\right)=1-P\left(E^{\prime }\right)=1-\frac{3}{16}=\frac{13}{16}$