A) \[\frac{11}{216}\]
B) \[\frac{5}{108}\]
C) \[\frac{9}{216}\]
D) \[12\frac{1}{18}\]
Correct Answer: B
Solution :
Total number of outcomes when three dice are thrown \[=6\times 6\times 6=216\] For sum of numbers to be 15, possible ways are, (6, 6, 3). (6, 3, 6), (3, 6, 6), (6, 5, 4), (6, 4, 5), (5, 4, 6), (5, 6, 4), (4, 5, 6), (4, 6, 5), (5, 5, 5) \[\therefore \] Number of favorable outcomes =10 \[\therefore \] Required probability \[=\frac{10}{216}=\frac{5}{108}\]
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