A circular racetrack is 3 miles in length and has signs post

Look at the diagram:



The car ends within a half mile of the sign indicating 2.5 miles means that the car should end in one mile interval, between the signs indicating 2 (2.5-0.5=2) and 3 miles (2+0.5=3), so within the red segment on the diagram.

Now if the car appears somewhere between the blue dots, between 1.5 and 2 miles signs then after traveling 0.5 miles the car will be in the red segment. So in order after traveling 0.5 miles the car to end within the red segment it should appear between 1.5 and 2.5 miles, so within 1 mile interval, as the circumference of the track is 3 miles then the probability of that will be P=favorable/total=1/3. As I mentioned in my previous post actually it doesn't matter where the car appears or what distance it travel, as long as favorable interval in the end is 1 mile and total interval is 3 miles then the probability will be 1/3 miles.

Hope it's clear.



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The car ends within a half mile of the sign indicating 2 1/2 miles means that the car will end in one mile interval, between the signs indicating 2 and 3 miles.

Now, it doesn't matter where the car starts or what distance it travels, the probability will be P=(favorable outcome)/(total # of outcomes)=1/3 (as the car starts at random point end travels some distance afterwards we can consider its end point as the point where he randomly appeared, so the probability that the car appeared within 1 mile interval out of total 3 miles will be 1/3).

Answer: C.

Hope it's clear.
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Stem gives us the information about the signs only to fix the point of 2.5 miles on the track and thus fix the 1 mile interval the car should end within. 1/10 mile increment is totally irrelevant.

Consider the following: A circular racetrack is 3 miles in length. If a race car starts at a random location on the track and travels exactly Y miles, what is the probability that the car ends within a half mile of some point X on the track?

The answer will be the same: as the interval for the endpoint of the car is 1 mile (from x-0.5 to x+0.5) then the probability will be 1/3.

Note that a car starts at a random location on the track and travels exactly Y miles means that the endpoint of the car will also be at random location on the track (travel part is also to confuse us: random location plus Y miles=random location). So the question basically ask what is the probability that the car ends within the particular 1 mile interval on the track of 3 miles.

Hope it's clear.
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Here is a rewording of the original question:
What is the probability that the car starts between the 1.5 mile and 2.5 mile mark on a 3 mi racetrack?

2.5 - 1.5 = 1
that is 1/3 of the racetrack

Answer is C

Everything else in this question is irrelevant.
Although, if you really wanted you could rephrase the question to account for the 1/10th mile sign posts:
What is the probability that the car stars between the 15th and 25th sign posts if there are 30 sign posts total.
But that is unnecessary extra work.

We are given that a circular racetrack is 3 miles in length and has signs posted to indicate each 1/10 mile increment. We are also given that a race car starts at a random location on the track and travels exactly one half mile, and we need to determine the probability that the car ends within a half mile of the sign indicating 2 1/2 (or 2.5) miles.

If the car ends within a half mile of the 2.5-mile sign, that means the car can end as far as the 2.0-mile sign or the 3.0-mile sign. However, since the car travels exactly one half mile, the starting point of the car can be anywhere from the 1.5 mile sign to the 2.5-mile sign. In other words, the car can be anywhere in this 2.5 - 1.5 = 1 mile stretch. Since the racetrack is 3 miles long, the probability that the car is in this 1 mile stretch is ⅓.

Answer: C
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Bunuel
A small doubt here to be precise the correct ranges would be 2.1,2.2,2.3,2.4,2.5,2.6,2.7,2.8,2.9 as 2.5 miles -/+ 0.4 miles (within 0.5 miles i.e, <=0.5-> <0.4) is between 2.1 and 2.9
probability would be 9/31. (31- 0,0.1,....3.0)
I know the above is wrong but I am having confusion with distance vs exact point transition.(In other words I am calculating exact point like 2.1,2.2,... as in question given 2.5 (within half mile) but unable to know how 2-3-> as OS is capturing the exact condition of question)
Pls help me to figure out where I faltered.

Ihave solved it this way.
If the vehicle starts at a random location and travels 1/2 mile, it is still at some arbit location.
Hence , I would try to find out the range of desired outcomes.
If the vehicle is between the mileposts indicating 2 miles and 3 miles, it is within a half-mile of the 2 and a half milepost. That's the desired range of 1 mile, out of a possible range of 3 miles length of the track. The probability, then, is 1 / 3.
The answer is C

Bunuel,
Sorry, I could not understand the language of the question and hence probably the answer is not clear to me. If possible , could you reframe the question or explain as to what the question is asking?

Don't we somehow need to use the information 1/10 mile increment signposts in calculating the probability?
Approach mentioned above is logical but there must be simple mathematical solution using the increment signposts?

+1 for C.

Thanks Bunuel

Thankyou Bunuel for the extra efforts in explaining. I have now understood your explanation.

Hi Bunnel,

What about if car starts between 0.5 and 2.5 miles?

So this will make probability = 2*1/3 ?

Bunuel
Shouldn't the answer be 2/3?

The question doesn't specify in which direction the car has to travel, so it could be in either direction. Or may be I'm overthinking

Shouldn't the question say "one half miles" instead of "one half mile"? Because "one half mile" can be read as one of 1/2 mile.

The question exactly says: 1/2 miles.

So, I solved it like this, How many half miles (0.50) miles in in 3 miles, 6 so the probability that car ends within a half mile is 1/6, but, the car could have moved in any direction; clockwise or counter-clockwise so multiply the probability of 1/6 by 2 hence, 1/3.

One doubt about the question. It is only given that a car starts in the circular race track. Doesnt direction come into play in such scenarios? Shouldn't it be mentioned that what direction the car is travelling in. If we imagine it can be either way the answer changes to the question. Just a thought. Looking at the question, I wasnt sure if we are supposed to take into consideration that the car could have travelled both ways.

I got a little mixed up on the wording of question. I interpreted "Within" a half mile to mean "less than a half a mile" not "A half mile or less." Could you rephrase the question.