A) \[\frac{5}{36}\]
B) \[\frac{1}{12}\]
C) \[\frac{1}{18}\]
D) \[\frac{1}{9}\]
Correct Answer: D
Solution :
Total case\[=6\times 6=36\] Number of ways in which the sum 5 on both dice are (1, 4) (4,1), (2, 3), (3, 2) \[\therefore \]Total number of favourable case = 4 \[\therefore \]Required probability\[=\frac{4}{36}=\frac{1}{9}\]
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