$\begin{array}{1 1} \large\frac{3}{16} \\ \large\frac{5}{16} \\ \large\frac{7}{16} \\ \large\frac{9}{16} \end{array} $

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Given that a second order determinant has 4 entries, which may be 0 or 1. Total number of determinants = $2^4 = 16$

We need to find the probability that the determinant is positive.

The only positive determinants are $\begin{vmatrix} 1 &0 \\ 0& 1 \end{vmatrix}, \begin{vmatrix} 1 &1 \\ 0& 1 \end{vmatrix} \;\text{and}\; \begin{vmatrix} 1 &0 \\ 1& 1 \end{vmatrix}$

Therefore the required probability = $\large\frac{3}{16}$

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