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 value of the determinant chosen is positive is

• A. $\frac{5}{16}$
• B. $\frac{7}{16}$
• C. $\frac{3}{16}$
• D. None of these

Answer 1

Answer: (C)

$\frac{3}{16}$

Since each element of the determinant can be placed in two wats $0$ or $1$, total number of ways $=2^4=16$
Since value of the determinant is +ve, so we have only $3$ cases:
$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right],\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]$
hence the required probability $=\frac{3}{16}$