A) 5/36
B) 1/9
C) 2/9
D) 19/36
Correct Answer: C
Solution :
3 can be thrown as (1, 2), (2,1). 5 can be thrown as (1,4), (4,1), (2,3), (3,2). an be thrown as (5, 6), (6,11). \[\therefore \]Required probability\[=\frac{8}{36}=2/9\]
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