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

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?

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.

+1 for C.

Thanks Bunuel

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

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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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.

Hi Bunnel,

What about if car starts between 0.5 and 2.5 miles?

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

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

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

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.

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.

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

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.