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!
Stopwatch probability
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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: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!
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]
Last edited by sanju09 on Fri May 21, 2010 1:14 am, edited 1 time in total.
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Why did u excluded the marks 51,52,54,55,56,57,58,59,0 ?sanju09 wrote: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: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!
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]
"If you don't know where you are going, any road will get you there."
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- sanju09
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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 seeliferocks wrote:Why did u excluded the marks 51,52,54,55,56,57,58,59,0 ?sanju09 wrote: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: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!
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]
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com