KarunMendiratta wrote:
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 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. \(\frac{1}{2}\)
OA Explanation:less than 10 marks from the 53-second mark: means it can go 9 marks ahead (3 second mark) or 9 marks before (44 second mark)
So, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 01, 02 (19 Marks)
Therefore, the required probability = \(\frac{19}{60}\)
Hence, option B