A) \[\frac{1}{9}\]
B) \[\frac{1}{4}\]
C) \[\frac{7}{12}\]
D) \[\frac{5}{12}\]
Correct Answer: D
Solution :
Sum of \[7\] can be obtained when\[(2,\,\,6),\,\,(3,\,\,5)(3,\,\,6),\,\,(4,\,\,4),\]\[(4,\,\,5),\,\,(4,\,\,6)(5,\,\,3)\]\[(5,\,\,4)(5,\,\,5)(5,\,\,6)(6,\,\,2)(6,\,\,3)\]\[(6,\,\,4)(6,\,\,5)(6,\,\,6)\] \[\therefore \]Probability of sum\[>7=\frac{15}{36}=\frac{5}{12}\]
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