GRE Question - What's wrong with my reasoning?

[Tim67]

Question:

Last night, Dave and Kathy both arrived at Pizza Palace at two different random times between 10:00 p.m. and midnight. They had agreed to wait exactly 15 minutes for each other to arrive before leaving. What is the probability that Dave and Kathy were together at Pizza Palace last night between 10:00 p.m. and midnight?

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 what's 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

the right answer is .234

 

[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.

 

[Ray Vickson]

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 what's 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

the right answer is .234

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.

 

[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.

 

[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.

 

[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.

 

[Pawan Sran]

Here we have a window of 1 -120 minutes. We will split the problem in two cases:
case 1 : Dave always comes first
Case 2 : kathy always comes first
Combining both the cases will give us total probability when they will be together in given window.

Case 1 : Let's say dave comes at 10:00 now kathy has exactly 15 minutes to come. for dave to come at 10:00 probability is 1/120 and for kathy to come from 10:00 to 10:15 probability is 15/120 . similarly at 10:01 , 10:02 ... 10:45 . Total probability in this case is (1/120 * 15/120 * 105 ) [we are leaving last 15 minutes which gives us 105 minutes , as, if dave comes at 10:46 , kathy will only have 14 minutes to arrive ] . Now, we will consider last 15 minutes , let's say dave arrives at 10:46 kathy has exactly 14 minutes to arrive and if he arrives at 10:47 kathy has 13 minutes .. which gives us the probability of (1/120 * [14 + 13 +12+ .. + 1]/120 ) . => (1/120 * 105/120 )
using sum of sequence 1 to n : n (n +1) /2

Total probability : (1/120 * 15/120 * 105 ) + (1/120 * 105/120) = 105 /(120 * 120) [15 + 1] => 16 * 105 / 120 * 120 => 7/60
similarly for case 2 : we will get 7 /60

Total probability is 2 * 7/60 = 7/30 = 0.2333 which is correct answer!

 

[PeroK]

Here we have a window of 1 -120 minutes. We will split the problem in two cases:
case 1 : Dave always comes first
Case 2 : kathy always comes first
Combining both the cases will give us total probability when they will be together in given window.

Case 1 : Let's say dave comes at 10:00 now kathy has exactly 15 minutes to come. for dave to come at 10:00 probability is 1/120 and for kathy to come from 10:00 to 10:15 probability is 15/120 . similarly at 10:01 , 10:02 ... 10:45 . Total probability in this case is (1/120 * 15/120 * 105 ) [we are leaving last 15 minutes which gives us 105 minutes , as, if dave comes at 10:46 , kathy will only have 14 minutes to arrive ] . Now, we will consider last 15 minutes , let's say dave arrives at 10:46 kathy has exactly 14 minutes to arrive and if he arrives at 10:47 kathy has 13 minutes .. which gives us the probability of (1/120 * [14 + 13 +12+ .. + 1]/120 ) . => (1/120 * 105/120 )
using sum of sequence 1 to n : n (n +1) /2

Total probability : (1/120 * 15/120 * 105 ) + (1/120 * 105/120) = 105 /(120 * 120) [15 + 1] => 16 * 105 / 120 * 120 => 7/60
similarly for case 2 : we will get 7 /60

Total probability is 2 * 7/60 = 7/30 = 0.2333 which is correct answer!

Better late than never! The Original Post was 6 years ago.

 

[ehild]

Usual solution is with Geometric Probability: The arrival times of Kathy and Dave are represented by the (x,y) points of the square. Dave and Kathy meet if their times of arrival are in the yellow strip.
The probability is the ratio of the area of the yellow strip to that of the total square.