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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 determinant chosen is non-zero is
$\frac{3}{{16}}$
$\frac{3}{8}$
$\frac{1}{4}$
None of these
Solution
(b) A determinant of order $2$ is of the form $\Delta = \left| {\,\begin{array}{*{20}{c}}a&b\\c&d\end{array}\,} \right|$
It is equal to $ad – bc.$
The total number of ways of choosing $a,\,\,b,\,\,c$ and $d$ is $2 \times 2 \times 2 \times 2 = 16.$
Now $\Delta \ne 0$ if and only if either $ad = 1,$ $bc = 0$ or $ad = 0,$ $bc = 1.$ But $ad = 1,$ $bc = 0$ if $a = d = 1$ and one of $b,\,c$ is zero.
Therefore $ad = 1,$ $bc = 0$ in three cases,
similarly $ad = 0,$ $bc = 1$ in three cases.
Therefore the required probability $ = \frac{6}{{16}} = \frac{3}{8}.$
