8

I've started studying geometric probability and I am having some difficulty with this version of problem :

Two integers are chosen at random between $0$ and $10$ inclusive. What is the probability that they differ by no more than $5$ ?

The integers restriction really makes it harder for me,without this restriction I would tackle the problem like this(I am not sure it is correct):

Given two numbers $x,y$ we want $0\le y-x \le 5$ or $x \le y \le 5+x $

From the last restriction I have to satisfy the following inequalities $y \le 5+x$ and $y \ge x $ where $ 0 \le y,x \le 10$ (look image below)

Therefore the area I want is $75$ Thus the probability would be $\cfrac{75}{100} $

Now with the integers restriction I would have enter image description here

where the red filled circles indicate the integers which satisfy the restriction.

How do I count them now ?I can't simply count the dots as that would lead me to a probability higher than the previous one,which is impossible (I would get $\cfrac{42}{100}$)..

Mr. Y
  • 2,637
  • With repetition or without repetition – Archis Welankar Jan 20 '16 at 10:44
  • ${1,2,3,4,5,6,7,8,9,10}$ or ${0,1,2,3,4,5,6,7,8,9,10}$ ? – Henry Jan 20 '16 at 10:47
  • the second one,I've mistaken with the graphs...(the conceptual question still remains).Let me edit... – Mr. Y Jan 20 '16 at 10:49
  • If you want a continuous approximation, then you would need to do some kind of continuity correction. Slightly extend the square to $(-0,5, 10.5) \times (-0.5, 10.5)$. And the points shift to $P = (-0.5, 5), Q = (5, 10.5), B = (10.5, 10), D = (0, -0.5)$. The region is no longer a trapezium, and the area becomes $\displaystyle 121 - \frac {5.5^2} {2} - \frac {10.5^2} {2} = 50.75$. The approximated answer is $\displaystyle \frac {50.75} {121}$, which is very close to the true answer by counting which yields $\displaystyle \frac {51} {121}$. – BGM Jan 20 '16 at 12:30
  • @BGM I still don't get the logic,I should have a probabily less than $\frac{35}{100}$.(Btw your answer matches with mine,but again is the logic that I am missing here) – Mr. Y Jan 20 '16 at 12:57
  • Wouldn't the point $(5,0)$ be good? It is not in your plot. I read "differ by no more than $5$" to be $\left|x-y\right|\le5$ – robjohn Jan 20 '16 at 13:42
  • Your computation of the area of the trapezoid is incorrect. The area of the square is $10^2$ the area of the lower triangle is $\frac12\cdot10^2$ and the area of the upper triangle is $\frac12\cdot5^2$. Thus, the area of the trapezoid is $100-50-12.5=37.5$. That is out of a total area of $100$, so the probability of being in the trapezoid is $0.375$ – robjohn Jan 20 '16 at 15:14
  • @Mr.Y: No, the sides are $10$ long. – robjohn Jan 20 '16 at 15:17
  • Oh ,you're right..my bad – Mr. Y Jan 20 '16 at 15:18
  • @Mr.Y: No. The distance from $0$ to $10$ is $\left|,0-10,\right|=10$ – robjohn Jan 20 '16 at 15:18
  • @robjohn sorry,that was dumb from my part.. – Mr. Y Jan 20 '16 at 15:23
  • @Mr.Y: you were just caught up in the formula for counting integers which is $\text{high}-\text{low}+1$. An understandable mistake. – robjohn Jan 20 '16 at 15:25
  • @Mr. Y The idea is treat every integral point as the centre of a unit square, and based on that we do try to divide the region accordingly. The last comment is draw with respect to the shaded region before edit. – BGM Jan 20 '16 at 16:00
  • BETWEEN 0 AND 10 ... does not include 0 or 10. – wolfies Jan 20 '16 at 16:37

3 Answers3

3

There are $\overbrace{6+7+\dots+10+11}^{\frac{17}2\cdot6}+\overbrace{10+\dots+7+6}^{\frac{16}2\cdot5}=91$ ordered pairs that differ by at most $5$ out of $11^2$ ordered pairs. That gives a probability of $$ \frac{91}{121} $$

Clarification: I read "differ by no more than $5$" to mean $\left|x-y\right|\le5$. Then for $0\le x\le5$, there are $x+6$ choices for $y$ and for $6\le x\le10$, there are $16-x$ choices for $y$.

