# GRE Question - What's wrong with my reasoning?

Tags:
1. Dec 29, 2012

### Tim67

Question:

I saw this on a blog, with the solution, so I'm not really asking what the answer is or how to get there, but I'm just wondering whats wrong with my reasoning, which gets me close to, but not at, the right answer:

Think of a number line: 1, 2, 3, .... 120, representing all minutes either can enter the place. So for any minute A enters, B can enter at any of 120 different minutes, so the total combinations of 'ways' both can arrive throughout the night is 120^2.

So, I think of three cases: A arrives in first 15, last 15, or any of the other 90.

Arriving in any of the middle 90 minute marks: B can arrive at any of the 15 minute marks within A's 15-minute interval, or can arrive any of 14 minutes before A arrives and they will still meet each other. So for each 90 minute marks we have 15+14 opportunities, so 90*29.

Either first 15 or last 15: A arrives at minute 1, B has 15 minutes to arrive; A arrives at minute 2, B has A's 15 minute interval or 1 back = 16 opportunities, etc. Same reasoning for last 15 minutes. So: 2(15 + 16 + 17 + .... 29)

so [2(15+..+29) + 90*29]/120^2 = .227

2. Dec 29, 2012

### LCKurtz

The random variables representing the arrival times are continuous random variables. You are treating them as discrete and getting an approximate answer. One wouldn't expect it to be exact.

3. Dec 29, 2012

### Ray Vickson

To expand on the response of LCKurtz: if you split the interval into 1200 (1/10)-minute intervals and repeat the type of computation you did, you will get something nearer to the 0.234 figure. You get even closer if you use 12000 (1/100)-minute intervals, etc.

4. Dec 29, 2012

### Tim67

You're right, I somehow got it in my mind that they could only arrive on the minute even though that was never stated! Thanks.

5. May 5, 2014

### bepardini

I am sorry, but I do not think that is where you went wrong.

You should be multiplying 90 by 30, instead of 29. When you split the interval into bigger number, all you are doing is getting closer to multiply 90 by 30, you will be considering, for example, when person B arrives 14min 59secs after person A.

My point is, person B can arrive within 15 mins before or within 15 mins later, not 14. Therefore you multiply by 30 (15+15).

Another thing I would do differently is to multiply 2(15)(22). If you count how many numbers there are between 15 and 29, you will find 15. The average of these numbers, as they grow in a linear function, can be described as the last term(29), added to the first(15) and divided by 2. Therefore, you have 22 as an average and the calculation would get much easier.

I am sorry if I did not write correctly, I am brazilian and I struggle a bit with english.

6. May 5, 2014

### jbunniii

Let $X$ and $Y$ denote the arrival times of the two people, in minutes. It isn't stated explicitly, but I assume these are uniformly distributed and independent. You need to compute the probability that $|X-Y| \leq 15$. This is straightforward if you know that the pdf of $X-Y$ is the convolution of the pdfs of $X$ and $Y$, hence a triangle with height = $1/120$ and base spanning the interval $[-120,120]$. Find the area of the portion of this triangle over the interval $[-15,15]$. I calculated it to be $1 - (105/120)^2 = 0.234375$, which matches the desired answer. I'll let you work out the details.