Stopwatch probability

Tanjello wrote:A certain circular stopwatch has exactly 60 sec 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 secs, what is the probability that the hand will stop less than 10 marks from the 53-sec mark?

A) 1/6
B) 19/60
C) 1/3
D) 29/60
E) 1/2

I think the answer is B (53+/- 9 means 18 ways, plus stopping at 53), thus 19/60. But it took me a while to realize the answer. What is the best way to approach this problem during a timed test? Thanks!

sanju09 wrote:

Tanjello wrote:A certain circular stopwatch has exactly 60 sec 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 secs, what is the probability that the hand will stop less than 10 marks from the 53-sec mark?

A) 1/6
B) 19/60
C) 1/3
D) 29/60
E) 1/2

I think the answer is B (53+/- 9 means 18 ways, plus stopping at 53), thus 19/60. But it took me a while to realize the answer. What is the best way to approach this problem during a timed test? Thanks!

Less than 10 marks from the 53-sec mark are the marks that are more than 43 and less than 3 in the continued cycle. Hence, following are the possible second-marks where the hand could stop to meet the inevitability:

44, 45, 46, 47, 48, 49, 50, 1, 2, and 3 ten in total, and so are the various possibilities for the hand to start from.

Required probability = [spoiler]10/60 = 1/6.

A[/spoiler]

liferocks wrote:

sanju09 wrote:

Tanjello wrote:A certain circular stopwatch has exactly 60 sec 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 secs, what is the probability that the hand will stop less than 10 marks from the 53-sec mark?

A) 1/6
B) 19/60
C) 1/3
D) 29/60
E) 1/2

I think the answer is B (53+/- 9 means 18 ways, plus stopping at 53), thus 19/60. But it took me a while to realize the answer. What is the best way to approach this problem during a timed test? Thanks!

Less than 10 marks from the 53-sec mark are the marks that are more than 43 and less than 3 in the continued cycle. Hence, following are the possible second-marks where the hand could stop to meet the inevitability:

44, 45, 46, 47, 48, 49, 50, 1, 2, and 3 ten in total, and so are the various possibilities for the hand to start from.

Required probability = [spoiler]10/60 = 1/6.

A[/spoiler]

Why did u excluded the marks 51,52,54,55,56,57,58,59,0 ?