Why The Discrete Probability Might be Greater than the Continuous

enter image description here

In the image above, the red squares hang down and to the right of the associated points. Thus, the red area, $91$, divided by $121$ represents the discrete probability. The area inside the black hexagon, $75$, divided by $100$ represents the continuous probability.

The red area is definitely greater than the area of the hexagon, but it is being divided by a larger number, so it is hard to tell which will be greater.

robjohn
  • 345,667
  • While I see that this must the correct probability I find this result counterintuitive.Why do I have that this probability between two integers is greater than the probability between two numbers not necessarily integers ?Geometrically this doesn't look like the case,the area of the region with red points (I've edited with the right image) contains infinitely many points which satisfy the restriction...what am I mistaking here ? – Mr. Y Jan 20 '16 at 15:56
  • @Mr. Y For the answer here, the continuous approximation required will be $121 - 5.5^2 = 90.75$. Draw the unit square grid with integral points as the centre, and you will see this continuity correction. – BGM Jan 20 '16 at 16:14
  • @Mr.Y: $\frac{91}{121}=0.752066$ whereas the "continuous" approximation gives $0.75$ – robjohn Jan 20 '16 at 16:24
  • @robjohn oh I see..but then how would we adress the case for any number if ,by calculating the area,we essentially do the same as counting the lattice points contained in the area ? – Mr. Y Jan 20 '16 at 16:50
  • @Mr.Y: I am not sure what you are asking. Are you interested in computing the discrete probability or the continuous? – robjohn Jan 20 '16 at 16:57
  • Sorry for the confusion.What I am asking is precisely this:how would the problem change if ,instead of integers,I would let $x,y$ to be any real number so that $0 \le x,y \le10$ ? – Mr. Y Jan 20 '16 at 17:02
  • 1
    @Mr.Y: The probability that $\left|,x-y,\right|\le5$ changes from $\frac{91}{121}=0.752066$ to $\frac{75}{100}=0.75$. – robjohn Jan 20 '16 at 17:05
0

There is no a priori reason why the probability should be less than the "continuous" value $37.5.$ When counting points you can use the general formula for the sum of the first $n$ nonzero natural numbers:

$$\sum_{i=1}^ni=\frac{n(n+1)}2$$

Also remember that the total number of grid points is 121 (not 100).

Justpassingby
  • 10,029
  • @Mr.Y There are $11$ points on each side of the square: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10$. – N. F. Taussig Jan 20 '16 at 11:09
  • I am an idiot..,thanks @N.F.Taussig – Mr. Y Jan 20 '16 at 11:11
  • @Justpassingby Why shouldn't it be less ? Aren't there more continuous values than integers in the interval $[0,10]$ – Mr. Y Jan 20 '16 at 11:14
  • There are "more" (i.e., infinitely many) continuous values both in the numerator and in the denominator of the fraction. The crucial number is not the count of the possible values but their total measure. – Justpassingby Jan 20 '16 at 11:17
  • If I would drop a ball,of the size of a point ,I would have that the probability that it lands on an integer point is less than the probability of landing on the entire region $PQBD$(in this context).This can't be false... – Mr. Y Jan 20 '16 at 11:32
0

https://english.stackexchange.com/questions/118402/when-is-between-inclusive-and-when-exclusive

'Between 0 and 10' to me means the integers 1 through 9 So in that case

p = 69/81

Community
  • 1
user306631
  • 9
  • I agree with your reading of the question ! And solution as: $\frac{23}{27}$ – wolfies Jan 20 '16 at 16:39
  • Yes ,I've pointed out it was a mistake ,forgot to edit to adjust the wording.Congratulations for spotting that out and making an answer for this. – Mr. Y Jan 20 '16 at 16:44
  • Is it more proper to reduce the fraction? In the unreduced form it expresses more information, i.e. the total count and total number of correct accounts for the problem at hand. The reduction is correct also but has a 'mystery' to the numbers. The decimal approximation can be useful for number sense but is not as complete as the fraction. – user306631 Jan 20 '16 at 17:03