The stop watch always travels 10 seconds or 10 divisions on the watch. So the final position has to be 10 away from initial position.

Set of all possible initial positions=\(\{1,2,3,.....60\}\) 60 possible positions

If the watch starts at 42 it will stop at 52. Which is within 10 divisions from 53. Start at 41 ends at 51 and so on

So as long as the watch stops at 44 (9 divisions before) or 02 (9 divisions after) the conditions are satisfied. Number of divisions between 44 and 02 is 19.

So for 19 starting positions the condition set in the problem is satisfied!

Probability =\(\frac{19}{60}\)

A certain circular stopwatch has exactly 60 second marks and a single hand. If the hand of the watch is randomly set to one of the marks and allowed to count exactly 10 seconds, what is the probability that the hand will stop less than 10 marks from the 53-second mark?

A. \(\frac{1}{6}\)

B. \(\frac{19}{60}\)

C. \(\frac{1}{3}\)

D. \(\frac{29}{60}\)

E. \(\frac{1}{2}\)