A) \[\frac{17}{18}\]
B) \[\frac{1}{12}\]
C) \[\frac{11}{12}\]
D) None of these
Correct Answer: C
Solution :
Favourable cases to get the sum not less than 11 are \[\{(5,\,\,6),\,(6,\,\,6),\,(6,\,\,5)\}=3\] Hence favourable cases to get the sum less than 11 are \[(36-3)=33\]. So required probability \[=\frac{33}{36}=\frac{11}{12}.\]
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