kaushiksin 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 exactly 10 seconds, what is the probability that the hand will stop less than 10 marks from the 53-second mark?

Note: I looked at the explanation as well but didnt quite get it. Can you help me in understanding how do we need to solve it? Thanks for your help.

P = good outcomes/total possible outcomes

Total possible outcomes:
The watch could be set so that the second hand stops on any of the 60 marks. Could it stop on the 5-second mark? Sure. Could it stop on the 27-second mark? Sure. Could it stop on the 53-second mark? Sure. It could stop on any of the 60 marks. So there are 60 total possible outcomes.

Good outcomes:
The 9 marks before the 53-second mark = 9 good outcomes
The 53-second mark itself = 1 good outcome
The 9 marks beyond the 53-second mark = 9 good outcomes
9+1+9 = 19 good outcomes

So P = good/total = 19/60.

Beautiful. Thanks.

The questions is - stop "less than" 10 marks from the 53-second mark --- in that case do we need to include the 9 marks beyond the 53rd mark... the clock is circular in shape --- but doesn't "less than" mean that you only consider from the 44th sec mark to 53rd sec mark - 10/60?

Nice explanation. However, I got answer as 1/3. I got 20 good outcomes and 60 totaol outcomes. I calculated 43 second to 03 second (clockwise). Please explain what is wrong in it?

GMATGuruNY wrote:

kaushiksin 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 exactly 10 seconds, what is the probability that the hand will stop less than 10 marks from the 53-second mark?

Note: I looked at the explanation as well but didnt quite get it. Can you help me in understanding how do we need to solve it? Thanks for your help.

P = good outcomes/total possible outcomes

Total possible outcomes:
The watch could be set so that the second hand stops on any of the 60 marks. Could it stop on the 5-second mark? Sure. Could it stop on the 27-second mark? Sure. Could it stop on the 53-second mark? Sure. It could stop on any of the 60 marks. So there are 60 total possible outcomes.

Good outcomes:
The 9 marks before the 53-second mark = 9 good outcomes
The 53-second mark itself = 1 good outcome
The 9 marks beyond the 53-second mark = 9 good outcomes
9+1+9 = 19 good outcomes

So P = good/total = 19/60.

gmat1011 wrote:The questions is - stop "less than" 10 marks from the 53-second mark --- in that case do we need to include the 9 marks beyond the 53rd mark... the clock is circular in shape --- but doesn't "less than" mean that you only consider from the 44th sec mark to 53rd sec mark - 10/60?

Less than 10 marks is in terms of distance and not time. So we will have to count intervals on either side of 53 second mark.

reply2spg wrote:Nice explanation. However, I got answer as 1/3. I got 20 good outcomes and 60 totaol outcomes. I calculated 43 second to 03 second (clockwise). Please explain what is wrong in it?

A good outcome is if the second hand stops less than 10 marks from the 53rd-second mark.

The 9 marks before the 53rd-second mark are: 44, 45, 46, 47, 48, 49, 50, 51, 52. The 43rd-second mark is 10 marks from 53rd-second mark and is not a good outcome.

The 9 marks beyond the 53rd-second mark are: 54, 55, 56, 57, 58, 59, 00, 01, 02. The 3rd-second mark is 10 marks from the 53rd-second mark and is not a good outcome.

9 marks before + 53rd-second mark + 9 marks beyond = 9+1+9 = 19 good outcomes.

Also, the 43rd-second mark to the 3rd-second mark = 21 marks (not 20). The formula for counting consecutive integers is:

total number of consecutive integers = biggest - smallest + 1

So from the 43rd-second mark to the 63rd-second mark there are 63-43+1 = 21 marks.