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# 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 determinant chosen is +ve is

$\begin{array}{1 1} \frac{3}{16} \\ 3 \\ 4 \\ 5 \end{array}$

With 0 and 1 as elements there are $2\times 2\times 2\times 2=16$ determinants of order $2\times 2$ out of which only.
$\begin{vmatrix}1 &0\\0&1\end{vmatrix}1\begin{vmatrix}1 &1\\0&1\end{vmatrix}1\begin{vmatrix}1 &0\\1&1\end{vmatrix}$ are the three determinant whose value is +ve.
$\therefore$ Required probability =$\large\frac{3}{16}$
Hence (a) is the correct answer.