A) \[\frac{1}{36}\]
B) \[\frac{2}{36}\]
C) \[\frac{5}{36}\]
D) \[\frac{31}{36}\]
Correct Answer: C
Solution :
Probability of getting 5 in first attempt\[=\frac{1}{6}\] Probability that no digit appear instead of\[5=\frac{5}{6}\] \[\therefore \]Required probability\[=\frac{1}{6}\times \frac{5}{6}=\frac{5}{36}\]
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