A) \[\frac{1}{9}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{4}\]
D) \[\frac{5}{9}\]
Correct Answer: C
Solution :
\[S=\{(3,\,1),\,(2,\,2),\,(1,\,3),\,(6,\,2),\,(5,\,3),\,(4,\,4),\,(3,\,5),\,(2,\,6),\,(6,\,6)\}\] Hence required probability \[=\frac{9}{36}=\frac{1}{4}.\]
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