# Problem in Geometric Probability

1. Jul 2, 2015

### Arka

The normal algebric probability is easy to understand but I find the geometric probabilities less understandable. Can you please help me with a few problem related to this area of probability so that I can understand it better. Here are my problems,
1. Two persons A and B agree to meet at a place between 11 to 12 noon . The first one to arrive waits for 20 min and then leaves. If the time of their arrival be independent and at random, what is the probability that A and b meet?(Ans :5/9)

2.Consider the cartesian plane R^2 and let X denote the subset of points for which both co-ordinates are integers, A coin of diameter 1/2 is tossed randomly onto the plane. Find the probability that the coin covers a point of X.(Ans: 0.2(approx))

Please give me detailed stepwise solution to the two problems with explanation for the steps. Thanks.

2. Jul 2, 2015

### Arka

By the way it's not some homework problem. The problem was given in a math olympiad in my country.They say that these sums are quite easy but I can't even understand how to proceed with the first step.It's not like those normal probability sums I have encountered before. Please give me some ideas to solve them.

3. Jul 3, 2015

### WWGD

I assume in problem one , that the 2nd person just arrived at some point and left, without waiting, and arrived at 11:40 at the latest (since 11:40+20 min =12:00)? Could we also assume the 1st person stayed for 20 minutes, did not meet the 1st and then left? Sorry, maybe it is obvious.

And, sorry, but it is against the policy of this site to give out explicit, detailed answers, even if it is not HW, but we will do our best to guide you through to the answer.

Last edited: Jul 3, 2015
4. Jul 3, 2015

### WWGD

For the 2nd, I would do something like this: consider a single square with vertices in $\mathbb N \times \mathbb N$. What region around any of the vertices would be a "winning region", i.e., a region where a coin falling would intersect the vertices?

For the 1st, I would suggest you consider the relation $|A_1-A_2| \leq 20$ , where $A_1, A_2$ are the arrival times, within a $60 \times 60$ square, or one going from 12 to 1..

Last edited: Jul 3, 2015
5. Jul 4, 2015

### Arka

Can you explain the graphical representation of these two sums. Thats what I'm not getting. The detailed solution wont matter as long as I understand the graph.Thanks.

6. Jul 4, 2015

### WWGD

Sorry I don't have too much time to go into detail, but I will come back later. The coin falls any where in the plane with equal probability . Note that I am making some simplifying assumptions and my work may not hold if these assumptions are false. The idea is that the corner of the coin farthest from a pair of the integers must land within a region that is 1/4 units (radius of coin) or less away from the pair of integers for the coin to overlap this pair of points. Notice this is not physically realistic, in that the coin may flip up-and-down before settling, but again, this is a Math problem ( I assume) and not a Physics problem, so I assume if the point in the coin farthest from the vertex lands in this region, it will intersect the vertex in question.

For 2, x,y will meet iff y arrives within 20 minutes of x's arrival for the two to meet. This gives ypu the equation $|A(x)-A(y)| \leq 20$, for $A(x),A(y)$ arrival times of $x,y$ . This describes a region in the unit square, and its area is equivalent to the probability of meeting. i.e., person 2 comes in, stays 20 minutes and leaves. This means person 1 must come in either (within) 20 minutes before or withn 20 minutes (after) the arrival of person 2 for the two to be able to meet. Now you need to consider the possible arrival times of person 2.

Will be out for a while now, but will return later on tonight.

Last edited: Jul 4, 2015
7. Jul 9, 2015

### Arka

Thanks for the tips .They were quite helpful.

8. Jul 9, 2015