A) \[\frac{1}{6}\]
B) \[\frac{1}{36}\]
C) \[\frac{1}{12}\]
D) \[\frac{1}{2}\]
Correct Answer: C
Solution :
\[1=\frac{1}{2}\,\,\left| \begin{matrix} x & y & 1 \\ 1 & 1 & 1 \\ 2 & 0 & 1 \\ \end{matrix} \right|\] \[\Rightarrow \] \[x+y=4\] |
\[\therefore \] number of solutions = 3 |
\[\therefore \] the required probability\[=\frac{3}{36}=\frac{1}{12}\] |
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