Problem on Probability based on co-ordinate points

Question

Let S be the array of integral points (x, y, z) with x=0,1,2 ; y=0,1,2,3 ; z=0,1,2,3,4. What is the probability that if two points are chosen from S, their midpoint is in S?

Part where I am stuck: Now for filling x we 3 choices for y we have 4 and z we have 5 so total points that are part of A are 3*4*5= 60 Selecting 2 of 60 is 60C2=1770 possible numbers can be selected from S. Now I am stuck on the part where we have to find number of pairs of such points whose midpoint is in S itself. I cannot think of a way to do so.

Answer 1

I get a smaller answer (or two)

• There are $60^2=3600$ ways of choosing two points, in order and allowing duplication. This reduces to $3540$ if you exclude the $60$ duplicates (and would then reduce to your $1770$ by ignoring order)

• For a midpoint to be a point on the array, the parities of the co-ordinates of the two picked points must be the same in each dimension so that their averages are integers

  • In the $x$ co-ordinate there are $2^2+1^2 = 5$ ways of matching the parities
  • In the $y$ co-ordinate there are $2^2+2^2 = 8$ ways of matching the parities
  • In the $z$ co-ordinate there are $3^2+2^2 = 13$ ways of matching the parities

• So overall there are $5 \times 8 \times 13 = 520$ ways of matching the parities. If you ignore the $60$ overall duplicates this reduces to $460$

So the probability looks to me like $\frac{520}{3600} =\frac{13}{90}$ if you allow duplicate picks, or $\frac{460}{3540} =\frac{23}{177}$ if you exclude duplicate picks