A) \[\frac{1}{16}\]
B) \[\frac{1}{8}\]
C) \[\frac{3}{16}\]
D) \[\frac{1}{4}\]
E) \[\frac{1}{2}\]
Correct Answer: C
Solution :
Since, each entry element of a\[2\times 2\]matrix, with elements 0 and 1 only, can be filled in 2 ways \[\therefore \]Total\[2\times 2\]matrix\[={{2}^{4}}=16\] There are three\[2\times 2\]matrices whose determinants are positive i.e., \[\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right],\left[ \begin{matrix} 1 & 0 \\ 1 & 1 \\ \end{matrix} \right]\] and \[\left[ \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right]\]. Hence, required probability\[=\frac{3}{16}\].
You need to login to perform this action.
You will be redirected in
3 sec
