A) \[\frac{11}{36}\]
B) \[\frac{22}{36}\]
C) \[\frac{15}{36}\]
D) \[\frac{29}{36}\]
Correct Answer: A
Solution :
(a) With a six on the first dice, there are 6 possibilities of outcomes that can appear on the other dice (6 &1, 6 & 2, 6 & 3, 6 & 4, 6 & 5 and 6&6). At the same time with 6 on the second dice there are 5 more possibilities for outcomes on the first dice: (1 & 6, 2 & 6, 3 & 6, 4 & 6, 5 & 6). Also, the total outcomes are 36. Hence, the required probability is 11/36.
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