Stopwatch probability

[Tanjello]

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]

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, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 1, and 2; nineteen in total, and so are the various possibilities for the hand to start from.

Required probability = [spoiler]19/60.

B[/spoiler]

[liferocks]

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 ?

[sanju09]

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 ?

edited my friend, edited, you didn't give me enough time dear, earlier I stopped my watch at 50 in place of 60, mistakenly, normally happens after the fifth one, you see

[gmatjedi]

my approach:

10 up and 10 down from 53
=20
but note that you include 53 twice going up and going down
so you need to eliminate a duplicate 53
=19
which is out of 60 options