A) \[\frac{3}{16}\]
B) \[\frac{3}{8}\]
C) \[\frac{1}{4}\]
D) None of these
Correct Answer: B
Solution :
A determinant of order 2 is of the form \[\Delta =\left| \,\begin{matrix} a & b \\ c & d \\ \end{matrix}\, \right|\] 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 \[ad=1,\] \[bc=0\] iff \[a=d=1\] 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 \[=\frac{6}{16}=\frac{3}{8}.\]
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